Understanding Mechanics for JEE (Main Advanced) Vol 2

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BOOKS
Dr. P.K. Sharma
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all other Engineering Entrance Examinations
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UNDERSTANDING
F IYSICS 
MECHANICS 
(Part-B)
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L—J MUZAFFARNAGAR
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Dr. P. K. Sharma
UNDERSTANDING 
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MECHANICS
(Part-B)
Dedicated to my mother 
Smt. Mukta Manjari Sharma 
and my wife Usha.
(Part-B)
2010
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GlRB
For IIT-JEE
& all other Engineering Entrance Examinations
By:
Dr. Pradeep Kumar Sharma
(M.Sc., Ph.D.)
UNDERSTANDING PHYSICS
MECHANICS
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PREFACE
It goes without saying that a teacher is worth of millions of books. The teachers are the evolutes of 
the mental energy of potential students. A teacher without students is lame and a student without teachers 
is blind. The teachers and students are interrelated, interconnected and interdependent in the world of 
knowledge. The association of teachers and students chum the ocean of knowledge to explore the ’ 
riddles of nature which forms the basis of our lives. It is not always expected to enjoy the personal 
presence of a teacher. In that time, an ideal book which personifies thousands of the brains of students 
and enviable experience and expertise of many teachers takes the credit of impersonal guidance to the 
students. Let me present my effort in this regard as following :
Expressing my deepest gratitude to my revered teachers and the.directors of the organizations for 
giving me the chance to cultivate my knowledge, extending my love and sincere thanks to all my 
students, unhesitatingly I can proclaim that by the raft of unswering devotion to teaching and philosophy, 
the unalloyed motivation and support of my teachers, students, colleagues and well wishers, I am able 
to present this text named as Understanding Physics. It has four parts; Part-1: Particle Mechanics; 
Part-2: Rigid Body Mechanics, Fluid Mechanics, Gravitation, Properties of Matter and Simple Harmonic 
Motion; Part-3: Waves, Optics, Heat and Thermodynamics; Part-4: Electromagnetism and Modem 
Physics. Each part has nearly ten chapters. Thus, the total General Physics is divided into forty chapters 
approximately.
Understanding Physics Mechanics (Part-B) has twelve Chapters (11 -22). In Chapter-11, mechanics 
of system of particle is described. The theory of collision of particles is explained in Chapter-12. The . 
mechanics of rigid body is divided into five Chapters (13-17). In Chapter-13, kinematics of rigid body 
is described. Then I explained the concepts of torque and angular momentum and applied these concepts 
for a particle in Chapter-14, for a system of particle (rigid body) in Chapter-15. The application of
• work-energy theorem and conservation of energy in the motion of rigid bodies are described in Chapter- 
16. In Chapter-17, I introduced the concept of angular impulse and discussed the application of 
conservation of angular momentum of rigid bodies in general problems and the problems related to the 
collision of rigid bodies.
After the completion of rigid body mechanics, the laws of gravitation of Kepler and Newton are 
described in Chapter-18. The applications of conservation laws (momentum and energy) and .work­
energy theorem in the problems related to planets and satellites are explained in the chapter-18 “gravitation”.
Using the theories described in the Chapter-11 (system of particles), I explained the mechanics of 
fluid which is divided into two chapters i.e. fluid hydrostatics and fluid (hydro) dynamics. The Chapter- 
19 deals with fluid (hydro) statics. The concepts of hydrostatic force, hydrostatic pressure, Archimedes’ 
principle, Pascal’s law and their applications are described in hydrostatics. In Chapter-20 (hydrodynamics), 
I discussed the equation of continuity (conservation of mass) and Bernoulli’s equation (conservation of 
energy) and their applications in many practical cases such as siphon, venturimeter etc.
In Chapter-21, I discussed various properties of matter i.e. elasticity, surface tension, capillary 
action and viscosity. In Chapter-22, using the concept of elasticity I explained the principle of oscillation 
of different mechanical systems such as spring-mass system, simple pendulum, physical pendulum 
torsional, pendulum, two particle oscillation, oscillation of liquid in U-tube etc.
. (Vi)
Pradeep Kumar SharmaMay, 2009
Hyderabad
Each chapter of this book starts with a lucid introduction justifying the significance of the chapter 
in practical life and its relation with the previous chapters. The contents of each chapter are divided into 
well defined sections. Each section builds a new concept from the scratch. The student friendly discussion 
and labelled diagrams in the sections make the things easy to grasp. After the explanation in each section, 
the main points are highlighted for your quick reference. The mathematical formulae related with physical 
concept are derived from the basics. I emphasized on the physical interpretations of each derived result 
(formula) rather than just getting the answer. Sometimes micro-interpretations are provided to understand 
the subtlety of macroscopic phenomena. At the end of an example, few questions as the “student task” 
for deep understanding of the example are asked. Furthermore, some finer points of the examples are 
tabulated by critically analyzing them which may help the students in broadening their understanding. 
After studying all sections of a chapter, you are expected to learn different tricks and. techniques of 
problem solving, with a-suitable level of analytical ability.
Then you are advised to refine your understanding by doing the exercises. To make the book 
balanced with theories and problems^ I have tried to put varieties of thought provoking questions in 
different forms such as discussion questions, MCQ, fill in the blanks, match the columns, true or false, 
assertion and reason, comprehensions and numerical problems. While doing the exercises, you may find 
some difficult ideas conveyed repeatedly through different types of questions. This may help you to 
remember them longer and understand them deeper. Some difficult sections may seem to be bit elaborative 
for the sake of understanding. Thus, an average student will enjoy a methodical self study without an 
extra labour in browsing different sources.
At last, I strongly suggest the students to read the book without skipping intermediate sections 
because all sections are interrelated and continuous. You should not try to design your own method of 
solution without mastering the concepts and methods discussed in the book. Always try to understand 
the underlying concepts rather than just finding an answer which is the prime basis of this book. You 
should not directly attack the problems without referring to the relevant sections. You may lose the 
confidence if you will get wrong answer. Always try to get an answer qualitatively in terms of the given 
parameters; then put the given values to get the numerical answer. In this process, not only you can save 
your time without repeated numerical calculations butand external forces and KE of system of particles. Applying work-energy theorem on the 
system of particles, we have
^int + ^ex. =
F. F MVC
Ajnt “A ,
. .. , (3m)(v0) + (2m)(-2v0) v0
where M = (3m + 2m) and vc =-------------- -----------= - —3m 4- 5
r 27 2,^2£int = ~mv0 +~x
35
mi and m2. If you
m2f
Fig. 11.48
Then,
Ans.
2
Then, we have Ans.
Student Task^
slide. Ans.
in work-energy theorem
Sol.
and
Substituting
we have 
This gives
~^2g‘
•x
= ^Fds2 + ' ds2 »
When the forces (external and internal) involved in the group of particles are conservative, the 
total mechanical energy of the system remains conserved.
Ex. 21. Discuss the effect of stopping a gas jar filled with an ideal gas of mass M and moving with a . 
speed v.
When the gas jar stops, the CM of the gas will stop(vCM)2 = 0
Since, the collision of gas particles with the wall of the gas jar is elastic, we can conserve the
^ex.
fs2
ds2 = dx
--------►F
System of Particles
(ii) We have two external forces, i.e., F and the frictions acting on 
assume m[ stationary, friction on m\ cannot do work. Then the total work done by 
external force is,
|At/ +AK = 0| 
The above expression tells us that;
;xt = Fx - \im2gx
where ds^ = dxi and fa =
wcxt = Fiodx~^m2s\Q dx
This gives W'ext = Fx - pm2gx
(iii) The total work done is W = lKjnt + JFext, where JFjnl = - y kx2 and 
kx2W = - -y- + Fx - p.m2gx
"'int
> 7n the above example, find the maximum elongation of the spring, assuming that mt does not 
2{F - pn2g) 
k 
Conservation of mechanical energy : If the external and internal forces acting on the system 
of particles are conservative, we can write
^ext+^nt =-A(7, 
where JFext = total work done by all externa! forces, 
PFjnt = total work done by all internal forces 
At/ = change in PE of the system.
-At/ = IFext+^nt 
+ ^ext = **•
-At/ = \K
36
+ (n2,+ (n =
This gives
where p =
vrel =1 vl" *2 I ■and
Student Task
Vo
> In the above example, can you conserve (i) KE (z‘z) momentum, of the gas particles relative 
to CM frame? Ans. (i) No (ii) Yes.
x
I
m 2m
Fig. 11.49
Total mechanical energy of a system of particles : Once we have the KE and PE of a system 
of particles, just add them to obtain the total energy which can be expressed as :
E = U + K,
where U = potential energy and K = kinetic energy, of the system of particles
G.R.B. Understanding Physics MECHANICS (Part-B)
KE of the system (gas but not gas jar). As there is no attraction between the ideal gas particles, 
the potential energy of interaction is zero. Neglecting the effect of gravity of the earth, the total 
internal energy is equal to the internal KE of the gas. Hence, we can conserve the KE of the 
system of gas particles to obtain
Wc)i
2
and (v£.)2 = 0
mpn2 
mi + m-— (called reduced mass because p Find the KE of a system of two particles of mass m and 2m moving 
with velocities voz and voj respectively, relative to its CM.
2 2Ans. ~mvo
This gives;
System of Particles 37
Total internal energy : When we subtract the KE of CM from the total mechanical energy of 
the system, we will get the total internal energy, because KE of the CM is controlled by external forces. 
Hence, the total internal energy is given as :
P = mv
Newton describes the momentum as quantity of motion.
Single force acting on a particle : In other words, linear momentum of a particle is defined as 
the product of mass and velocity of the particle. Since, mass is a scalar and velocity is a vector quantity, 
momentum p is a vector quantity. Therefore, we can express the “force as a rate of change in linear 
momentum”.
velocity at a rate a
§ 11.6 Impulse and Momentum Equation for a System of Particles
Momentum of a particle :Newton’s second law for a particle tells us that a particle changes its
dv ~►— in the direction of a force F applied on it. According to Newton’s second 
dt )
law, applied for a particle of mass m, we have
r -p MvC 
^int ~ ~
-» dP 
F = — 
____ dt_
The above expression tells us that force is equal to the rate of change of linear momentum, which 
is a standard expression of Newton’s second law of motion.
?(= = = 
at
Many forces acting on a particle : When number of forces F(, F2,..., Fn, say, act on a particle, 
each force will tend to change the linear momentum of the particle. In consequence, the net change in 
momentum is caused by the combined effect (action) of all forces. Then, the force F in right hand side 
of the above expression is the net (or resultant) force acting on the particle
-> mdvF = ma =------
dt
Since, “m” is a constant, take it into the derivative.
d -» F = — (mv) 
dt
This expression tells us that the force changes the vector quantity “mv ” which is defined as 
“momentum of the particle” denoted by the symbol p .
38
Sol.
where p = fii + t2 j
F = 3t2i + 2tjThen, we have
Since, F = F}+ F2, F2 can be given as :
Substituting
and
we have Ans.
Student Task
—►
F2 will lose its y-component completely. -i-3j
G.R.B. Understanding Physics MECHANICS (Part-B)
Hence, Newton’s second law can be expressed as :
X = (2/+3j-).
F2 = (3t2 - 2)/ + (2Z - 3)J
F = 3t2i + 2tj
The net force acting on a particle is equal to the rate of change of linear momentum of the 
particle, where linear momentum of the particle is defined as the product of its mass and 
velocity. Newton calls “Linear momentum”, “quantity of motion of matter”. In this way, force 
causes the change in quantity of motion. The total change in momentum is attributed to the net 
(total) force but not any individual (component) force.
Let us see how the net force can change the momentum in the following example.
Ex. 22. A particle experiences two forces, one of them is a constant force = (2i + 3f)N ■ If the 
particle changes its momentum with time as per the relation P = r37 + t2j, find the other force
->
Fl-
The net forces are given as :
-> dPF = — 
dt
dPF = — 
dt
where F = net force and p = mv
> In the above example, find the change in momentum due to F2 over the time t during which
3 fl • _
Ans. 7 ~ 3;4 —>
/71, v( = (Dnz)vc >Since,
we have
Substituting
we have
Sol. System Mvc >
Then, we have Ans.
Student Task
> In the above example, if the gas jar is stopped, what will be the momentum of the system?
Ans. Zero
When the CM of a system of particle (gas + gas jar, in last example) does not move, straight 
way we can say that the momentum of the system is zero, even though each gas particle has 
some velocity. You should remember that the total momentum is equal to vector sum of indi­
vidual momenta, but not the sum of the magnitudes of momenta of all particles.
Momentum of system of n particles 
is equal to sum of momenta of all 
particles; P = E Pi 
Fig. 11.50
-» i=n -»
P = fm, v, 
i=l
The total momentum of a system of particles is equal to the vector sum (or resultant) of 
momenta of all particles of the system, which can also be equal to the product of total mass and 
the velocity of CM of the system.
Ex. 23. What is the momentum of a system comprising a gas jar of mass Mo moving with velocity v 
and the gas of mass m?
P = (jLmj}vc 
Xm, = M,
P = MVq,
where M = Mo + m and
■^system (^0 + /w) v
= v
Momentum of a system of particles : For a system of n particles, the momentum P of the 
system is equal to vector sum of the momenta of all particles of the system.
^system (= = ’
where M = mass of the system
—>
and vc = velocity of CM of the system.
The above discussion tells us that;
where Pt = w, vt (momentum of zth particle)
G.R.B. Understanding Physics MECHANICS(Part-B)40.
Symbolically,
When
Fig. 11.51
Ex. 24. Discuss the possibility of conservation of linear momentum of a block 
moving on a rough inclined plane if |i = tan 0.
Hence, p = constant
That is what we call law of conservation of linear momentum.
In other words,
Whenever, no net force acts on a system of one or more particles, the linear momentum of the 
system remains constant (conserved). Since, the net internal force is zero, it (the total internal 
force) cannot change the linear momentum of the system, even though the individual momen­
tum of the particles of the system may change.
but not |P|=E|Xl
However, the KE of the system need not be zero even though the centre of mass of the system 
does not move. For instance, in the last example, we have
|P| = |E^|
-» dPF = — 
dt
F = have — = 0 
dt
—> —> —> —>
where P = m v and F = Fnet 
For a group of particles, Newton’s 2nd law can be given as :
K = K’+^,
2
where vc = 0 after stopping the gas jar. Then, we have K = K'; K' = sum of KE of all 
particles relative to CM, which can be equal to the sum of KE of all particles relative to ground 
because vc = 0 (gas jar is stopped).
Conservation of linear momentum : For a particle of mass m moving with a velocity v, 
Newton’s second law can be written as :
F-".
dt
where P = Im,- v, = Mvc and F = /^et = ’LFi\ ~F\ = external force acting at Ith point.
If there is “no net force” acting on a particle or a system of particles, its momentum remains 
constant.
41
yf
mg sin0Fx = >ngsin0-/,
x
Vo
Sol.
yN
ma
—•.a
x
and
> Can we conserve the block's momentum when it is pushed upwards along the plane in nega­
tive x-direction? Justify your answer. Ans. No, because Fnet = 2mgsin0\
The above example tells us that;
If the net force acting on a particle in a certain direction is zero, the momentum of the particle 
will remain constant in that direction.
System of Particles
Sol.
—> —>
In accelerating (non-inertial) frame, if the resultant of all forces (Fpseudo and Frea!) acting on 
a particle is zero (in any direction), the momentum of the particle remains constant relative to 
that frame (in that direction).
Imposing pseudo-force on the block as observed from 
the accelerating wedge, we have the net force
F = mg + N+Fpi,
Sometinjes we observe particle in an accelerating frame. In that case, in addition to the real 
forces we must take the pseudo-forces into account. If the net effect of Frea[ and Fpseudo is zero, in 
that case we can conserve the momentum of the particle (relative to the accelerating or non-inertial 
frame).
In other words,
Let us use the above idea in the following example.
Ex. 25. Can you conserve the linear momentum of a block placed on a smooth 
accelerating wedge? Explain.
Let us consider the motion of the block in x-direction. The net force 
acting on the block is,
where f = png cos 0
This gives; Fx = mg(sin0-pcos0) Fig. 11.52
Since, p = tan 0, Fx = 0 . Hence, we can conserve the momentum of the block down the 
plane.
Student Task^
where r = -ma
Resolving the forces in x and y-axes, we have
Fx =mgsin0-macos0
Fy =-mg cosQ ~ masinQ + N
~*"a oK
Fig. 11.53
/et 
/ mg____________ efxx
777777777777777777777777777/
Fig. 11.54
42
2v0
(6) The motion of a smooth block on a smooth wedge.
Sol.
m
A
2m
FbIWWW
Fig. 11.55 (a)
/^zzzzzzzzzzzzzzzzzzzzzzzzzzzz
Fig. 11.55(b)
opi mg ivitw cuv k/i|uai anu vpjjvoiiv. i ivuw, F F
of the system remains constant, which can |~A~pAfWWOWP-[~B~|
Fig.11.56
(a) The net external force acting on~the system A + B is zero 
because the internal spring forces are equal and opposite. Hence, 
the momentum P In the above example :
(а) Canyon conserve the momentum of the block relative to ground when the wedge moves 
with an acceleration a = gtanQ?
(б) Can you conserve the momentum of the block relative to the wedge in any direction in 
the plane of the slant, when a = gtanQI
Ans. (a) No, because Fb|Ock = mg tan 9 ->
(b) Yes, because Fblock = 0 relative to the wedge.
Let us now look at some examples of conservation of linear momentum of system containing 
two or more particles.
Ex. 26. Can you conserve the linear momentum of the following system?
(a) Two particles interconnected by a light spring. vp p
G.R.B. Understanding Physics MECHANICS (Part-B)
Since, the block does not lose contact with the wedge, we can say that the block does not move 
relative to the wedge in ^-direction (normal to the wedge); In other words we can say that the 
acceleration of the block is zero relative to the wedge in ^-direction. In this way, we have 
Fy - 0 on the block so as to remain with zero momentum along y-axis relative to the wedge. . 
That means, we can conserve the ‘‘zero momentum” of the block relative to the wedge in y- 
direction (normal to the slant of the wedge) for any acceleration (including zero acceleration) of 
the wedge. Now let us analyse the effect of forces inx-direction. Since, Fx(= zngsinG -macos9), 
Frwill be zero only when a = gtan9. So, if the wedge accelerates towards right having 
acceleration a = gtanG , the net force acting on the block relative to the wedge will be zero (in 
both x and ^-direction). Hence, the momentum of the block remains constant relative to the 
wedge when a = gsinOz . If we push the block 
wedge along x-direction, it will go on moving with same momentum “mv” relative to the 
wedge.
Student Task^
43
N
y
x
mg 0
Mg Fig. 11.57
> Can you conserve the linear momentum of the following system?
(a) Particles interconnected by a spring projected arbitrarily.
WPX
dP = Fdt
(ft) The particle m is projected on the curved surface of the 
smooth wedge.
////////////////////////////////
Fig. 11.59
Fig. 11.58
Ni 
" -N
Ans. (a) Yes, because Fnet = 0.
= c because Fx = 0 but Py * c because Fy * 0.
a= accel-
The right hand term “ Fdt ” is called “linear impulse” of force F during time dt. Impulse is a 
vector quantity because a vector F is multiplied with a scalar dt.
If we want to find the impulse of the force F over a finite time interval At = t2 -t], we will have 
to integrate (sum up) all elementary impulses.
-> dp
Linear impulse : As discussed earlier, effect of a force F can be given as : F = — 
dt
Then, the change in momentum of a particle under the action of a single force is given as :
System of Particles
(b) The normal reactions N and -N between the wedge 
and block are equal and opposite. Hence, there will be 
no effect of the internal forces of the system 
(M + m). Then, we have the vertical external forces 
such as gravity forces -Mgj and -mgj and norma! ///)/}// '////////////y//f/////// 
reaction AtJ offered by ground on the wedge. As there 
is no horizontal external force acting on the system
(M + m), we can conserve the “momentum of the wedge + block system” horizontally.
Since, the block accelerates vertically downwards with ay =asin0, where
eration of m relative to M, the net force acting on the system (Af + m) is not zero. Then 
the net vertical force acting on the system (Af + m) is Fy = [N{ - (M + z„)g] * 0 • 
Hence, we cannot conserve the vertical momentum of the wedge-block system.
Student Task/f
44
where Fdt = dP
This gives
Imp = |f dt = APThen, we have
is numerically equal to the change in
Work = W
The impulse of a force F during a time interval A/ = z2 — 0 is given as :
Imp = p Fdt,
When many forces act on a particle, impulse of the net force during any time interval is equal to 
the change in momentum of the particle during the given time interval.
Please remember that the “impulse-momentum” equation is just a time integral form of Newton’s 
2nd law. Similar to this, “work-energy theorem” is a distance integral of force, derived from Newton’s 
2nd law. Hence, we should not treat “impulse-momentum” and “work-energy” relation as totally differ­
ent concepts. The basis of each concept lies in Newton’s 2nd law. We recast the integral and differentialforms of Newton’s 2nd law and work energy theorem as following :
XF,dt = J(E^)
Imp = ^F dt which represents the area under F-t graph with time-axis.
If the area lies above /-axis, it is positive. It , 
means, the change in momentum AP (caused by F 1 
only) is positive. The “positive AP ” points in positive 
directions of the axes but you should not misunder­
stand that the magnitude of final linear momentum is "o 
greater than the magnitude of initial linear momentum.
Similarly, if the area lies below the time axis, the 
impulse is negative. Then, AP (due to F only) is 
negative which signifies that AP points in negative x, 
y and z directions in coordinate axes.
Sometimes, a force may reverse its direction from posi­
tive to negative. A positive force gives a positive impulse, 
changing the momentum in positive direction. Likewise a 
negative force causes a negative impulse which changes the 
momentum in negative direction. In this way, during a time 
interval A/(=/2 -/j.say), we can have both positive impulse 
during some time intervals and negative impulse for the other 
time intervals. In other words, the area under F-t graph will 
be sometimes positive and sometimes negative.
In this case, let us sum up all areas algebraically. The 
total impulse is,
Imp = Total area, A (say)
The area under F-t graph gives the 
impulse of the force F, which is 
equal to change in linear momentum 
caused by the force F
Fig. 11.60
where A = area generated by F-t curve with time-axis between the times / = /] and t = t2 .
The above discussion tells us that;
46 G.R.B. Understanding Physics MECHANICS (Part-B)
1.
2
t2 3 51 4
Fig. 11.63
Sol.
F
where
and
t
Since, Fig. 11.64Fdt = tsP,
we have
Substituting
we have
&P = P-P0,
P = ^-2/,
o 
1
0 
1
jFdt = 2-4 = -2
Ex. 27. A particle of mass I kg moves with a velocity v0 = 2 mis in positive 
x-direction at t = 0. The particle experiences aforce Fading along 
x-axis which varies with time as shown in the figure. Find the mo­
mentum of the particle at the end of Sth second.
The net impulse = $Fdt = Area under F-t graph
= - A2
A] = 2x 1 = 2
A2 = 1 x 4 = 4
Then, we have
Then, A = 040 + (-^2)
= (Magnitude of positive area) - (Magnitude of negative area)
= (Magnitude of area lying above /-axis) - (Magnitude of area lying below /-axis)
—>
If the total area A is positive we have a net positive impulse and hence a positive A P. If the net 
area A is zero, A P = 0; there is no net change in linear momentum in the given time interval. However, 
A P = 0 does not tell us that the momentum of the particle remains constant for all instants between the 
time interval. When A is negative, we have a net negative impulse. This signifies a change in momentum 
in negative directions.
Recapitulating,
Area under F-t graph gives the impulse of the force F which is equal to the change in 
momentum “due to that force”. Hence, area under any force-time graph can be equated 
with the total change in momentum of a particle due to that force.
2. The net impulse = net area of Fnet -/ curve = (Magnitude of area of F-t curve above /-axis)
- (Magnitude of area of F-t curve below /-axis).
3. If impulse is positive the change in momentum due to that impulse [but not necessarily the 
total (or actual) change in momentum] is directed in a positive direction and vice-versa. 
 ->.... . . . -»4. If net impulse is zero, A P = 0 but this does not conclude that p remains conserved.
F
txP = -2i
47
Ans.P =0
>
i
t0
F
A
Sol.
System of Particles 
where PQ = mv0 
Substituting
Since, the net internal force is equal to zero according to Newton’s 3rd law, when the net 
external force acting on a system is zero, the momentum of the system is conserved in accor­
dance with Newton’s 2nd law.
F
Fo"
♦T***T-
Fig. 11.65
Ex. 28. Considering the entire universe as a closed system, discuss the plausibility of conservation of 
linear momentum of the universe.
Since each particle interact with the other, action-reaction pairs cannot produce any net im­
pulse. Newton’s law deals only with the interaction between material particles. Sometimes, 
photons interact with matter. In that case, we need to modify Newton’s laws using the modem 
ideas of relativity and quantum theories and try to conserve the momentum of the universe as a 
vector sum of momenta of all material and photon particles. Needless to mention, no net force 
is acting on the universe when we assume it as an isolated system. Hence, the linear momentum 
of the universe is a ‘constant quantity”. Please note that, “Conservation of linear momen­
tum is an independent principle”.
m = I kg and v0 = 2 m/s,
A F0TAns.
m
Role of Newton’s 3rd law in conservation of linear momentum of system of particles : We 
have discussed how Newton’s 2nd law takes part in explaining the principle of conservation oflinear 
momentum of a system of particles. Now we will discuss the role of Newton’s 3rd law in explaining the 
principle of conservation of linear momentum of a system of particles.
Newton’s 3rd law tells us that, internal forces appear as 
action-reaction pairs. Let us consider a pair of two interacting ----- ». „-----
particles A and B experiencing forces F and -F respectively. A B
Then the sum of impulses of these two forces will be equal to Action-reaction pair contributes no net 
zero. It means that, according to Newton’s 3rd law, the equal imPu|se, causing no change in linear 
, .’ ■ ■ cc u i .-u momentum of the system (A + B)and opposite action-reaction pairs of forces as a whole contrib­
ute no net impulse. Then adding the impulses of all possible ac- ‘ ;
tion-reaction pairs of internal forces, we have no net impulse. As a result, no change in momentum takes 
place due to internal forces as a whole. Hence, when the net effect of external forces acting on the 
system of particles is zero, we can conserve the linear momentum of the system.
You can summarize the above matter in the following way;
we have
Student Task
A particle of mass m experiences a force that varies with time as shown 
in the figure. If the particle was at rest at t = Q, find the velocity of the 
particle at the end of time t = 2T.
GR.B. Understanding Physics MECHANICS (Part-B)
1.
2.
6.
13.
| 2= - p.vre| relative to CM, where H = reduced mass of14.
15.
16.
19.
20.
17.
18.
9.
0.
1.
12.
3.
4.
5.
7.
8.
48
§ 11.7 Advantages of CM Frame
The following properties of CM makes the problem solving easier:
Net force acting on the systemrelative to centre of mass is zero, whereas the net force acting 
on CM (relative to ground) is equal to the resultant of all external forces acting on the system. 
Momentum of the system relative to CM is zero. That means, neither internal forces nor exter­
nal forces can change the momentum of the system relative to CM.
Kinetic energy of a system is minimum relative to CM.
The work done by pseudo (inertial) force in CM frame is zero for a translating system.
The work done by gravity is zero relative to CM (for the bodies having the size negligible 
compared to the radius of the earth).
The torques about CM point and about any other moving or stationary points coinciding with 
the CM are equal. This will be dealt in chapter-15.
The net torque produced by the internal forces is zero about CM.
Gravity cannot produce a torque about CM of a body whose size is negligible compared to the 
radius of the earth.
The torque about the CM causes rotation of a system of particles (rigid bodies).
The angular momentum about CM defines the rotation of a system of particles.
Linear momentum of a system is equal to the linear momentum of CM of the system.
Internal angular momentum of a system is equal to angular momentum of the system relative to 
its CM.
Kinetic energy of a system is (generally) greater than KE of the system about the CM of the 
system. However, both are equal in CM frame.
For a two particle system £inI = ^system ■■ \
the system and vre| = relative velocities between the particles.
For a two particle system, Psystem = ^-pvre) relative to CM.
For a two (or many) particle system Ezn, rlC = 0, Xm, viC = 0 and ajC = 0 . That means, 
sum of (i) moment of masses (ii) linear momenta (iii) product of mass and acceleration of the 
particles relative to CM is zero individually.
For a two particle system, momenta of the particles are equal and opposite relative to CM.
For a two particle system, ratio of KE of the particles relative to the CM is equal to inverse ratio 
of their masses.
For a two particle system, the ratio of velocities, displacements and accelerations of the par­
ticles is equal to the inverse ratio of their masses.
If FexI = 0, the CM moves with constant velocity. Hence, the CM (but not necessarily other 
points of the system which may be accelerating) can be treated as an inertial frame (point).
21.
22.
23.
24.
25.
26.
27.
28.
F
'2-
This gives
we have and dr = dx
This gives
ri
where f = internal force and 
the particles m, and m
(a) If f = spring force,
we have
System of Particles
The net effect of internal forces is zero on the CM whereas it is not zero (in general) for other 
points of the system. Inconsequence, the CM moves with a constant velocity whereas other 
points may move with same accelerations when 7*’ext — 0 .
Angular momentum (and rate of change in angular momentum) of a system of particles relative 
to CM point and of any stationary or moving point coinciding with the CM is equal.
For any system, the internal KE is equal to KE of the system relative to CM which becomes 
internal property of the system.
For rigid bodies the KE of rotation is equal to the KE of the particles due to their motion 
perpendicular to the line joining the CM with the corresponding particles of the system.
The KE of vibration of a two particle system is the KE of the particles due to their motion along 
the line joining the CM of the system with the particles.
Internal forces as a whole cannot change the KE of the CM of the system whereas external 
force can change the KE of CM of the system.
When the net work done by the internal forces is non-zero, the KE of the system about the CM 
changes due to the internal forces.
Internal forces cannot change the momentum of CM of the system.
Some points regarding the CM will be covered in the forthcoming chapters. Anyway, keeping all 
the above points in mind, we will proceed to next section where we will clear some of the above points 
(concepts) through suitable examples.
§ 11.8 Application of Impulse-Momentum Equation and Work-Energy Theorem for a 
Two-Particle System
Work done by internal forces : In the previous sections we learnt - ----- x
that, for two particles system, work done by internal force is given as :
Wf=-\fdr,
r = distance of separation, between Tota| work done by an in. 
ternal force pair can be 
given as : W( = -f f dr, which 
does not depend on the 
reference frame
f = kx (by Hooke’s law) and dr = dx. Fig. 11.67
(b) If f = gravitational force,
-F \ 
r—► /
f _ Gm\ml
r2
( I
W = Gmyir^ —
■ 17
50 G.R.B. Understanding Physics MECHANICS (Part-B)
(c) If f = electrostatic force,
we have and dr = dr
1Then, we have
(d) If f = static friction,
^ext =
’ vrel =l vl“v2 I and Af = (wj[+m2)
ri
we have f = fs{ F
by the pseudo-forces ~m} ac and m2 ac > where ac = —. Then we find the IFjnt and IFext 
relative to CM'. Finally, find the change in kinetic energy (AX') relative to CM : 
JFext pl driC and F^int =_E]/-dr . Please note that pseudo-forces as a whole cannot 
perform work relative to CM. Hence, you may disregard the effect of pseudo-force when 
solving the two-particle system relative to their centre of mass.
Ex. 29. Two constant forces F} and F2 acts on two smooth par­
ticles (blocks') of masses m} and m2 interconnected by a f, 
light spring of stiffness k. Find the maximum elongation 
of the spring using W-E theorem.
Let us assume that after some time, jq and x2 are the position vectors of the particles m} and 
m2, respectively. During a small (elementary) time interval dt, let the particles move through
—►dx,
1-»—
52 GR.B. Understanding Physics MECHANICS (Part-B)
.2
...(0
...(H)and
Substituting xic =
Fl -F,
This gives Ans.
Aliter (CM frame):
Since,
X7H] 
7M] + 7M2
mtx
+ m2
-m2x 
7H| + /n2
2 1 , . 2' = -(m1+m2)v
£ 2 
F2x2C-F]X1(?--x =0
Work done by external force F2 is,
^f2 ~ +F2x2
Work done by the internal force (springforces) is,
Ws. =--xl, sp 2
where x = elongation of the spring
Then, the total work done is,
(-F1+F2)xc=i(mI+m2)v2
vrei^. =0 and vre) =0, we have
= l*x2 
2
x^2(ff>iF2+m2F|)
k(mx + m2)
W-WF]iWFjiWSf
„ k= -F1x1+F2x2--x'
The change in kinetic energy of the system till the spring has maximum elongation is,
&K = ~(zni +w2)v2
where v = speed of the particles because both and m2 will move with same velocity at the 
time of maximum separation between them which corresponds to maximum elongation of the 
spring.
Now applying W-E theorem W = AX’, we have
k ■-F1x1+F2x2--x'
Substituting X| = (xic +xc) and x2 =x2C + xc,
1 2 1 2we have (-^*i + F2 )xc + F2x2C - F]XIC --kx = -(m^ + m2 )vc
Comparing the terms in both sides, we have
-zn2x------— and x2C =
+m2
in eq. (i), we have
System of Particles 53
Putting
we have
Then, we have
Arie - and Ar2C =we have
+ ^2^1
Ans.This gives
Student Task
and zero
where
> In the foregoing example if F| = 0 and F2 =.F, find the maximum and minimum elongation 
of the spring.
T 
Fl
m2x 
mx + m2
mxx 
mx + m2
__ ^r2C 
r~*F2
2mxF
(wi] + m2)h
Conservation of linear momentum : If Fext = 0 on the two-particle system, external impulse
—> —> ~►
is zero. Then, we can conserve the linear momentum of the system; Pj=Pf. It means that vc =
constant and txKc = 0. Hence, AX' = AX' ■
Substituting txKc =0 in the W-E theorem IFint + IText = AXC + AX’',
we have
Ans. xmax
+ 'LFi-driC = 0^.4
-> -> kx2
lFrdriC=—,
k 2
— X
2
0int+^xt=Ar,
4K' = 5rf(»re|)/-(vrel)?l
where Z [?rdK = jX«Mc+ jM, In the above example, find the maximum height h if the wedge is pushed initially towards left 
2w2vq 
(™1 + m2)g
§ 11.9 Application of Impulse-Momentum Equation and Work-Energy Theorem for 
Many Particle System
(a) Constant mass system : Suppose there are two holes, 
I and 2 in a container filled with a liquid. In hole 1 and 2 equal 
amount of liquid flows such that the mass of liquid in the container 
remains constant. Let the velocities of incoming water at hole 1 and 
outgoing water at hole 2 be V] and v2 respectively. If "'dm” amount 
ofliquid enters into the container with velocity V| and same amount 
of liquid escapes from the container with velocity v2, the change 
in momentum of the system is equal to change in momentum of the 
liquid, which can be given as :
P = dm v2 - dm V| = dm{y2 - V])
The impulse JXxt^ of th® external forces acting on the 
system (container + liquid) must be equal to change in momentum 
of the .system. Writing impulse-momentum equation, we have
v2
The net impulse of the external 
forces is equal to change in 
momentum of the system 
fFextdt = AP
Fig. 11.72
h =
2|l + —|g
System of Particles
(because, finally, at the maximum height, both m{ and m2 have same velocity as the relative 
sliding between them stops)
Then, we have
W-E theorem :
-7W|g/i = -(m| + m2)v -
Substituting v from eq. (i) in eq. (ii),
(/M] +m2)v = m|v0 
^ext+Jfint=AK,
’ 2 „ 2 \ 2wi[V m2v w,Vq 
~2~ + ~2 ~
I
2
v0
m\
m2
56. GR.B. Understanding Physics MECHANICS (Part-B)
Fext dt = d P
This gives
where d P = dm(v2-v})
Then, we have
v 
X
Sol.
we have
Ex. 31. An L-shaped tube of uniform cross-section "a" carries a flowing 
liquid. If the liquid moves with a speed v and changes its direction 
by 90°, find the external force F required to keep the tube in equi­
librium. Neglect friction.
11 
11 
11 
11
T 
V 
Fig. 11.73
y
and substituting = fi, = -vj
^ext
at
_> _> _>
Following the derived expression Fext = —(v2 - vj) 
dt
dm
Please note that, here — is not the rate of gain or rate of loss of mass, it is just a rate of flow;
dm
— - ^iPV] = ^Pv2 according to the equation of continuity (mass conservation).
at
M - di ■
r
\Q______
Fig.11.74
* , dm>i ana — = apv, 
dt
Fig. 11.75
—* 0
Fext =apv[(-Vj)-(v/)]
= -apv2(f + j)
The magnitude of Fext = V2apv2 and applied at an angle of 135° with positive x-direction.
Student Task^
> A non-viscous liquid enters the tube of uniform cross-sectional 
area a with a speed v and leaves the bent tube with same speed. 
Find the force exerted by the liquid on the tube.
0
Ans. F = 2pav2sin- applied at an angle of = =
0
n — with +ve x-direction.
2
Ex. 32. A smooth piston of area A is pushed by a constant horizontal 
force with a velocity V]. This results in escaping the incompress­
ible non-viscous liquid of density p, with a velocity v2. Find 
the magnitude of F.
57
dm
Since, the mass “dm” changes its KE from —v2 to
we have •
Then, we have
Substituting
Ans.we have
System of Particles
Sol. When we push an elementary segment of liquid of mass dm 
with a velocity v, through a distance In the above example, if you proceed with applying impulse-momentum equation we will 
equate the impulse F dt with the change in momentum dm(v2 - vt). Then the force applied 
F can be given as : F = —(v2-v(), where — = Apv{. This gives, F = Apvfy2 -v}) which 
dt dt
is different altogether! Where is the discrepancy? Explain.
Ans. The net force cannot be the applied force which is opposed by the 
components of horizontal reaction forces given by the tapering tube.
(b) Variable mass system : Constant mass approach : Let us take an example of a rocket 
which moves with a velocity after ejecting a gas of mass m2. Let us assume that the ejected gas 
■moveswith a velocity v2 . If the mass of the rocket + remaining gas inside the rocket is equal to ,
dm
2
58
Fext
This gives
Since, (m]+m2) = constant,
...(ii)we have
...(HD
Hence,
Substituting
-(iv)
,...(v)we have
dP _ d(mxv\ +m-jV-]) 
dt dt
dP dm} V| m}dvt dm2v2 m2dv2
dt dt dt dt dt
dt
^21 = 0 in eq. (hi), we have 
dt
dPSubstituting — from eq. (iv) in eq. (v), we have, 
dt —>
dm. dv —> —>
where (v2-V])(= vre|,say) is the velocity of the escaping gas relative to the variable mass 
system “ ” (rocket + residual gas).
—> —>
dP -*dm. dv. - = + m,_
Since, the change in momentum of the system (m, +m2) is caused by the external force Fext, 
using Newton’s 2nd law of motion for system of particles,
ext dt
G.R.B. Understanding Physics MECHANICS (Part-B) 
(ni] +m2), that is, the total mass of the system (rocket + gas) Fext zl^.v2 ,~vp % 
remains constant. The momentum of the above system is, —[^2] | m 1 | '
P = mlvl + m2v2 The rocket of mass mv velocity
The rate at which the momentum of the system changes is *1 forms a closed system with its 
ejected fuel of mass m2 which moves 
given as . with ve|0Cjfy v2i (mi + m2j = c
Fig. 11.77
dP dm. dv. dv2
-----= HVi-V-J + ZHi—- + m->—- dt dt 1 2 1 dt 2 dt
After leaving the “system m} ” (rocket + residual gas), the “system m2” that is (ejected gas) 
moves with constant velocity v2,
dm\ _ dm2
~dt~~~~dT
dm2 dm\
Substituting -— = ——
at at
from eq. (ii) in eq. (i), we have
-> 
dv2
dt
59
where “ v,
where
Applying Newton’s 2nd law
we have
-> 
dv = m—> 
dt
* dm„ 
rel dt
^imp
^ext+ ^imp
-» dm
Let us try to write Newton’s 2nd law for Variable Mass System (VMS) in following way :
—>
Suppose, the VMS has a mass m and velocity of its CM is Vj at any instant. Then, the momen­
tum of VMS can be given as :
-* dm dv
ext rel dt dt
is called reaction or impact force ^mp
Then, we can write,
P = mvx
d , 
^ext =^-('”vl)
dt
dv\ .v\dm
ex‘ dt dt
or ^reaction
System of Particles
In general, if m is the mass of variable mass system moving with a velocity v at any instant and 
Fext is the net external force acting on it ^excluding the reaction force of the incoming or outgoing 
mass), putting vj- v2 = - vre|, W] = m and vj = v, Newton’s 2nd law can be written as :
This gives
Which is totally different from the obtained expression, 
=> dm dv 
F„.. = -VreI----- + m----- 'ext reI dt dt
Then, where do we commit mistake? Probably ignoring the rate of change in momentum of
-> dm2
,little escaping stuff, that is V2~T~> makes all the difference! However, you may consider the dt
above (generally a wrong) formula, correct only when v2 = 0 (the special case when the mass 
leaves VMS with zero velocity).
Impact force : We have derived the expression for the reaction or impact force little ago. We can 
also find this by using the following method.
-» dm
dt
60
VMS 
□ dm
VMS
Substituting v2~vl =vrel>
we have
This gives
Sol.
4 =
nil 
ni l
The elementary mass “dm" changes 
its momentum from dm v5 to "dm v2 
during time dt
Fig. 11.78
V1
. t = tj 
F
□
t = t, + dt 
vi
G.R.B. Understanding Physics MECHANICS (Part-B)
Method - 1 : Let us take an elementary segment of mass 
dm fixed with the VMS. By applying suitable force (may be a 
variable force) somehow we keep the velocity of VMS constant, 
that is, (say). Suppose, the elemc.tary mass leaves the VMS 
during time dt with a relative velocity vre| and starts moving 
with a velocity v2, (say).
Hence, v2 = v( + vrel
Then the change in momentum of the elementary seg­
ment dm during time of impact dt is,
dP = dm v2 -dmv\ = dm(y2 - v,)
During a time t, the impulse of Fext is,
Imp = J^xl dt = Fti
Impact force (on an elementary mass or on VMS) depends on the—(i) rate at which the VMS 
gains or loses mass (ii) velocity (magnitude and direction) of the incoming or outgoing mass
relative to the VMS. If the VMS loses mass, — is -ve, if it gains mass, — is +ve.
dt dt
—>
Ex. 33. A trolley car of mass Mo initially moving in the rain with a velocity Vq towards right, experi­
ences a constant horizontal force F. If the rain falls vertically such that the rainwater col­
lected by the trolley car per unit time is P, find the velocity of the trolley car after a time t 
measuredfrom the instant of beginning the interaction (collection') of rainwater with the trolley 
car. Neglect friction at the ground.
Method-1 (Constant mass method) :
Let us take total rainwater plus trolley car as a system. The net horizontal 
force acting on the system is,
v0
„ -► F
nmmimnmnn
Fig. 11.79
dP = dmvte\
If the elementary segment experiences a force ^mp to change its momentum during time dt, by 
using Newton’s 2nd law or impulse-momentum equation, we can write,
^imp dl ~ vrel
r* -> dm
^imp - vrel
61
(Mo+ Am)
...(ii)we have
This gives Ans.
Am = gt
Here,
Then, we have vrel=-vj
iiimiiiuiiiiiiniti
Fig. 11.81
Fig. 11.80
>/M0
*■ Fext
—>
^imp
-> dm
-’"■T
dm 
dt
Method-2 (Impact force method):
We can solve the above example by impact force method as following. As 
the raindrops are continuously colliding with the trolley car, it keeps on exchang- Fimp
because the VMS gains mass and (vre|) = horizontal velocity of raindrops relative to the trolley 
car. If the trolley car has a velocity v at any instant,
--------F
—*■ v
vrel=(vr)x-(v/)x.
where (vr)x = 0 as the rain is falling vertically and (vz)x = vi
Equating impulse from eq. (i) and AP from eq. (ii) using impulse-momentum equation, 
we have Fti = (Af0 + pi)vi - Movoi
Ft + Movo
V —
Af0 + pr
ing momentum with the trolley car. Due to the change in horizontal momentum of 
the raindrops, we can observe a forward impact (reaction) force on each collid­
ing raindrop. In consequence, an equal and opposite (backward) impact force 
pfmp acts on the trolley car. Let us calculate it by using the derived formula,
System of Particles
Let the rainwater of mass Am collected in the trolley car during time 
/ move horizontally with a velocity v.
This causes a change in momentum of the system from P{ to P2 in 
horizontal during time /, which can be given as :
AP = P2-Pl,
where P2 = (Mo + Am)vz" and P} = Movoi
----A A
Then, we have AP = (Af0 + &m)vi - Movoi
Since, the rainwater is collected at a constant rate of “ p ”, Am = p/ 
Substituting Am = p/,
AP = (Af0 + p/)vz - MqVqi
62 G.R.B. Understanding Physics MECHANICS (Part-B)
in the formula
we have
^net ( ^ext ^imp) ~ ma->
This gives
Substituting
Evaluating the integration, we have Ans.
constant mass approach, you will get v and then
Student Task
Vrel = -vi
> In the foregoing example, if F = 0, can you apply the principle of conservation of linear 
momentum? Explain.
Ans. Yes; Px = 0 on the system (rain + trolley car) because Fx = 0 
but not on the trolley car because (^net)x = impact-
Substituting — 
dt
force ^nP
= p and
—> dm —
In impact force method, find Fimp by considering the signs of — and vrel; then add the impact 
with the given applied forces on VMS. Finally, the net force Fnet(= Fapp + ^imp) *s 
-> —> d v
equated with ma, where a =— and m = mass of VMS = + pt. Then do the mathematical
dt ->
work by solving the differential equation. In impact force method you will get a, then v , but in 
->
-> dv a =—.
dt
rr dt
J0A/0 + pt
„ dv F -pv = m— 
dt
m = M0 + pt and rearranging the terms, we have
dt dv —
Mo + pt F - pv
If the velocity of the trolley car changes from v0 to v during time t, integrating both sides, we 
have
Negative signifies that acts backward on the trolley car.
After finding ft , apply Newton’s 2nd law on the VMS to obtain.
rv dv
Jv0 F - pv
Ft + Movo
V — '
M0+\it
dvwhere a = — ,Fext =F and FJ = -pv 
dt r
dm 
'“imp ~ vrcl ’ 
dt
^imp = —
Sol. Fig. 11.82
Hence,
If the velocity of falling part is v (=-vf) its momentum is,
(>)
...(ii)
using eq. (i) and (ii), we have
Substituting
we have
F = -(mg+ 2\igy)j,we have
v2— + mg IJ
dm' 
dy
System of Particles
Ex. 34. A chain of uniform linear mass density P is releasedfrom rest from verticalposition 
as shown in the figure. Find the reaction force offered by the ground after the chain 
falls through a distance y.
’ ->dm 
=------ + v —
dt
'♦m + m‘ = M 
' = const
Nt|Z 
m’g
Fig. 11.83
m + m' = M. 
dm 
dy
dP_ 
dt dt 
Since, the falling part has a free fall acceleration,
dP— = -{\a(2gy) + mg}j 
at 
mass of the falling part.
d~P "*
Equating — with the net force F, 
dt
Then, 
where m =
A
/j = -mvj
Since, the fallen part does not move, its momentum P2 is zero. Then, the total momentum of
the system (total chain) is, P = P{ + P2 = -mvj
The rate of change in momentum of the system is,
Method-1 (Constant mass approach) :
Let us consider the total chain as a system of constant mass M. If the , - N 
mass of the chain fallen is th* and mass of the failing part of the chain / i 
is m, q
we have - ■ 1 ■ ®
( dm_ v . —
I dt
dm dy 
-------- = pv, 
dy dt
dv 
dt
dP 
dt 
dm 
~dt 
dP 2 
— = pv + mg
Since, the chain falls freely through a distance y, 
v2 = 2gy
d(mv) mdv
dt
64 G.R.B. Understanding Physics MECHANICS (Part-B)
Ans.
where
where
Hence, we have
0 in the formula Fnet = m'a to
N-m'g- 2^gy = 0,
Ans.N = 3pgy
Since, each elementary segment dm of the falling mass changes its velocity from -vj to zero, 
the change in momenta of dm can be given as : dP -dmvj during time dt.
Hence, the reaction (impact) force acting on the mass dm is,
where
This gives
Then, we have
where m = py
This gives V = 3pgy
Method-2 (Impact force method) :
Since, each falling link collides with the fallen links with a relative velocity
where m' = py 
This gives
~F - {-{mg + m'g) + N} j (referring to FBD)
- m'g + N = -{mg + 2jj.gy)
N = m'g-2\kgy,
m'g |N
Fig. 11.84
the net force on m' is, Fnet -{-m'g-2\kgy + N)j
Since, the acceleration of the fallen part is zero, substitute a = 
obtain
vrel = ~VJ> 
v = velocity of the falling chain, we have the impact formula for the
VMS “m” given as: ^mp=vre|^-,
dt
where vre) = -v j and ---- = pv (positive, as mass m' is increasing)
dt
where — - uv. 
dt
dP dm -
This gives 7
(impulsive or impact) force Fjmp . However, the interacting (dark) 
link will receive an impact force ^mp from the fallen part and gives 
an equal and opposite impact force to the fallen part. In consequence 
the falling part moves with free fall acceleration g.
66
1.
13.
1.
2.
(a) vc =
(b) KE of the system (M + m) about CM is equal to
(c) The maximum work done by the spring is equal to -
(d) x,
11.
12.
9.
10.
G.R.B. Understanding Physics MECHANICS (Part-B)
Assignments
2.
3.
4.
5.
6.
7.
8.
Mm 
(M + m)k
Two particles of masses m and 2m move with accelerations 2ai and -ai respectively :
(a) The acceleration CM is zero (b) The net force acting on (m + 2m) is zero
(c) The velocity of CM must be zero (d) The bodies must interact with each other
Two blocks of masses m 
Two particles of masses mx and m2 in projectile motion have velocities V] and v2 respectively 
at t = 0. They collide at time t = t0. Their velocities become and v2 at time t = 2r0 while still 
moving in air. The value of | (wj v[ + v2) - (m2 vj + m2 v2) | is : 
(a) zero (b) 2(m1+w2)gr0
(d) “("'i + "’2)2'0
V *
-- - - J 1 . • ..1 1 •
2’ IVlUliVV W VI IV UVUV
and strikes the block B which moves with a velocity as shown in the figure. If the hose pipe 
moves with a velocity vj and the cross-sectional area of hose pipe is a, the impact force of 
water (assuming that water falls dead after colliding with the block) is :
(a) -^apv2 (b) aPv2
(c) japvA block of mass m is placed gently onto a long plank of mass M fm-]
moving with a velocity v0 on a smooth horizontal floor. If friction is m ~*vo
... . , . 77777777777777777777777777777777777777777"
present between M and m :
Mv0
M + m
69
1.
2.
3.
4.
5.
6.
7.
8.
System of Particles
15. When a chain is lowered onto the ground such that it will be heaped :
(a) the fallen part does not experience reaction force due to the falling part
(b) the falling part does not experience reaction force due to the fallen part
(c) the fallen part experiences reaction force due to the falling part
(d) the reaction force offered by ground must be lesser than the weight of the chain 
Assertion-Reason Type Questions
(a) Both assertion (A) and reason (R) are true and R is the correct explanation of A.
(b) Both assertion and reason are true but R is not the correct explanation of A.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
(e) Both assertion and reason are false.
A : Even though no real mass is present at the CM, its acceleration cannot be defined.
—> X p —>
R : aCM = —" = act> f°r uniform gravitational field.
A : The net work done by internal force is zero.
R : The net internal force is equal to zero.
A : The net work done by friction and spring forces does not depend on the choice of reference 
frame.
R : The net internal force is zero.
A : The locus of many particles projected in all possible directions with same speed is spherical 
relative to another freely falling body.
R : The particles move equal distances in all directions relative to the CM which falls freely.
A : When there is no net external force acting on a system of interacting particles, the momentum 
of the system is conserved.
R : According to Newton’s law, the internal forces are action-reaction pairs leading to zero net 
force.
A : The total KE of a system of interacting particle is conserved when no external force acts on 
the system.
R : The net force of interaction is zero.
-> F
R : flCM - ~ 
Lm
A : The CM does not accelerate due to internal forces.
R : The total internal force is zero.
A : The centre of mass and centre of gravity must coincide.
G.R.B. Understanding Physics MECHANICS (Part-B)
(P)
(q)
(r)
(s)
andvc =
where
The maximum acceleration of the man is:1.
(a) pg
(c) Kc
 mv(b) — 
M
(c) zero
M — m , , . , . , ,------- v, where v= velocity of man relative to the boat.
M + m
The recoil velocity of the boat is : 
m — v 
M 
m ---v 
M + m
M
(c) ------- vv 7 M + m
(d) v, where
ml
M + m 
, ml* (d) —tM
72
phhywwwtrR
7/77777777777777777777777777777777777777,
(b)
(c)
2.
.2
(d) none of these(c)
3.
(a) F
(d) none of these(c)
4.
(b)(a)
(d)(c)
5.
(b)(a)
(d) none of these(c)
6.
(b)(a) zero
(d)(c)
2Fm2 
k(mx + tm2) 
2Fmx 
k(mx+m2)
G.R.B. Understanding Physics MECHANICS (Part-B) 
Using the foregoing theories answer the following questions.
Two blocks of masses 7?i| and m2 are connected by an 
undeformed ideal spring of stiffness k. A horizontal force F 
acts on the block mx. Neglect friction between the blocks and 
ground.
Sum of work done by pseudo force (relative to the CM) and external force F till the spring is 
elongated by a length x, is :
(a) (b)
2 mx + m2
(d) ifc? 
ffj] + m2 2
Upto the maximum elongation, the total work done by all forces (relative to ground) is :
F2
(b) ---------------2(zM] + m2)k
mxF2 
(mx + m2)k
_2 j?2 It r
2mxm2k
m\ 
m2(mx + m2)k
F 
y]k(mx +m2)
The maximum elongation of the spring is :
2Fmx 
km2 
2Fm2 
kmx
The maximum KE relative to the CM is :
mxF2
2(m] + m-^k
mjF2
2(tt1] + m2)2 k
Till the maximum elongation of the spring, the total work done by all forces is : 
n2F2k 
2mxm2
% F mxm2 
8(/M] + th2)2£
(a) zero
n2mxm2F2 
2(t71] + m2)2 k
The maximum relative velocity between the blocks is :
(b) — +m^k
TH] mxm2
73
6.
12.
, where m = massK2. Then the relative velocity between the reference frame is
16.
9.
7.
8.
The total work done by static friction on a system of particles is
The net work done by kinetic friction is
The net work done by constraint forces is
In a cracker, the work done by internal forces is
The work done by gravity relative to the CM of a small object is
If two surfaces move relative to each other by a distance x being pressed by the contact forces 
N, assuming p = coefficient of friction, the net work done by friction is 
The gravitational potential energy of a cube of length /, density p is
The minimum kinetic energy of a closed two-particle system of masses m} and m2 moving with 
velocities V| and v2 relative to an inertial frame is
If a bullet of mass m leaves the free gun of mass M with a horizontal muzzle velocity u, the recoil 
velocity of the gun is
13.
14.
15.
8.
9.
10.
11.
1.
2.
3.
4.
5.
2|*|-*2I
m
The CM does not move if Fnet = 0.
The net effect of internal force is zero on any particle of the system.
The KE of a system is minimum about the CM of the system.
The gravitational potential energy between the particles does not depend on the coordinate system. 
The momentum of a system about the CM is zero even though the system experiences a net 
external force.
The CM and CG coincide for an object.
7. . The CM must lie inside the body.
The momentum of a system is equal to the momentum of CM of the system.
The net work done by the internal force is zero in shifting the CM.
The net work done by the internal forces may be zero, positive or negative.
Impulse and impulsive forces are same.
KE of a system in a reference frame is K\, the KE of the system in another reference frame is
System of Particles
True Or False
of the system.
To conserve the linear momentum of a system, each particle must experience no net force.
F = ma cannot be applied on a variable mass system.
The work done by internal forces is always equal to the change in KE of the system about the 
CM.
The internal forces as a whole cannot change the KE of the CM.
Fill In The Blanks
1.
2.
3.
4.
5.
6.
G.R.B.Understanding Physics MECHANICS (Part-B)
1. P
x
y
2.
2
Find the CM of the annular disc of radii R and 2R.3.
Find the CM of the shaded region of a circular plate.4.
2R.
5.
Find the CM of a water molecule.6.
7.
-—M
74
Problems
A constant force F acts vertically at the mid-point of a massless 
inextensible string which connects two particles each of mass m. If 
the particles do not lose contact with the smooth horizontal floor, 
find the acceleration of the particles and CM of the system of par­
ticles at the given position.
Find the CM of the system of a solid hemisphere capped by a solid cone 
of same material.
8. Two particles of mass and m2 are connected by an inextensible string 
which passes through a fixed rough pulley. The coefficient of friction between 
the string and fixed pulley is If the particles move with a speed v, find the (i) 
velocity of CM (ii) acceleration of CM, of the system of particles (mi + m2) ■
CM Calculation
Find the CM of a stick whose density varies linearly with distance from one of 01
its end O as X = Xn -
0 I
Find the CM of the shaded area of a uniform plate.
75
9.
— F
//'P//7//////7P////////Z/P//Z///////////////////W
10.
11.
F
12. m
v
v
13.
14.
•■v 
M
T1
I
R
Im
are connected with the central bead of (m)
l
(M>
/
(m)
Two identical particles each of mass m 
mass M by two identical inextensible strings each of length /. The middle mass 
M is pulled by a constant horizontal force F. Describe the motion of the particles 
and CM of the system.
I
A smooth horizontal tube of mass M and radius R is pushed towards 
left with a velocity v. Simultaneouly another two particles each of 
mass m placed at the opposite ends of the vertical diameter of the 
tube, are pushed towards right, with velocities v. Find the : /
(i) velocity of CM of the system (Af+2m). \
(ii) velocity of the tube just before the particles meet
(iii) relative velocity with which the particles meet.
A uniform chain of mass m and length I is placed on a smooth vertical
semicircular tube of radius Find the : (i) velocity of the chain if
2nJ
half of the chain falls from the tube, (ii) velocity of the chain just after 
leaving the tube.
A uniform chain of mass m and length Z hangs from a table top by a length x.
(i) Find the acceleration of the chain.
(ii) What is the speed of the chain at the time of leaving the table top?
(iii) Find the work done by gravity in pulling the entire chain down from 
the table.
(iv) Find the work done by an external agent in slowly pulling the entire 
hanging chain onto the table.
Three insects of masses ?r?|, w2 and m3 move along a light inextensible string 
a i 
with accelerations alta2 and a3 as shown in the figure. The string passes over | 
a smooth pulley. Find the : (a) acceleration of (i) CM of the system (ii) each 
insect (b) tension in the string (c) net force acting on the system (zn, + /h2 + zn3).
a2 it
System of Particles
A smooth block of mass m slides on a wedge of mass A£ with 
a velocity v at the lowest position. The wedge moves towards 
left with a velocity v. If the coefficient of friction between the 
wedge and ground is p and a constant force F acts horizon- v describe the motion of 
the CM of the blocks.
In the problem 17, if the particles are connected by an inextensible string, , 
find the acceleration of CM of the system of two particles. //
A spring is compressed by a thread connecting two particles (blocks) of masses 
W] and m2 ■ The blocks are rigidly connected with the spring. If the string is cut, 
the blocks move vertically.
(i) Find the minimum compression x of the spring so that the block wj will lose 
contact with the floor.
Two particles of masses and m2 slides down the smooth inclined 
planes as shown in the figure. Find the acceleration of the CM of the 
system ml+m2.
In the above problem, let the wedge slide along the smooth horizontal surface. 
Find the acceleration of CM of the system of particles 1, 2 and 3.
^Zzzzzz/zzzzzzzzzzzzzzz
1
X
T
1
A uniform smooth chain of length l = nR slides in a vertical plane 
from the given position along a smooth tube.
(i) Find the speed of the chain when the entire chain slides on the 
horizontal floor.
(ii) What is the total displacement of CM of the system when the 
chain just leaves the curved track?
(iii) At the given position find the : (a) magnitude of acceleration of 
each element of the chain (b) acceleration of CM of the chain.
(iv) What is the (a) KE (b) t/gr of the chain if half of the chain slides along the curved path?
/9
'7777/777777777777777777777777777777777777
m2
/$ 29 0\
/////////77/////////////////7//
G.R.B. Understanding Physics MECHANICS (Part-B)
IQ
77
21. %
M
22.
M
xo'
23.
24.
m.
25.
26. m
R
I 
h 
1
y
J-
System of Particles
A block of mass m is released from rest from the top of a smooth 
wedge of mass M. Find the : (i) displacement (ii) velocity (iii) 
acceleration of the wedge M, just before the block touches the ground.
m
‘ &
7777777777/ V I /
1 M X__ y
^777777777777777777777777777777777777777777777777
A man of mass m runs from rest from one end of a boat of 
mass M and length I with an acceleration a relative to the 
boat. If the viscosity of water is neglected, find the:
(i) acceleration of CM of the system (M + m).
(ii) acceleration of the man and boat.
(iii) displacement of M and m.
In the above problem, if the man does not slide relative to the boat, find the : (i) frictional force 
(ii) work done by friction on man and boat (iii) total work done by friction (iv) work done by man 
(v) velocities of man and boat (vi) velocity of CM of the system (A/ + m) (vii) displacement of 
CM of the system (Af + m).
A small smooth ring of mass m which is free to slide along a rigid hori- j
zontal wire, is connected to a bead of mass M by an inextensible string 
of length I. The system (M + m) is released from rest when the string is 
horizontal as shown in the figure. Describe the motion of the particles M and 
Two blocks of masses m} and m2 are connected by an inextensible 
spring of stiffness k. The block is pushed with a speed v0 to 
left when the spring is undeformed when the block m2 is just ......
touches the wall. If we ignore the friction between all contacting surfaces.
(i) Find the variation of velocity of CM of the system (zm, +m2) as a function of time.
(ii) How does normal reaction on the block m2 offered by the vertical wall vary with time?
(iii) How does the KE of the system relative to CM change with time?
(iv) Find the maximum deformation of the spring when m2 loses contact with the vertical wall.
(v) Describe the subsequent motion of the blocks.
A bead of mass m is released from the top of a smooth 
wedge of mass M. In consequence, the bead slides down 
the semicircular surface of the wedge.
(a) Find the:
(i) variation of normal reaction offered by the vertical 
wall of the wedge.
(ii) variation of vertical reaction given by ground onto the wedge, till the wedge loses contact 
with the vertical wall.
m
/
k -v°
'^7777777777777777777777777777777/7777/777777777
78
27.
M2
28.
M
29.
m
k
30.
■
T 
h
T 
h 
i M I
’777777777777777777777777777777777777777777/7777777777777
W
7/7/7/7777/7777777/77/777777777/7
G.R.B. Understanding Physics MECHANICS (Part-B)
(b) How high will the bead rise after the wedge loses contact with the vertical wall?
(c) Describe the motion of m, M and the CM of the system (M + m).
A bead of mass m is released from the top of a wedgeof t 
mass . Just after reaching the ground, the bead climbs I 
onto another wedge of mass Af2 • The friction between all I 
contacting surfaces are neglected. A
(i) Find the speed of m when it reaches the ground level.
(ii) What fraction of the total mechanical energy is carried by the wedge ?
(iii) How high will the bead rise onto the wedge M2
(iv) Find the variation of horizontal component of vCM for the system + m)} (M2+m) and 
(A/] + M2 + m) with time.
A bead of mass m is projected with a velocity v0 so that it climbs 
onto a smooth primatic wedge of mass M without bouncing.
 Find: . . .
(i) the change in velocity of the bead and the impulse associ- 
ated with its sudden change in velocity.
(ii) the velocity with which the bead starts climbing up onto the wedge.
(iii) the velocity of the wedge when the bead starts climbing onto the wedge.
(iv) the maximum height attained by the bead.
(v) the magnitude of v0 so that the bead climbs upto the top of the wedge.
(b) In (v), let v0 = vc; if v0 > vc what will happen to the bead? Explain.
A block of mass m is released from rest from a height h onto a smooth sledge of mass M fitted 
with an ideal spring of stiffness k.
(a) Find the :
(i) velocity of the block and sledge just before the 
block touches the spring.
(ii) maximum compression of the spring.
(iii) time of contact of the block with the spring.
(iv) maximum work done by the spring on the block m and sledge.
(b) Describe the subsequent motion of m, M and CM of the system (M + m).
(c) Explain the critical case when M -> oo.
A block of mass m is projected up along the slant of a smooth wedge of mass M with a speed v0. 
Find:
(i) the velocity of CM of the system (Af + m).
(ii) time after which the block reaches the highest position.
(iii) acceleration of CM of the system (Af + m).
(iv) maximum height attained by the block.
79System of Particles
31.
v2 h
32. I
0
M
7777777777777777777777777/7777
33.
M P
vo
34.
jti
m -o
R—?O 
■P m 
................................... .........
77777777777777777777777777777777777777777777777777777777
0^2
A particle of mass m\ is projected to right with a speed V|
onto a smooth wedge of mass m2 which is simultaneously
projected due left with a speed v2 ■
(i) If the particle attains the highest point of the wedge, n—X1
find h. 77ff77777777777777777777777777777777777777777777?7777/77
(ii) If the given height h' of the wedge is less than h the particle leaves the wedge, find the 
velocity of the particle at the time of losing contact with the wedge.
(iii) Referring to (ii), find the maximum height attained by m{.
(iv) Will W] reach the wedge again? Explain.
(v) Describe the subsequent motion of m{.
(vi) Discuss the case when /n(V] = m2v2.
A particle of mass m is connected to a block of mass M by an 
inextensible string of length /. The block is free to slide smoothly on 
the ground. When the particle is released from the horizontal position 
of the string as shown in the figure, find the:
(i) angular speed of the particle relative to 0 and the linear speed of 
the block when the string makes an angle 0 with horizontal.
(ii) tension in the string as the function of 6.
(iii) maximum speed of m relative to O.
(iv) speed of M when the particle is about to hit the wedge.
(v) maximum tension in the string.
(vi) normal reaction offered by ground onto the wedge as a function of 0.
A small ball of mass m is projected with a minimum horizon- , t
tai velocity v0 on a smooth wedge of mass M so that it will 
reach the highest point P of the wedge.
(i) Find the value of v0.
(ii) With what velocity will the wedge recoil when the ball 
leaves the point PI
(iii) When and where will the ball strike the surface of the wedge after leaving the wedge at P1
(iv) Discuss the case when M ->oo.
A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R, is 
gently pushed towards right. As a result, the wedge slides due left. Find the:
(i) speed of the wedge.
(ii) magnitude of velocity of the bead relative to the wedge. Xi
(iii) normal reaction between M and m, as the function of the angular posi- [ M|0,R \
tion 0 of the bead with respect to the centre 0. mwr/wM/w/war,
(iv) position where the bead loses contact with the wedge.
80
35.
36. m2
vo
ro
37.
38.
39.
40.
41.
Vq
S'" M,,, I r
777777777777^/777777'
G.R.B. Understanding Physics MECHANICS (Part-B)
A block of mass m kept on a smooth wedge of mass M is interconnected 
by an ideal spring of stiffness k. If we release the system from rest, find 
the:
(i) displacement, velocity and acceleration of CM in horizontal.
(ii) velocity of m relative to M as the function of deformation of the spring.
(iii) maximum velocity of m relative to M.
(iv) maximum compression of the spring.
(v) velocity of the wedge when m moves most rapidly with respect to the wedge.
(vi) maximum acceleration of the block relative to the wedge and ground.
A particle is kept at a distance rQ from the particle m2, m2 is
projected with a velocity v0 towards right. Taking the gravitational
interaction into account, find the:
(i) velocity of CM of the system (mj +m2).
(ii) acceleration of the system (m|+m2).
(iii) maximum distance of separation.
Two smooth blocks of masses m and M are pressed against a com- k
pressed spring of stiffness k. If the maximum speed of m is v, find 
, "77/777/7777777777777777777777777777777777
the:
(i) maximum speed of M.
(ii) maximum compression of the spring.
(iii) velocity and acceleration of CM of the system (M + tn).
(iv) maximum distance moved by the blocks.
Variable Mass System
A trolley car of mass M is pushed with a velocity v0 on a smooth horizontal ,
floor. The car collects rain water at a constant rate P- If the speed of each A H | | 
raindrop is v, find the:
(i) horizontal force of colliding rain drops with the trolley car.
(ii) velocity and acceleration of the trolley car as the functions of time.
In the above problem, let the trolley car be pulled from rest towards right with a constant force 
F. Find the velocity and acceleration of the trolley car as the functions of time.
A smooth container filled with water moves with a velocity v0. If water leaks
at the base with a constant rate p kg/s, find the velocity and acceleration of — r Vq
the container as the functions of time. Assume M= initial mass of the system k > 
(container + water). zzzz>)zzz)jzzzmw
A rocket has initial velocity v0 and mass m0. If it travels in free space, find the velocity of the 
rocket as the function of its existing mass M. Assume that the rocket 
expels the fuel with a velocity wre] with respect to itself.
81
42.
43.
44.
fi
x
45.
46.
w\\\\\\
I ------- in
A chain of mass tn and length / is hanging from a ceiling so as to 
just touch the upper surface of a long trolley car of mass M 
moving with a velocity v0 of a smooth horizontal surface. If 
the chain is released from rest from the ceiling,
(i) find the speed of the trolley car as the function of time.
(ii) what is the distance covered by the trolley car till the entire 
chain lands on it?
A chain of length I is heaped as A and B after passing over a smooth pulley. 
The vertical distance between^ and B is equal to h.
(i) If the chain is released from rest find the speed of the chain as the func­
tion of time till the heap at A will get exhausted.
(ii) Find the value of h so that the chain moves with uniform speed v.
^hooper
—-F
tion a (b) velocity v, as the function of the length x of the chain pulled up.
(ii) Find the work done by the driving force and power delivered by the external 
agent.
(iii) What is the normal reaction offered by ground?
(iv) Find the acceleration of CM of the chain.
A chain of mass m and length / is heaped on a smooth ground.
(i) Find the vertical force F required to pull the chain with constant (a) accelera- 111
System of Particles
A cart is pulled with a constant force F with a constant velocity v on a smooth 
horizontal ground. A hooper loads the cart with sands at a rate of P kg/s.also can check the relevant steps more easily. - 
Even though some students are deprived of getting proper guidance, sincere reading of this book may 
guide them step by step to improve their analytical skills to the highest level. The prime motto of this 
book is to make the General Physics more interesting and accessible to those who think that physics is 
tough and rigorous. If my effort will stimulate your interest in General Physics, I think it as the greatest 
reward for me.
I extend my sincere thanks to Mr. Prakash Chand Bathla, proprietor of Prakash Publications for 
undertaking this publishing work. Furthermore, I am grateful to my students and colleagues to review 
the matter scrupulously. I humbly request all of you to draw my attention to all sorts of errors which will 
help me to revise the book in next edition. I will remain indebted to all my readers for their sincere 
suggestions and positive criticism to improve the quality of the book.
CONTENTS
CHAPTER -12CHAPTER -11
COLUSIONS 87-132'SYSTEM OF PARTICLES 1-86
11
28
49
114
55
37
48
1
2
6
122
122
124
125
125
130
118
118
94
95
99
103
66
66
69
70
70
73
73
74
82
W 
i#8
91
107 
lastic
12.1 Introduction
12.2 Definition.
12.3 Impulsive Force and Impulse
. 12.4 Types of Collision
12.5 Conservation of Linear Momentum 
in a Collision
1?.6 Energy Consideration in Collision
12.7 Elastic Collision
12.8 Elastic-Oblique Collision
12.9 Complete Inelastic Collision (e = 0) 104
12.10 General Solution for Head-on
Collision between two Bodies
12.11 General Solution for Oblique Inel; 
Impact of two Smooth Bodies '
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 121 
O Match the Columns
O Comprehensions 
O True or False
O Fill in the Blanks 
O Problems 
O Answers
11.1 Introduction
11.2 Definition of Centre of Mass
11.3 Characteristics of Centre of Mass
11.4 Calculation of Position of Centre of 
Mass (Finding CM)
11.5 Work-Energy Theorem for System 
of Particles
11.6 Impulse and Momentum Equation 
for a System of Particles
11.7 Advantages of CM Frame
11.8 Application of Impulse-Momentum 
Equation and Work-Energy Theorem for 
a Two particle System
. 11.9 Application of Impulse-Momentum 
Equation and Work-Energy Theorem for 
a Many Particle System
Assignments
O Discussion Type Questions 
O Multiple Choice Questions
O Assertion-Reason Type Questions
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks '
O Problems
O Answers
( Viii )
CHAPTER-13
ROTATIONAL KINEMATICS 133-187
205
210
144
CHAPTER -15
224-319
231
232
CHAPTER -14
235
236
245
179
180
181
182
182
186
148
154
158
163
171
172.
174
176
177
188
188
DYNAMICS OF RIGID 
BODIES
193
194
197
201
215
216
217
218
219
219
220
222
224
225
227
14.1 Introduction
14.2 Torque about a Point
133
134
134
136
138
14.3 Torque due to Many Forces Acting 
at a Point
14.4 Torque about an Axis
14.5 Angular Momentum about a Point
14.6- Physical Significance of L
14.7 Angular Momentum about an Axis 202
14.8 Relation between T and L
14.9 Conservation of Angular
Momentum
Assignments
O Discussion Type Questions
Q Multiple Choice Questions
O Assertion-Reason Type Questions 217
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks
O Problems
O Answers
13.1 Introduction
13.2 Definition of a Rigid Body
13.3 Rotation (Angular Motion)
13.4 Angular (Rotational) Quantities
13.5 Types of Motion of a Rigid Body
13.6 Kinematical Equations for Rotation 141
13.7 Physical Interpretation of Angular 
Motion (Rotation)
13.8 Calculation of Velocity of a Rotating 
Body
13.9 Calculation of Acceleration
13.10 Concept of Rolling
13.11 Instantaneous Axis of Rotation
13.12 Radius of Curvature of the Path
Traced by a Point of Rigid Body
13.13 Rotation about a Point
13.14 Summary
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 179
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks
O Problems
O Answers
15.1 Introduction
15.2 Torque on a Group of Particles
15.3 Angular Momentum of a System
of Particles y
15.4 Relation between T and L
15.5 Conservation of Angular Momentum 
of a System of Particles
15.6 Newton’s Laws for a System of 
Particles
15.7 Angular Momentum of a Rigid Body 
about its CM
15.8 Calculation of Moment of Inertia of
a Rigid Body about Centroidal Axis 237
15.9 Newton’s Laws for Motion of Rigid 
Bodies
TORQUE ON A PARTICLE 188-223 
AND ANGULAR MOMENTUM 
OF A PARTICLE
Cix)
CHAPTER-17
287
359
CHAPTER-16
320-352
CHAPTER -18
GRAVITATION 403-474
15.13 Newton’s Equations for Rigid Body 
Motion Relative to a Non-Inertial 
(Accelerating) Frame
Assignments
O Discussion Type Questions 
O Multiple Choice Questions
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks
O Problems
O Answers
246
250
320
320
321
327
330
333
341
341
O Problems
O Answers
369
371
383
384
387
387
389
390
391
398
347
351
403
404
405
409
412
426
344
345
. 346
347
WORK AND ENERGY 
OF RIGID BODIES
18.1 Introduction
18.2 Kepler’s Laws
18.3 Newton’s Law of Universal
Gravitation
18.4 Inertial and Gravitational Mass
18.5 Gravitational Field and Field Intensity, 
Superposition of Gravitational Field 411
18.6 Calculation of Gravitational Field 
Intensity
18.7 Work done by Gravity
Body or System of Particles
17.6 Impact of Rigid Bodies
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 386
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks
O Problems
O Answers
IMPULSE AND MOMENTUM 353-402 
OF RIGID BODIES
17.4 Principle of Impulse and 
Momentum
16.1 Introduction
16.2 Gravitational Potential Energy
16.3 Kinetic Energy
16.4 Work done by a Force on a Rigid
Body
16.5 Work-Energy Theorem
16.6 Conservation of Energy
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 343
O Match the Columns
O Comprehensions
O True or False
O Fill in the Blanks
15.10 General Procedure to Solve the 
Problems of Motion of Rigid 
Bodies
15.11 Need of Friction in Rolling
15.12 Alternate Solution for Fixed Axis 
Rotation (or, it Rolls on a Fixed Surface)
279
17.1 Introduction 353
17.2 Linear Momentum of a Rigid Body 353
17.3 Angular Momentum of a Rigid Body
354
290
292
O Assertion-Reason Type Questions 296 17.5 Conservation of Momentum of Rigid
296
297
299
300
301
315
(X)
19.8
488
18.9 19.9
429
459
CHAPTER -19 CHAPTER -20
HYDROSTATICS 475-564 FLUID DYNAMICS
19.1
19.2
19.3
19.4
19.5
19.6
19.7
475
476
479
480
456
457
449
452
443
446
427
428
437
441
460
461
20.3 Streamline Motion
20.4 Equation of Continuity
20.5 Flux of ~v -Field (Find 
the:
(i) force F.
(ii) power delivered by the force.
(iii) rate of change of KE.
(iv) rate of energy loss and account for the loss of mechanical energy.
A chain of mass m and length / is kept vertical so that its lowest point just touches 
the ground. If the upper end of the chain is released, find the normal force exerted 
by the ground on the chain when the highest link of the chain falls down by a 
distance x.
I
I®
//////77///////////77
I M , J—^vq
77777777777777777^77777^7777777777777777
82
47.
T
48.
(T
M
777777777777777777777777777777
49. u
M
777777777777777777777777777777
50.
ANSWERS
8. (a,c)
12. (a,b,c,d) 13. (a,b,c,d) 14. (a)
4. (d) 5- (b) 6. (a) 8. (d)7. (a)
4. (b)2. (c) 3. (b) 5. (a) 6. (a)
5. (c)2. (c) (b) 4. (b) 6. (c)3.
9. (T)8. (T)2. (F)
11. (F)
(T)
(F)
4. (T)
13- (F)
5. (T)
14. (F)
6. (F)
15. (F)
7. (F)
16. (T)
m 
2000—-u
''/////////////
x
r
h
3.
12.
6. (a,b) 7. (c)
15. (b,c)
A cannon with shots of total mass A/o is kept on a rough horizontal surface. The coefficient of 
friction between the cannon and horizontal surface is P- If the 
cannon fires n shots per second with a velocity u relative to it, find 
the velocity of the cannon when it has a total mass M with the 
remaining shots, after a time /. Find the distance travelled by the 
cannon as the function of time t.
A wedge of mass M is kept on a smooth horizontal surface. A jet of 
water emerging from a nozzle of cross-section A strikes the wedge 
horizontally with a speed u. Assuming p as the density of water.
(i) Find the maximum thrust imparted by water jet on the wedge and 
the maximum power of the wedge.
(ii) Obtain an expression for the speed of the wedge as the function of time.
A smooth flexible chain is released from rest such that it slides over Ht 
the pulley and heaps at the ground which is situated at a depth h 
below the pulley. Find the velocity of the chain as the function of 
the distance x moved by its free end on the horizontal table.
Multiple Choice Questions
1. (a,b) 2. (a,b,c,d) 3. (a,b,c,d)4. (a,b,c,d) 5. (a)
9. (a,c,d)10. (c) 11. (a)
Assertion-Reason Type Questions
1. (d) 2. (a) 3. (d)
Match The Columns
1. a - r, s; b - q; c - p; d - q
Comprehension
Passage-1
1. (a)
Passage-2
1. (b,d)
True or False
I- (F)
10. (T)
G.R.B. Understanding Physics MECHANICS (Part-B)
A chain is heaped on a smooth table. If we pull one end of the chain 
slightly through the smooth hole made on the table, find the speed of the 
chain as the function of length x of the hanging portion of the chain.
83
2. negative 3. zero1. zero 4. positive 5. zero 6. (-jiACr)
8.7. 9.
A *
5. ~—J from the centre of the base of the cone
7. ac = 0, ax = — cotQ, if F if F > 2mg
| v, - v2 |2
4/nv2
R(M + m)
9. ac\ = —
Problems
21
1- ~3
28F
31 9n
F 
ac =----------
TH] + m2
1 /4 
-psi
f cote 
g' ax ~ ~ 2m
4mv‘
M + m
F
Part'c'es W'N accelerate towards the central particle and the central particle 
accelerates horizontally.
F
2m
R
6
gx
I
84 G.R.B. Understanding Physics MECHANICS (Part-B)
17. ax
a g
19. Qct = “(cos9 +cos20)zr
= ~~ (sin 20 - sin 0) j
21. (i) (iii)
23. (i) f = (») (*7)boat = (iii) zero
(v) vc = 0 (vi) =0 (vii) =0
25. (i) vc = VC
(iii) Kc =
/
0v) Xmax ~ VQ
and the blocks oscillate about the CM with
w =
J
= m\vo
M + m
(ii) N = (A//n1^)v0 sin
Mma
M + m
Mmal
M + m
I m\m2
S (mj + ot2)£
k(ml + m2) 
m^m2
M|Vq 
2(m| + m2)
/T
7?J]V0COS I---- 1
Nwi
+m2
; t>n 1^-
V1
; t>n 1^-
 mt sin 0 + m2 cos 0 
y ni| + m2
=of observation of the cosmic bodies as 
“Collision”.
The above example clarifies the fact that,
A process of strong and rapid interaction between two bodies which brings about a tremendous 
(quick) change in momenta of the interacting bodies is called collision.
Ex. 1. How do you explain the collision between two stars and galaxies?
Sol.
88 G.R.B. Understanding Physics MECHANICS (Part-B)
ot linear momentum and Newton’s impact formula to calculate the velocities of two colliding objects 
just after the collision, if their velocities just before the collision are known.
§ 12.2 Definition
If you throw a ball vertically up with a speed v = 10 m/s, the ball comes to rest instantaneously
V 2at its highest position, after a time t = — = I s; (assuming g = 10 m/s and ignoring air flow and air- 
g
resistance). It means, after a time of I s, the ball gives up all its momentum wv(= 10kg m/s) to earth, 
wv
Then, the average downward force acting on the ball is F = — = ION, that is what we call gravitational 
force (or weight) of the ball.
If you throw the ball vertically down with same speed just above the ground, you can see that the 
ball will lose its total momentum of 10 kg m/s much more rapidly than before. The ball comes to rest 
momentarily and then starts bouncing up. Since the ball gives off its entire momentum very rapidly 
within a short time of 0.01 s, say; it will experience an average upward force of one hundred times the 
weight of the ball! Then, who gives this force? Since the ball strikes the earth, it must be the earth which 
imparts this huge force on the ball decelerating it so quickly, but this force is totally different from 
gravitational force (pull) of earth.
Let us, now analyze the above two examples by comparing them. In first example, gravitational 
interaction between the earth and ball retards the ball to rest (momentarily) within Is. In second example, 
the earth and ball interact much more (100 times) strongly during a short time (1/100th of that in the 
first example) for the same change in momentum of the ball. In other words, a short, strong and rapid 
interaction between earth and ball which is totally different gravity from, and in general, much more 
greater than gravitational interaction, is called “COLLISION” or “IMPACT”. Since the time of interaction 
(0.01 s) between the bodies is much lesser than the time of observation (Is) of the body in air, the 
reaction force acting on the colliding bodies is very large.
Hence we define a collision as following.
Collisions 89
Sol.
The time of interaction should be very small compared to the time of motion (observation) of the 
interacting bodies, and during this time of interaction, the bodies must experience considerably 
larger forces compared to the forces acting on them in their general (or normal) state of motion.
In the previous discussion we have taken the example of a collision between the ball and ground. 
In this collision, physical contact takes place between the colliding bodies. It means, the neutral bodies, 
in general, touch each other during their collision.
Ex. 2. Is it always essential that the bodies would touch each other during a collision? Justify your 
answer.
The famous Gold Foil Experiment performed by Rutherford tells us that, a-particles do not 
touch the nucleus as both a-particles and nucleus are positively charged. Since electric field is 
zero outside the atom, the a-particles experience no force outside the gold atom. However the 
a-particles experience tremendous electrostatic field (force) due to the interaction of nucleus 
inside the atom. Since a-particles move very fast, they remain inside the atom for a very short 
time. During the short time of journey inside the atom, a-particles change their momenta very 
rapidly. Some a-particles get deflected by large angles, few of them retrace their path by 
bouncing back after stopping at the lowest distance from the nucleus. Hence we state that 
a-particles collide with the gold atoms (nuclei) experiencing a large force during a short time, 
without touching the nuclei. When one planet (or any huge heavenly body) moves close to 
another stars or planet without touching each other, we can also call it collision. For instance, 
our solar system came into existence by the motion of a huge body by the side of sun.
The above example clarifies the fact that,
In a collision the colliding bodies need not touch physically and during the collision the bodies 
experience a very large force compared to the force associated in their general motion. The 
time of collision is negligible compared to the time of observation of the bodies.
Now a question arises; is there any limit of interaction which can define the impact of collision? 
You may say a short time but how much short? Well, you can say a fraction of a second, one 
tenth, hundredth ... millionth of a second ... still smaller. On the other hand, the huge interaction of 
galaxies takes considerably long time of several years which is much lesser than the time of their normal 
or general motion.
§ 12.3 Impulsive Force and Impulse
Let us assume that a colliding body changes its momentum by A P during the time Ar of collision.->
A P Then the average change in momentum of the body is---- , which is equal to the average impact force
(force acting on the body averaged over the time of impact).
Impulsive force : According to Newton’s 2nd law, we have
i,='zr —>
Since the collision time Ar is very small and the change in momentum A P is finite, the body 
must experience a “huge average force” during the collision. That is what we call “average impact (or 
impulsive) force”.
90
F
J
which
mg
tl ‘2 t
Impulse of the impulsive force F and 
Fdt = area of F-t graph
Fig. 12.1
impulse of Fgr, that is, F& dt(= F&&t), is very small compared 
to \'2 F dt, we can neglect the impulse of gravity.
J/,
F”t=
^ext
0
G.R.B. Understanding Physics MECHANICS (Part-B)
Ex. 3. During a collision why is it convenient to use average force than the instantaneous force? 
Sol.
(
-» -> A POnce we have A P and A/, we can find Fav =----
k kt )
tells us about the “strength of collision”. Since the time of impact is
very small, the average impact (or impulsive) force is very large. jFdt
If you look at the F -1 graph more carefully, a non-impulsive Body B hits A vertically 
gravitational force continues to act during the collision. In general, the downward giving an impulse
P2 Fdt, where F is the impulsive
J‘i
force between A and B during
the time of collision r = t2 -1|
Fig. 12.2
When two bodies collide, due to elasticity of matter, millions of molecules of the colliding 
bodies are pressed against each other. Hence the impulsive force during the collision is a sum of 
millions of strong molecular interactions. As we know, variation of molecular force between
two isolated molecules follows the law given as F = . However, the net effect of
rb r
millions of molecular forces is extremely complex during a collision. In other words, the exact 
variation of impulsive force with time is incredibly complex. This limits us to use the differential
-> d p 
expression of instantaneous force in Newton’s 2nd law, given as F =---- . Now a question
-> d 
arises; when we do not know the exact variations of F as the function of time during collision, 
how can we find the final velocities (velocities of the bodies just after the collision)? The 
answer is to use integral form of Newton’s 2nd law “the Impulse-momentum equation”.
Impulse-momentum equation: When a body A moves in 
gravity, if we ignore viscosity of air, only a force Fcxt(= mg) acts 
on the body. If another body B hits the body A, the impulsive force N 
acts on A for a very short time Ar(= t2 - ft). As the impulsive 
(reaction) force F varies rapidly without following any definite rule, 
it is neither essential nor practical to use the instantaneous force to 
track any change in velocity or momentum of the body during a very 
short time of collision.Hence it is important to use the idea of impulse 
of F (instead of instantaneous impulsive force F), that is, j 2 F dt 
-->
which is equal to the net change in momentum AP of the body 
during the collision. Using impulse-momentum equation, we have
91
Common tangenti
C2
tangent
C2
Cinormal
Collisions
Recapitulating,
line of 
impact
Common tangent, line of impact and 
line of motion of the colliding 
Fig. 12.3
Common normal 
(line of impact)
Lines of motion
\ J line joining
l the centres
I C1 and C2
In central impact, the line of impact and 
line joining the geometrical centres are same 
Fig. 12.4
The net impulse on a colliding body is approximately equal to the impulse of the impulsive force 
during the impact, which is equal to the change in momentum of the colliding bodies during the -> _> a p
collision. The impulsive force averaged over the time of impact is given as F = —-. Since 
_> -> av Ar
| AP | is finite and A/ is very small, generally Fav is a large force. The time period of collision 
depends on shape, size and velocities of the colliding bodies and their elastic properties of the 
bodies. Hence the exact variation of impulsive forces with time is extremely uncommon. Then, 
the idea of average impulsive force is generally more useful.
§ 12.4 Types of Collision
Line of impact: Let two smooth bodies 1 and 2 start 
colliding at a common point P. Now draw a tangent at P and 
then draw a normal at P. Since the bodies are assumed to be 
smooth, no impulsive reaction acts along the tangent. Then 
the bodies must experience impulsive forces F} and F2 along 
the normal. In other words, we can say that impact takes 
place along the normal. Hence the normal line is known as 
“line of impact” of smooth colliding bodies. As the action­
reaction pairs, F\ and F2 are equal and opposite, 
F\ + F2 = 0. It means that the net impulse of the forces F} 
and F2 is zero.
Tangential impact force
If the bodies are rough, an action-reaction pair of tangential impulsive forces may act on the 
bodies along the tangent.
Line of motion : The line in which the centre of masses Ct and C2 of the bodies move is called 
the line of motion.
Let us now categorise the collision into various types depending on the position of the CM of the 
bodies relative to the line of impact as following.
Central impact: When the line joining the CM of the bodies lies on the line of impact, we call the 
impact “central impact”.
yp
\ z
'Vc?
92
line of impact
Ex. 4. Name the following impacts :
(/v)(»0
Sol.
Head-on impact (collision): When the both lines of motion 
and lines of impact of the colliding bodies coincide, the impact is 
called direct-central impact or head-on collision. For instance, 
collision of two identical balls moving in same line can have head-on 
collision.
Eccentric impact: When the line joining the 
CM of the colliding bodies does not coincide with 
the line of impact, this is known as eccentric impact. 
Impact of cricket ball and bat is the familiar example 
of an eccentric impact.
(i) Line of impact, line of motion and line joining the CM are collinear. Hence the collision is 
head-on (direct-central).
In head-on collision, line of motion, 
line of impact coincide 
Fig. 12.7
line of 
impact 
lines of 
motion
In oblique collision, line of motion 
and line of impact are different
Fig. 12.5
line joining the 
CM C, and C2
G.R.B. Understanding Physics MECHANICS (Part-B)
Oblique impact: When the line of motion of the bodies does 
not coincide with the line of impact, we call it “oblique impact”.
In eccentric impact, line of impact and the line 
joining the CM of the colliding bodies are different 
Fig. 12.6
93Collisions
(>>)
(iii)
(iv)
1.
The colliding bodies change in reaction 
Fig. 12.9
2.
3.
4.
Impact forces between two colliding bodies are equal and opposite even though surfaces of 
the colliding bodies are rough. In addition to the normal impulsive forces, tangential impact 
forces also come into play.
The line of impact and line of motion must be parallel in direct impact.
The line of motion and line of impact are different in the oblique impact.
The line of impact and line joining the CM of the colliding bodies are different in the eccentric 
impact.
The above classification is done based on the geometry of the colliding bodies. Let us now 
classify the collisions based upon the nature of the colliding bodies.
Scattering : When the composition and mass of the colliding particles (or bodies) remain unchanged 
during collision so that, before and after the collision, each colliding body remains identical, 
this type of collision is called scattering. For instance, collision between ideal gas molecules 
is an ideal example of scattering.
The line of motion is different from the line of impact and line joining the CM of the 
bodies. However, CM of the bodies lies on the line of impact. So this impact is said to be 
oblique-central.
The line joining the CM is different from the line of impact and line of motion. The lines of 
motion and line of impact are parallel. Hence this collision is said to be direct-eccentric. 
Since the line joining the CM of the bodies does not coincide with the line of impact, this 
is an eccentric impact. Since the line of motion of the body does not coincide with (or 
parallel to) the line of impact, we call it oblique. In combination of these two, this is called 
oblique-eccentric impact.
Recapitulating,
In scattering, the colliding bodies m1 and m2 remain the same
Fig. 12.8
Reaction : Many times, a collision between an atom A and molecule B -C yields a molecule A - B 
and an atom C. That means, the final particles (or bodies) of the colliding system are not 
identical with the initial particles. We call it “reaction”. In this way, chemical and nuclear 
reactions are the consequences of collisions.
(Pa>.
(PA) A block A strikes the block B which is connected 
with a light spring. Discuss the possibility of 
conservation of momentum.
Ans. Since the impulse of spring force and friction 
are very small during the time of collision, we can
conserve the linear momentum of the system (A + 5) just before and after the collision.
94 G.R..B. Understanding Physics MECHANICS (Part-B)
§ 12.5 Conservation of Linear Momentum in a Collision
Let us assume an isolated system of two bodies A and B.
If collision takes place between A and B, the impulse of impulsive 
force FA changes the momentum of A from (PA)j to (PA)y 
during the collision
(?,), + {£line shows the region of collision 
In which the particles change their velocities 
with considerable interaction
Fig. 12.12
1. The internal impulses can change the momentum of the individual (colliding) bodies. However 
for an isolated system (external impulse = 0), the net internal impulse is zero. Hence the 
internal impulse cannot change the momentum of the system.
2. Even though an external force acts on a system, if the forces are non-impulsive or its 
impulse is very small during the time of impact, we can still conserve the momentum of the 
system (approximately) during the impact. The momentum just before the impact is nearly 
equal to momentum just after the impact. In this way, the displacement, velocity of CM of 
the system during the collision can be ignored compared to that of the colliding bodies (or 
particles).
§ 12.6 Energy Consideration in Collision
Recast the familiar example of “collision of a falling ball”. When we drop a ball from certain 
height, it collides with the ground, bounces back and forth and finally stops. This experiment tells us 
that the body (ball) must lose some fraction of its mechanical energy during each collision in the form of 
heat, sound and light, etc. However, the total energy (translational + rotational + vibrational + heat + light 
... + others) must be a constant quantity in a “crude sense”. Sometimes there is a system of particles, 
that collide with tremendous speeds. For instance, in nuclear reactions, for the fast moving particles, we 
have to take the relativistic effect into account which tells that mass increases with velocity.
Generally, in “all” kinds of collisions between the bodies, due to the effect of huge impulsive 
forces, the colliding bodies may change their internal structures, composition, configuration of particles 
within them. This may cause a change in “internal energy” which depends on the configuration of the 
particles inside the colliding bodies and the kinetic energy 
of the particles (electrons, atoms, molecules, etc) of the 
colliding bodies relative to their centres of mass.
When no external force acts on the system of 
colliding bodies, the total energy, that is, kinetic energy 
of the CM of the bodies plus their internal energy, remain 
conserved. In collisions, generally internal energy of the 
colliding system changes. Hence the kinetic energy of 
the CM bf the colliding bodies must change.
Let us consider two systems of masses and 
m2 colliding in a region (shown as dotted circle) such 
that they interact by direct touching or no physical
Ex. 6. What is the need of energy conservation to find the velocities of the colliding bodies just after 
the collision?
Let us assume that two balls moving with momenta Fj and P2 collide with each other in a 
smooth horizontal plane. Just after the collision, if their momenta are and P2 respectively, 
ideally we can conserve the momentum of the closed system. Then we have,
P{ + P2 =P]'+P{
Now we have one equation and two unknown variables, i.e.,P}' and P2. Then, we need another 
equation. Here we need the second equation as the conservation of energy. Let us now talk 
about it.
Since
-.(i)we have
Since
we have Jf
2/77|
Substituting
we have
96 G.R.B. Understanding Physics MECHANICS (Part-B)
—> —> —> —>
contact. Let the masses of the colliding systems are m}i m2. velocities Vj, v2 and momenta Ph P2 
respectively. Similarly, let the masses of the systems just after the collision are and m2 respectively 
and their velocities and momenta are vj.vo and P{,P2 respectively.
The sum of total energy of the systems just before the collision is
E, = Kj + Ut, 
where Kt and U, are the sum of kinetic energies of the CM and internal energy of the colliding 
systems just before collision respectively.
When Q = 0, the collision is said to be perfectly elastic; hence no change in KE takes place.
When Q # 0, the collision is said to be inelastic as KE changes after collision.
When Q 0, KE increases and hence the internal energy decreases. That means, the KE of the 
particles of system increases at the expense of internal energy. Then this inelastic collision is 
said to be “Exorgic”. Nuclear fission and some exothermic chemical reactions including 
explosions, etc are the familiar examples of exorgic collision.
Ef
Kf 2m{ 2/„2 ’
2m\ 2m2
Using the principle of conservation of total energy just before and after the collision we have
E, = Ef
p’l
= -!- + + Uf
2m[ 2m'2
K - + P2
2m1| 2in2
2??Z| 2zzz2
Similarly, the sum of total energy of the system just after the collision is
Ef = Kf+Uf,
where Ay and Uj are the sum of kinetic energies of the CM and internal energy of the colliding 
system just after the collision respectively.
P2
2m2 2m\ 2m'2
Uj - Uf = Kf - Kj = Q,
2znj 2m2 2 nt) 2m2
Pi = P, say;
and
...(ii)
From eq. (i), we have
...(iii)
-Q
Substituting a>
COS0p
Ans.
Substituting p. and Pj, we have, 
Conservation of energy yields
and
we have
1 + 4 -K. 
m2 )
2W|'
P2 
— = 
2m\ 
and simplifying the factors, we have
= Kp
2y]m}m{KaKP 
m'2
where m} = ma, m2 = mN, m{ - mP and m2 = m0, 0 = 60' 
Substituting the values of masses and kinetic energies,
Ka = 4 MeV 
KP = 2.09 MeV, 
Q = -1.2 MeV
-Vje \
■^017 ■■zr’
final system
Q = Ka
i3i2=ip-X'i2
This gives P2'2 = P2 •+ P}'2 - 2P/]'cos 0
Substituting P2'2 from eq. (iii) in eq. (ii), we have
P2 _ ^'2 f2 + ^2-2^,cos9
2m} 2m{ 2m2
pf = />'+ pf respectively 
Conserving the initial and final momentum of the system, we have 
Pi = Pf • 
p = Pf+ P2
pL = ^L,±.q 
2/7J| 2t71| 2/712
i-4 -
ATI 2 )
Collisions 97
Ex. 7. Find the Q (energy of reaction) of the collision between an a-particle and a stationary jV14 
nucleus. The kinetic energy of bombarding of a-particle is Ka = 4 MeV and proton is deflected 
with a kinetic energy KP = 2.09 MeV at an angle 0 = 60° to the direction of motion of 
a-particle.
Sol. Following the procedure as described earlier, let us first proton 
conserve the momentum of the system. The initial and final p'_ a-particle '■> 
momentum of the system can be given as N’V
initial
system
Fig. 12.13
98 G.R.B. Understanding Physics MECHANICS (Part-B)
Sol.
...(ii)
Ans.
Sol.
Negative Q signifies that, kinetic energy of a-particle is absorbed by the reaction so as to 
produce the products proton and q17 having more internal energy.
Student Task^f
> Using the equation of conservation of momentum and energy, find the energy of explosion if 
a stationary body explodes into two fragments of masses m\ and m2, each having the 
momentum of magnitude P.
Ex. 8. Two identical particles collide elastically (0 = 0). Find the angle of deviation between the 
deflected particles. Assume that one of the colliding particles is stationary.
Writing momentum conservation, we have
M'+3l=lXl
Since p}'. = o, either Pi = 0 or P2 =0 or cos 0 = 0
if cos 0 = 0, Pi 1 P2. It means, either the particles exchange their momenta during collision 
or the particles deflect making 90° with each other.
0 = - rad . If P{ = 0, Pi = Pp
2 J
(rri| + mf)P2 
2m^m2
Ex. 9. Using the principle of conservation of momentum and energy, find the expressions of velocity 
and kinetic energy of a particle of mass m} colliding with a momentum P} with the stationary 
mass m2, elastically. Assume head-on collision.
Following the usual notations for conservation of momentum, we have
/f+ r2. X
Then squaring both sides, we have,
Pi2 + Pi2 + 2PiPi cos 0 = P?
Squaring both sides, we have
Pi2 + Pi2 + 2 PiPi = P]2
Now following the equation of energy conservation, we have
2ni|' 2m2 2mt
Since the collision is elastic and the particles are identical, m\ = m'2 - m}
Then, we have P/2 + P2'2 = pf
Eq. (i) - eq. (ii) yields f'. = 0
The above expression tells us that the particles will be deflected at an angle of 90° with each 
other. **
...(H)
we have
This gives
^' = Ans.
2
Hence
=
Ans.where
Student Taskf
> Find P2 and K2 in the above example.
2
Ans. P2
 mj - m2
mt + m2
§12.7 Elastic Collision
As discussed earlier, collision is said to be elastic when the kinetic energy of a system of colliding 
particles (or bodies) just before the collision remains the same just after the collision. For instance, if a 
ball collides with a wall elastically, it will bounce with the same speed, that is, same kinetic energy. As
Collisions 99
If we assume that the particles move after the collision in the direction of the colliding particle 
m}, we have cos 0 = 1. Then
2zw2/]
/M] + m2
2mt
/l'+ Pl = -••(>)
[You can also directly obtain the above equation {eq. (i)} by assuming p}\ and P} unidirectional]
Now conservation of energy gives
2/W| 2ffj| 2m2
Eliminating P2'2 from eq. (ii) by substituting P2'2 = (/] - Pf)2 from eq. (i)
p2 p;2 P? + /]'2 - 2PyP{ 
-----=------ + 
2/Wj 2zr?|------------ 2m2
[/{(ffl] - m2) - P\\m{ + m2)]2 = 0
Then we have
P2 f ml - m2 
2m\ m\ + m2
- m2 
m} + m2
^1 =^'”lvl2
I2 
I 2zr?|
2
K\
nt\ - m2
+ m2
K> _ ^'2 _ P\ (ml ~m2
2tfi| 2a?7| mt + m2
/2
• ••(ii)
...(Hi)
...(iv)
...(v)
m2 and v2 =W] +vl "1 «2
Sol.
_ ct( - m2 
ml + m2
(a) Just before collision, the 
bodies have velocities u1 and u2
! ™2 ~ ">l
CT] + CT2
CTjU] + CT2W2 = CT] V| + ct2v2 ...(i)
Conservation of kinetic energy of the system gives
2ct]
CT] + ct2
— CT|U|2 +
(b) Just after head-on collision, the 
bodies move with velocities v-| and v2
Fig. 12.14
Dividing eq. (iv) by eq. (iii), we have
W] + V| = v2 + u2
This gives u} - u2 = -(V| - v2) 
Solving eqs. (i) and (v), we have
2ct2
CT] + ct2
1 2 1 2 I 2 I 2-CT|U| +-ct2u2 =-ct,V] +-ct2v2
From eq. (i), we have
?«](«] - V]) = ct2(v2 - w2)
From eq. (ii), we have
CT,(W]2 - V2) = ct2(v2 -w2)
100 G.R.B. Understanding Physics MECHANICS (Part-B)
another example, in a collision of gas particles (molecules) with the container, since the collision is 
perfectly elastic, we can conserve the kinetic energy of each gas particle just before and after the 
collision. We categorise the elastic collision into two parts; (i) Direct central (head-on) and (ii) Indirect 
central (oblique).
Head-on collision : As explained earlier, in a head-on 
collision, the line of motion, line of impact and the CM of the 
colliding bodies lie on the same line.
Let us assume that two identical iron balls of mass CT] 
and ct2 collide elastically with velocities u} and u2 respectively. 
Let the particles move with velocities V| and v2 respectively.
Assuming all velocities are in positive x-direction, let us 
write the equation of momentum and energy conservation.
Conserving momentum of the system (ct] +ct2), we 
have
Ex. 10. Explain the physical significance of the result given by equation (v); Uj - u2 = -(V| - v2).
(«i - u2) is the velocity of m} relative to ct2. If you stand on the body ct2, seems to 
approach m2 with a relative velocity uapp = u} -u2. That means, («] - u2) is the velocity of 
approach, that is, relative velocity between the colliding particles along the line of impact just 
before collision. Similarly, (V] - v2) is the velocity with which moves away relative to ct2 
just after the collision. That is what we call velocity of separation Since,
ut - u2 = ~(V] - v2), we can write vapp = -v^p. It means that CT] seems to approach (move
■i
“12 V12
Student Task
Sol.
vi "2.
V2 = «l “2.
U| +vl
we have
we have
Then substituting = 0 in 
m2
7
+ m2
In a perfectly elastic collision ul2 = -Vj2; Inother £
words, u 5
velocity of approach = - velocity of separation. © ~** @
(Just before collision) (Just after collision) 
Fig. 12.15
Similarly, substituting — = 0 in v2 = 
m2
2
------^""2’
1 + ^L
m2
Collisions / )0l
towards m2) just before tl. collision with a speed wapp(= “i “ “2) and it will seem to depart 
(move away from /w2) just after the collision with the same speed vsep(= V] - v2).
---- « “2.
1 + ^-
m2
> Can you prove ul2 = -vI2 relative to CM frame of the system of colliding bodies?
Ans. Yes
Ex. 11. Discuss different cases in a head-on elastic collision between two free particles of masses m\ 
and m2 colliding with speeds u\ and u2 respectively.
Case 1. = m2 : Substituting m} = m2, in the formula
 m\ - m2 + 2m- 
m} + m2 1
we have v, = a2
Similarly substituting m} = m2 in the formula
rt, - m, 
------------- Mi + —----------
fflj + m2 m\ + m2
we have v2 = ut
Since v( = u2 and v2 = the colliding particles just exchange their momenta.
Case 2. m2 » mx-. When m2 » m^; — = 0-
m2 
za._i
_ m2____
’^L + l 
m2
V| = -U| + 2m2
2-Lm!
m2 |
^L + l 
m2
v2 =u2
The velocity of a massive body remains practically constant.
102
«i and v2 =vi
Find the final velocities of the body m in each case of head-on elastic collision.>
u MMM (h0GO(/)
M = mM >> mM >> m
u
MM (V)(fv)
M m2; V] is positive. Hence will proceed forward.
M « m
Ans. (i) 2u -> (ii) u -> (iii) u -> (iv) zero, (v) u «>.w2
Student Task
103
u
y
X
...(ii)
...(iii)
Sol.
... (a)
...(b)
...(c)
-(d)
V2 
2m v0 = 2mvl cos 30° + mv2 cos 02 
v2 cos 02 = 2v0 - V3v|
Collisions
§ 12.8 Elastic-Oblique Collision
For the sake of simplicity, let us consider an oblique 
collision between mj and m2 on a smooth horizontal 
surface when m2 is stationary. Since Fnet ~ 0, we can 
conserve the momentum of the system (m] + m2). Let 
the bodies move with velocities v( and v2 respectively just 
after the collision with angles 0] and 02 made relative to the 
x axis. Conserving the momentum of the system in x- 
direction, we have
This gives
Using eq. (ii), we have
-(2m)V] sin 30° + (m)v2 sin 02 = 0
This gives v2 sin 02 = V]
Using eq. (iii), we have
m\U = mjV| cos0| + m2v2 cos02
Conservation of momentum of the system in y-direction yields
-m^i sin 0! + m2v2 sin 02 = 0
Conservation of kinetic energy of the system yields 
1 2 1 2,1 2-m|M = -m1v1 + -m2v2
Now we have three equations and four unknown quantities, i.e., V],v2,0| and 02. Hence we 
need another equation which can be derived from the condition (data) given in the examples. To solve 
the equations let us look at the following example.
Ex. 12. Substituting m} = 2m2(= 2m), u = v0 and 0, = 30°, find vbv2 and 02 using the eqs. (J),(ii) 
and (iii), derived above.
Using eq. (i), we have
rrij
Oblique collision 
between m1 and m2
Fig. 12.16
i(2m)vo = |(2m)v2 + ^(m)vj
This gives 2v2 + v2 = 2vq
Squaring eq. (a) and eq. (b) and summing them, we have 
(2v0 - V3v])2 + v2' = v2Eliminating vj from eqs. (c) and (d), we have
2v2 + v2 + (2v0 - Tivj)2 = 2vq
Then, we have 6v,2 - 4^v0V] + 3vq = 0
104 G.R.B. Understanding Physics MECHANICS (Part-B)
This gives Ans.V1
Ans.
Ans. 90°
m., + m2 v
V2
m2
...(i)Then, v =
> In the foregoing example, if mj = m2 and the collision is elastic, find the maximum angle of 
deflection of m} relative to m2 just after the collision.
Just after inelastic collision 
the bodies move as a combined 
mass with velocity v
_ 4x/3v0 ± t/(4>/3vo)2 -(4)(6)(2v02)
12 73
This turns out to be a “unique number" in the given example. Substituting 
eq. (c), we have v2 = 2v0/>/3 
Then substituting v2 = 2v0/V3 and
Student Taskf
7H| V| + m2 v2 
rtt] + m2
§ 12.9 Complete Inelastic Collision (e = 0)
When both the colliding bodies move with same velocity just after the collision as a combined 
unit (single body), we call it perfectly inelastic (or plastic) collision. It means, there will not be any 
relative motion between the colliding bodies just after the impact.
In this case, we can conserve the linear momentum of the system (m( + m2). The initial 
momentum of the system is P, = m} Vj+ m2 v2, and final momentum of the system is 
Pj- = TMj v + m2 v .
V) = v0/>/3 in eq. (b), we have 02 = 30°
V| = v0/73 in
Ans.
Just before collision the bodies 
m1 and m2 move with velocities 
v1 and v2 respectively
Fig. 12.17 
—> —>
Velocity just after collision : Since, Pt = Pj- 
we have, (mi + m2) v = m} V]+ m2 v2
Collisions 105
,2
...(ii)
> 2
I vi “ v2 I , after simplifying the terms.
Sol.
Ans.
Student Task f
Ans. (a) (b) tan
—► —> 7 ? ?
where |v]-v2| =vi + v2 + 2V|V2 cos0
Substituting mt = m,m2 = M,V\ = u,v2=v and 0 = 90°, we have
Mm(u2+v2) 
2(M + m)
/\ M \
I „ I 
/777777777777//777777777W777777777777777
Fig. 12.18
Mv "*— with the vector u 
mu
Sometimes the combined mass just after an inelastic collision is constrained to move in a “definite 
line or plane”. In these cases, we need to conserve the momentum of the system (/h, + m2) in the line 
of motion of the combined unit. We consider the following example to demonstrate this.
Ex. 14. A particle, (a mud pallet, say) of mass m strikes a smooth
stationary wedge of mass M with a velocity v0, at an angle 
0 with horizontal. If the collision is perfectly inelastic, 
find the:
(а) velocity of the wedge jus! after the collision.
(б) change in KE of the system (M + m) in collision.
&K = -
2
The change in kinetic energy of the system, by substituting v from eq. (i) in eq. (ii), is :
m 
\*o.
> In the above example, find :
(a) time and distance of sliding of the combined truck and car, after the collision, if the 
coefficient of friction between truck, car and ground is M-
(b) direction of motion of the combined mass just after the collision.
./~2 2 , i/2,,2 2 2 , i/2 2. , v m u + M v m u + M v Ans. (a) ----------------- ,-------------—
p(A/ + m)g 2p( M + my g
k|-v2|2,
| (m, + m2 )v2 - (1 mrf + 1 m2v22
M = W|W2 
2(/?7| + m2)
M = ...
2(777] + m2)
Ex. 13. A car of mass m moving with a velocity u strikes inelastically with a truck of mass M moving 
with a velocity v perpendicular to the direction of motion of the car. If the truck and car stick 
together after the collision, find the energy loss in the collision.
We have the formula
106
Sol.
m M v
This gives
,2 ,2AX = — (M + /n)v----mvQ
where
AXThis gives
2
Ans.
Student Task^
wedge and ground? Ans. (a) ■» (b) No
The change in KE depends on the angle 0. At 0 = 0, AX = - (minimum AX). At
Pulley
x
Fig. 12.20
(M + fflsin2 0)?nvQ
" 2(M + m)
 (M + m) f mv0 cos 0
M + m
vo
> In the above example :
(o)
(b)
////////////////////////////////^^^
Fig. 12.19
m2
What is the impulse of the impact force on the wedge?
Can you conserve the horizontal momentum, if the friction is present between the 
Mmv0 cos0
M + m
Ex. 15. Two particles of masses m} and m2 are 
connected by a light and inextensible string 
which passes over a fixed pulley. Initially, the . 
particle m} moves with a velocity v0 when the 
string is not taut. Neglecting friction in all / 
contacting surfaces, find the velocities of the I 
particles m} and m2 just after the string is taut. \
0 = 90°, AX = - (maximum AX).
2
- mvo
2
cos2 0
MmvQ 
2(M + m)
2 1* 2| ~~^0
1-------—
M + m
1
2
G.R.B. Understanding Physics MECHANICS (Part-B)
(a) Let the system (M + m) moves as a single mass with a velocity v.
Conserving the momentum of the system in horizontal, we have 
mv0 cos 0 = (M + m)v 
mv0 cos 0 
M + m
(b) The change in KE of the system is 
„ 1 ...
2
mv0 cos 0
M + m
107
Jt dt
zzzzzzzzzz/z u2, the collision will take place. During 
collision the bodies are pressed against each other. Hence the bodies will experience huge impulsive 
forces -F and F in the form of action-reaction pairs. As we know, the relative velocity between the 
bodies is, wt - u2, just before collision. Since the bodies push each other during collision by their 
impulsive forces, the relative velocity between the bodies will rapidly decrease to zero during a time 
called “deformation time td ”. At the deformation time td, let the bodies will move with same velocity v, 
say.
----- ► V
J hl
--
J_ m2
Jt dt Jt2 dt*
Fig. 12.21
Jt, dtDuring the impact, a huge tension develops. The impulse 
of tension 7] is equal to the change in momentum of 
mt which is given as
-Jr, dt = m}(v- v0) ...(i)
Similarly, the impulse of tension T2 is equal to the 
change in momentum of m2 which is given as
- J?2 dt = m2(-v)
Since 7] = T2 equating ffdt with jF2dt, from eq. (i) and (ii), we have 
wi(vo - v) = m2v
V = -^VQ . 
mt + m2
> In the foregoing example :
(a) Find the change in momentum of the system (mt + m2).
(5) Can you conserve the momentum of the system (m} + m2 + pulley)?
An,.(a)_awk 
ffi| + m2
The student must remember that A + A = JtJ dt + jfydt = J(7J + 7^)- F' dt = z»|V| - W]V
Similarly, during the restitution phase, the body m2 accelerates from velocity v to v2, say, 
changing its momentum from mv to mv2. Then, the impulse of F' is
Jo F'dt = m2v2 - m2v ...(iv)
Coefficient of restitution : As stated earlier, the change in momentum during restitution (which 
is equal to impulse of the restitution forces) is always less than that during the deformation (which is 
equal to the impulse of the deformation forces). Then the ratio of magnitude of restitution impulse to the 
deformation impulse can be termed as the “coefficient of restitution” denoted by the letter “e".
109Collisions
...(v)Symbolically,
-(vi)
...(vii)
Using eqs. (vi) and (vii), we have
V) - v2 = -e(u, - w2)
£■
vsep Vapp
Just before the elastic collision, m1 
approaches to m2 with velocity, 
v«pp “(U1 -u2) relative to m2 and 
departs with a velocity vop = -(*1 - v2) 
relative to m2
Fig. 12.23
vsepa ration
• ••(A)
This formula is called Newton’s impact (or empirical) formula which can be remembered as
- e( velocity of approach) = velocity of separation
The coefficient of restitution depends on the nature of material, velocities, shape and size of the 
colliding bodies.
Perfectly elastic collision : The value of e lies between 0 and 1; 0 = (P),-,
where (P)z = (l)(2) + 2(-l) = 0
and (P)f = miV] + m2v2 = (l)(v,) + (2)(v2) = v, + 2v2
Then, we have v, + 2v2 = 0
Writing Newton’s empirical formula, we have
-e(ut - w2) = V] - v2 j
Substituting W| = 2, u2 = -1, v, = +vj and v2 = +v2 and e = —, we have 
-|[(2)-(-l)] = -v2
3v.-v2=--
1 .
Solving eqs. (i) and (ii), we have v, = -1 m/s and v2 = — m/s
1(2) + 2(-l) 
1 + 2
(c) = 1(1)[(-1)2 - (2)2] + l-2
2
I -(-I)2
C.R.B. Understanding Physics MECHANICS (Part-B)
Ex. 18. Two iron balls of masses 1 kg and 2 kg are colliding centrally with 2 m/s 1 m/s -•----
velocities 2 m/s and 1 m/s respectively. Find the :
(a) velocities of the bodies just after the collision.
(d) velocity of CM of the system of the balls.
Ex. 19. Two spherical bodies of mass m} and m2 fall freely through a distance h, 
before the body m2 collides with the ground. If the coefficient ofrestitution 
of all collisions is e, find the velocity of mI just after it collides with m2.
As the bodies m} and m2 fall freely through a distance h, m2 collides 
with ground with a velocity v0 = y/2gh. Hence m2 bounces up with a 
velocity ev0 just after the collision with the ground. Then it collides with ♦ 
h
the freely falling body m}. Just after the collision, let us assume that V]
and v2 are the velocities of mt and m2 respectively. Ignoring the impulses Fifl'12-29
1
2
113
Fig. 12.30 (b)
...(i)
•••(ii)
v0 +vl
Putting mt = m2,
Ans.we have vi
Student Taskf
> In the above example, discuss the case when
vi
Newton’s impact formula,
-e[W] -u2] = V| - v2, where »| = -v0 and u2 = ev0
Fig. 12.30 (a)
Conservation of linear momentum
P = - mt v0 + /n2ev0 = ?H|V| + m2v2
Then -e[(-v0) - (ev0)] = v, - v2 
Solving eqs. (i) and (ii), we have
= -(ffl, - em2) 
ml + m2 
e = -J- and v0 = y/2gh,
V 32
e = 1 and m\ = m2.
Ans. V] = y]2gh T; v2 = ^2gh i
In the above example, at first m2 collides with ground, then m2 collides with freely falling m} 
while bouncing up. In this way, two collisions occur one by one. While freely falling, the 
reaction force between m} and m2 is zero. However, during their collision, huge reaction 
(impulsive) forces act on them.
If m2 is stationary just before the collision and the other body collides with m2 with a speed 
u centrally. In this case, putting u2 = 0 in the general expressions of and v2, we have
~ emi j (1 + e)m\u= —------ - uj and v2 =---------—, where u = ut
m} + m2 m\ + m2
Collisions
due to the weights of the bodies during the period of collision, let us conserve the momentum 
of the system (^i + m2).
(! + e>2
evom) + m2
1
2
114
Fig. 12.31
Sol.
Fig. 12.32
..(ii)
Ans.
(b)
Ans.
Ni
>ku2n
U1
m2v = nj|V0 
-e(W] - u2) = V] - v2, 
-e(v0 - 0) = (0 - v) 
v = ev0
\N2
In oblique Impact of smooth bodies m1 and 
m2, normal impulsive reactions and N2 
come Into play; since the bodies are smooth, 
no tangential force will appear during collision 
Fig. 12.33
$N2dt = 0.
G.R.B. Understanding Physics MECHANICS (Part-B)
However we can do any problem of head-on collision from the basics by writing momentum 
conservation equation and Newton’s impirical formula.
Let us look at the following example.
Ex. 20. A particle of mass m} collides centrally with a stationary particle 
of mass m2 with a velocity v0. If the particle mx stops just after 
the collision, find the :
(а) coefficient of restitution.
(б) change in kinetic energy of the system.
(a) Conserving momentum, we have,
P, = Py, where = m}VQ and Pj- = m2v2 = m2v 
This gives 
Applying 
we have 
This gives
Using eqs. (i) and (ii), we have
~mi)
AK = ------------------
m2
§ 12.11 General Solution for Oblique Inelastic Impact of Two Smooth Bodies
In oblique impact of smooth bodies, first of all we 
draw a tangent at the point of collision. Then draw a normal 
at the point of collision. Since the bodies are smooth, no 
impact force acts along the tangent. It means, impact 
(impulsive) action-reaction forces act along the normal. 
Let the normal impulsive forces be N} and N2 
respectively. Since 7V| and N2 are action-reaction forces, 
N] + N2 = 0, hence, the sum of impulses of the normal 
impulsive forces is zero. Symbolically 
Jj?] J/ +
Case 1. Fext - 0 : If no external force acts on 
the system (m} + mf), jXxt = Even though some 
non-impulsive forces act on the system, the impulse of 
these forces during short time of impact is negligible. Since
/ ''
* U2*
m2
1 2 1 2The change in kinetic energy tsK = — m2v2 - ~m\VQ
Substituting v2 = ev0, we have
115
...(•)
...(ii)
M
R
Fig. 12.35
Sol.
Fig. 12.36
Then, ...(ii)
] R/2
v ' *2n
Since Ft=0on each body the 
tangential momentum of each body 
can be conserved. However, the total 
(iii) momentum of the system (rrv, + m2) 
along normal is conserved because 
ZFn=0
Collisions
the sum of impulses of all internal impulsive forces is zero, we can 
conserve the momentum of the system during the collision. It 
means,
Momentum just before collision = momentum during collision 
= momentum just after collision
Resolving the momenta along the normal, we can write
m2 u2n = m\ vl« + m2 *2n
Since no tangential force acts on the bodies as they are 
smooth, individually we can conserve their tangential momenta.
m\ Mu = m2 v2i
m2u2t = m2v2t
Now we have three equations and four unknown variables;
i.e., vu,v2l,vln and v2n ■ So, we need another equation which Fig. 12.34
should be none other than, Newton’s empirical formula which is always applied along the line of impact.
-e(uXn-u2n) = vXn-v2n -0v)
Now we have four equations and four unknown variables. You can solve these equations for all 
four unknown quantities specified earlier.
Ex. 21. A particle of mass m collides centrally with a smooth disc of
R
mass M at a distance ~ from the centre of the disc. The disc 
moves vertically up and the particle moves horizontally to right 
with equal speed v0, Just before the collision. Assuming the
coefficient of restitution e = -, and 2m = M, write the four 
relevant equations following the process as described in the 
foregoing section.
Directly following the procedure described above, we conserve 
the momentum of the system (m + M) along the normal.
+mv0 cos 9 - Mv0 sin 6 = +nivln + Mv2n
Substituting 2m = M and 0 = 30°. we have
. f (V3 - 2)v0
2v2„ + % =-------- “
Conserving tangential momentum of m, we have
mvy - mv0 sin 9, where 0 = 30°
vo
V,,—
11.6
...(iii)Then we have
• ••(iv)Then, we have vi» ~ v2n = vo
> In the above example find the speeds of the bodies just after collision.
Ans. V] vo
7
t
/u1t n
>2t
u2
uu = mt vlt
_2Jo
2
(73 tl)
4
= -^,v2 =
72
GR.B. Understanding Physics MECHANICS (Part-B)
Conserving tangential momentum of M, we have
Mv0 cos 0 = Mv2i where 0 = 30° 
73 v2/ = y vo
Then use Newton’s empirical formula;
-e(M|„ - w2„) = vln - v2„,
u 1 73v
where e = -,uln = + — ,u2„ =
'716-273'
4
mi z
NX
J V2 \v2n 
’///////
'^/vVln /l-N 14
\ N. m2Y
Z/ZZZ/ZW^
Just after collision
Case 2. External impulsive force acts on the system (W] + m2) :
If any object m2, say, of the system + m2) is constrained to move in horizontal plane, 
during the collision between and m2, another collision takes place between m2 and ground. In this 
process, m2 experiences an upward (normal) 
impulsive reaction force N' from the ground. This 
is an external force to the system + m2) in 
addition to the weights of the colliding bodies. Since 
the weights are non-impulsive forces, their impulses 
during the short time of collision can be neglected. 
However, the impulse of the impulsive force N', 
that is, ^N'dt cannot be ignored. Hence we cannot 
conserve the momentum of the system (rrj| + m2) 
in any direction like the previous cases. However 
we can conserve the momentum of the system (m} + m2) perpendicular to the external impulsive force 
N', that is, horizontally. We can conserve the momentum of along the tangent drawn at the point of 
collision between and m2 because no tangential impulsive force acts on the body However we 
cannot conserve the tangential momentum of m2 because, the component of the impulsive N' force 
along the tangent is not zero. Let us assume that the velocities of and m2 just after the collision are 
V] and v2 respectively, where V] = vlz + vln and v2 = v2, + v2„ .
Conservation of tangential momentum of mt :
m1 v’QN
Xs ui\
V m2 J
Just before collision
Fig. 12.37
117
Sol. 5
Vx
vocos0 N dt
...(ii)
...(iii)
...(v)
Ans.
Student Taskf
Collisions
Conservation of horizontal momentum of the system (mj + mi): 
Resolving the velocities along horizontal (x-axis), we have
> In the foregoing example, when the horizontal surface is smooth, find the total horizontal 
distance covered by the particle till it stops colliding (bouncing).
...(iv)
Let us
iiiiniini/uiHiiinHuiiniiiii/fiiiii
VpSine
iiiDiiiiuiiiiiniiiiiiiiiiiiiiiiiiiiin
m] u]x+ m2 u2x = m} vlx + m2 v2x ...(H)
where V|x = (V|,)x + (v)rt)x and v2x = v2 (because m2 is constrained to move horizontally) 
Coefficient of restitution along the normal:
~e(u\n~ = V\n-V2n -(Hi)
Now we have three equations and three unknown quantities, i.e., V|O vIn and v2. Let us apply 
the above procedure in the following example.
Ex. 22. A small rubber ball strikes a rough horizontal surface with a velocity v0 at an angle 0 with 
vertical. If the coefficient ofrestitution and coefficient of friction between the ball and horizontal 
surface are I and p respectively, find the velocity of the ball just after the impact.
As the particle collides with the rough horizontal 
surface, impulsive forces come into play in both 
tangent and normal to the horizontal surface. Let the 
normal impulse be jN dt and horizontal impulse be
where F = pN ...(i) Fig. 12.38 ’
Writing the impulse momentum equation for the particle in x-direction, we have,
mv0 sin 6 - |f dt = mvx
Similarly, the impulse - momentum equation iny-direction gives
-mvQ cos 0 + jw dt = mvy
The coefficient of restitution along y-direction (normal) gives
= -e(-v0 cos 0)
Then, vy = ev0 cos 0
We have four equations for four unknown quantities, i.e., ff dt, jhl dt, vx and vy.
solve them. Substituting f = in eq. (ii) and eliminating J bl dt from eqs. (ii) and (iii), we 
have
Ans.
(l-e)g
vx = v0[sin 0 - p(l + e) cos 0] 
From eqs. (iv) and (v), we have,
{v0 sin 0 - p(l + e) cos 0} i + ev0 cos 0/
118
10.
16.
17.
18.
1.
1
41
t
0
2 3
1.
2.
3.
4.
5.
11.
12.
13.
6.
7.
8.
9.
G.R.B. UnderstandingPhysics MECHANICS (Part-B)
Assignments
14.
15.
Discussion Type Questions
When we drop a ball onto a floor, can we conserve energy and momentum? Explain.
What do you mean by momentum delivered by a moving body?
When do a colliding body deliver maximum momentum, maximum velocity and maximum kinetic 
energy?
In explosion of a cracker in mid-air, what is/are conserved during collision, after collision?
In sailing of a boat, which principle is applied? Can you conserve momentum and energy of the 
boat?
Why does a rifleman receive a backward kick while triggering the rifle?
Do the relative distance and relative velocity between two particles remain constant? Explain.
Blade turbines are curved rather than flat, why?
In an oblique collision between two free particles, when do they get deflected by a maximum 
angle?
Why is hydrogen rich paraffin used in slowing down of the fast moving neutrons rather than 
lead, in nuclear reactors?
During the collision of two particles, which must be conserved; momentum or energy?
In which case does a ball exert larger impulse, if it sticks or it rebounds in head-on collision?
A bird suddenly jumps from the floor of a cage which is placed on a spring balance. What will 
happen to the reading of the spring balance if the cage is (i) open (netted) (ii) closed?
Can the coefficient of restitution be greater than one?
If the masses are equal, in a head on elastic collision, velocities are interchanged. If masses are 
unequal, is it possible to exchange the momenta in head on collision?
Does the principle of impact force work in (i) moving of a fish in water, (ii) flying of a bird, 
(iii) flying of a rocket, (iv) flying of an aeroplane?
In head-on collision is it possible for the colliding particles to move with a common velocity even 
for an instant in (i) head-on collision (ii) oblique collision?
Is it possible to have more kinetic energy of the colliding particle just after collision?
Multiple Choice Questions
The F-1 graph for a particle moving in a straight line is given. If the body of mass 2 kg projected 
towards right with a speed of 1 m/s is acted upon by a force F for four seconds in the line of its 
motion, its velocity after 4 s is : F
7
(a) — m/s towards right
1
(b) — m/s towards left
(c) 3.5 m/s towards right
(d) none of the above
119
2.
F
(b)(a)
t t70 0
FF
(d)(c) hAt t
00
3.
4. —~v
—heavy mass
5.
6.
7. (H)
777//7777777777777777777777^7777p
Collisions
Which of the following graphs represents a collision of two bodies in air ?
F
A smooth ball strikes obliquely with a velocity v0 at an angle 0 with vertical 
with a horizontal surface. If e = coefficient of restitution of collision, the 
change in momentum of the ball is :
(a) znvocos0(l + e)j (b) mvosin0(l-e)j
(c) -mv(l + e)sin 0 J (d) none of these
A ball of mass m moving with a speed 2v strikes a heavy wall 
elastically, which is moved with a velocity v. The work done 
by the heavy wall on the ball is;
1 2
(a) ~mv
(b) mv2
(c) 2mv2
(d) none of the above
An insulated moving jar containing an ideal gas is suddenly stopped. Which of the following 
remain(s) constant for the system (gas jar + gas)? (where KCM = KE of CM, K' = KE about 
the CM)
(a) Kcm and P (b) K'
(c) K' + Kcm (d) None of these
An elastic ball of mass m is dropped onto smooth ground. The average force offered by the 
ground over one cycle of its motion is:
(a) very large upward force (b) mg t
(c) mg I (d) zero
Two identical iron balls of masses mt and m2 are kept on a smooth 
horizontal floor. If the ball collides with the ball m2 with a speed v0, 
the maximum deformation energy is:
—2v
77777777777777777777^7777777777^
120
(a)
(c) (d)
8.
(b)(a)
(d) none of these
9. m
[
elevator
10.
(d) mg
11.
(c)
12.
T h 1
(a) e = — 
m2
0
I
m2
m
M + m
x
^2gA
A ball of mass m} collides with a hanging ball of mass m2 with a horizontal velocity 
v. If the ball m} stops just after collision [where T and To are the initial (just before 
the collision) and final (just after the collision) tensions in the inextensible string]:
2 2 
(b)r-r0=^
|K
Vm2
(c) t-T0=^- 
">1
(d) e =
vo
mi”»2v0 
2(m] + m2)
I 2 J
L. ..2
... (1
(b) mg - + 1 
lx
G.R.B. Understanding Physics MECHANICS (Part-B)
1 2(b) -mvQ
~m2
mi + m2
In Q.-7, the fraction of KE lost by for elastic collision is :
M
M + m
4 Mm
(c) (M + m)2
An elastic ball strikes a moving elevator with a velocity v. If the elevator has 
a speed v, the maximum height attained by the ball after the collision measured 
from the point of collision is :
v2 v2
w ~g —>
(m, v, + m2 v2)(m, -m2)
?W| + 
-> ->
(V] +V2)
m2 v2
ffj] + m2
mi + m2 m\ + m2
A particle of mass 2m moving with a velocity 2v0 collides centrally with a particle of mass m 
moving with a velocity v0. If the coefficient of restitution of collision between the bodies is
1 2m
> the value of | AX' | of the system (m + 2m) is : z-x
1 2(a) ~mv0
(0^
Assertion-Reason Type Questions
(a) Both assertion (A) and reason (R) are true and R is the correct explanation of A.
(b) Both assertion and reason are true but R is not the correct explanation of A.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
(e) Both assertion and reason are false.
A: An electron revolves in an atom in a very short time. It is said to be colliding with the nucleus.
R: A short timed interaction is called collision.
Collisions
Two identical particles each of mass m collide perpendicularly with equal speeds v. m 
The change in kinetic energy of the system for completely inelastic collision is : Q-
(a) mv2 (b) 2mv2
(c) -mv2
A bird of mass m alights onto a block of mass m with a velocity v0 at an angle 
60° with horizontal. If the block is placed on a smooth horizontal surface, the 
loss of kinetic energy of the system (bird + block) is :
1 2 7 2(a) -mv0 (b) — mv0
I lo
3 2(c) -mv0
The coefficient of restitution depends on :
(a) shape
(c) velocity
122
2.
3.
4.
5.
v0 just after the collision.
xO
(P)
(q)
(r)
m (s)
u2U1
V1o
m
AY
C.R.B. Understanding Physics MECHANICS (Part-B)
A: Maximum kinetic energy will be exchanged for head-on elastic collision when — = 1. 
m2 R : 100% momentum exchange takes place when ml = m2.
A: In an inelastic collision, maximum kinetic energy is delivered.
R: The moving body sticks to the other reducing the kinetic energy relative to the CM to zero.
A: An elastic ball is dropped from rest. After a long time, the average force exerted on the ball by 
-► 
the ground is equal to -mg.
R: The weight of the ball remains constant.
A: The particle is projected with a speed v0. Then the body m2 moves with a velocity
———v0 just after the collision. m. V
W| + m2
R: The momentum of the system (m, + zn2) is conserved 
along x-direction.
m
M + m
M
M + m
Comprehensions
Passage-1 (Central or head-on collision)
In central (head-on) collision we can conserve the momentum 
m}U\ + m2u2 - m}V\ + m2v2 (i)
Then we can use Newton’s empirical formula :
V]-v2 =-(U|-u2) (ii)
Now we have two equations (i) and (ii) and two unknown quantities 
such as V] and v2.
Match The Columns
(a) A ball of mass m strikes a station«>y ball of mass centrally.
The fraction of total mechanical energy which cannot be lost 
during collision is
(b) A bullet of mass m fired horizontally can penetrate a block of 
mass AY kept fixed, through a distance x. If the block is freeto move, the distance of penetration of the bullet is x'. Then 
x'
x
(c) A man of mass m jumps out of a boat of mass AYhorizontally 
with a speed u relative to the boat. In consequence the boat
v
recoils with a velocity v. The value of----- =u-v
(d) A ball of mass m strikes a smooth 
stationary, trolley car of mass AY. If the 
collision is head-on and elastic, the fraction 
of mechanical energy lost by the ball is
4 Mm
(M + w)2
* M
Z2 g
124 G.R.B. Understanding Physics MECHANICS (Part-B)
1.
2.
(b) V(a) 0
(d) vo
3.
(b) 7
(d) none of these
Bullet
6.
7.
8.
9.
10.
11.
1.
2.
3.
4.
5. Gun
zzz/zzzzzzzzzzzzz
The coefficient of restitution is :
1 
(a)-
1
(O-
The speed of the disc M is :
2vn
The fraction loss of mechanical energy is :
(a) y
True Or False
The colliding bodies must touch each other.
The gravitational force during supernova explosion of matter is a collision.
Impulse and impulsive forces are same.
In elastic collision, the total kinetic energy must be conserved.
When a stationary gun which is placed on a smooth horizontal surface T 
fires a bullet, kinetic energy of the gun is greater than the kinetic energy 
of the bullet. m»"
When we fire a bullet from the gun in Q-5, the momentum of the system (gun + bullet) remains 
constant.
A rough ball strikes obliquely with a rough horizontal surface, its horizontal momentum cannot be 
conserved.
A bullet of mass m strikes a block of mass A/placed on an inclined plane of angle 
of inclination 0 = tan-1 p, where p = coefficient of friction between the block 
and surface. During and after the collision, we can conserve the linear momentum 
of the system (Af + m).
If the colliding particles have equal mass, for an oblique collision, the maximum angle of deviation 
is equal to 90°.
Mechanical energy can be conserved in the case when momentum is conserved for any system. 
A truck and car, after pressing their brakes simultaneously; stop simultaneously after travelling 
equal distances.
-t
Gravitational Field
18.13 Variation of I geff I and Apparent 
Weight
18.14 The Motion of Planets
18.15 Motion of Planets and Satellites in 
Circular Orbit
18.16 Escape Velocity
18.17 Orbital Velocity and Nature of 
Orbits of a Satellite
18.18 Weightlessness
18.19 Earth as an Inertial Reference 
Frame
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 464
O Match the Columns 465
O Comprehensions 465
O True or False 467
O Fill in the Blanks 467
O Problems 468
O Answers 472
(Xi)
583
CHAPTER -22
667-779
CHAPTER -21
672
713
715
719
«
«
596
596
SIMPLE HARMONIC 
MOTION
655
655
657
658
658
664
667
668
669
695
700
704
PROPERTIES OF MATTER 608-666
608 
609 
614 
620 
625 
625 
628 
628 
630 
632
20.7 Equation of State of Fluid Motion • 576
20.8 Bernoulli’s Theorem 578
20.9 Applications of Bernoulli’s Theorem and
Equation of Continuity
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 598
O Match the Columns 599
O Comprehensions 599
O True or False 600
O Fill in the Blanks 600
O Problems t 601
O Answers 606
O Multiple Choice Questions 652 
O Assertion-Reason Type Questions 655 
O Match the Columns 
O Comprehensions 
O True or False 
O Fill in the Blanks 
O Problems 
O Answers
21.1 Introduction
21.2 Elastic Property
21.3 Hooke’s Law
21.4 Calcu lation of Deformation
.21.5 Elastic Energy
21.6 Calculation of Elastic Energy
21.7 Viscosity
21.8 Newton’s Law of Viscosity.
21:9 Stoke’s Law
21.10 Surface Tension
21.11 Molecular Theory of Surface Tension
633
21.12 Measuring Surface Tension 634
21.13 Surface Energy 637
21.14 Relation between Surface Tension and
Surface Energy 639
21.15 Pressure Difference Across a Liquid. .
Surface ’ 640
21.16 Angle of Contact 643
21.17 Capillary Action 644
Assignments
O Discussion Type Questions
22.1 Introduction
22.2 Definition of Oscillation
22.3 Simple Harmonic Oscillation
22.4 Oscillation of a Particle Connected 
with a Light Spring
22.5 Relation between SHM and UCM 680
22.6 Phase of a Particle in SHM 682
I
22.7 Energy of an Oscillating Spring-Mass
System 684i J
22.8 Calculation of k^ for a Complicated
Spring-Mass System . 687
22.-9 Oscillation of a Particle Connected with
Different Elastic Bodies 693
22.10 Energy Considerations of Oscillations 
without Springs
22.11 Oscillation of a Simple Pendulum
« Accelerating simple pendulum 
« Simple pendulum with large string 
708
« Simple pendulum connected with a 
spring ’ 709
Simple pendulum.in a liquid 711 
Variation of time period of a simple 
pendulum
22.12 Physical (or Compound) Pendulum 
(Oscillation of Rigid Bodies)
• 22.13 Oscillations of a Rigid Body
• 651 Connected with a Spring
( Xii )
747
731
733
753
756
760
760
763
764
765
777
22.23 Finding Angular Frequency of Oscillation 
by Energy Method
Assignments
O Discussion Type Questions
O Multiple Choice Questions
O Assertion-Reason Type Questions 759
O Match the Columns
O Comprehensions
O Tnw^r False
O Fill in the Blanks
O Problems
O Answ
22.14 Oscillation of Symmetric Rolling Bodies
on a Flat Surface . 721
22.15 Oscillation of an Asymmetric Rolling
Body on a Horizontal Surface 727
22.16 Oscillation bf a Uniform Rolling Body on
a Curved Surface 729
22.17 Oscillation of Rolling Asymmetric Body 
on a Curved Surface
22.18 Torsional Pendulum (Oscillation)
22.19 Oscillations of a Body Inside Earth 736
22.20 Oscillations of a Liquid in a Tube 739
22.21 • Oscillation of a Floating Body in a Liquid
742
22.22 Oscillations of a Two Particle System
744
System of
Particles
Study^j^Points
1 1 .9
11.1
1 1.2
1 1.3
1 1.4
1 1.5
1 1.6
1 1 .7
1 1.8
Introduction
Definition of Centre of Mass
Characteristics of Centre of Mass
Calculation of Position of Centre of Mass (Finding CM)
Work-energy Theorem for System of Particles
Impulse and Momentum Equation for a System of Particles
Advantages of CM Frame
Application of Impulse-momentum Equation and Work-energy Theorem for a Two- 
particle System
Application of Impulse-momentum Equation and Work-energy Theorem for a Many 
Particle System
CHAPTER 11
§ 11.1 Introduction
So far we discussed the kinematics and dynamics of a particle. We measured position, displace­
ment, velocity and acceleration of a particle by using Newton’s laws of motion and work-energy theo­
rem.
In general, we draw “ free-body diagram” of a particle and measure the net force acting on it. 
Then by dividing the net force F by the mass of the particle, we find the acceleration given as :
->
-> Fa = — m
Integrating acceleration, we find velocity; integrating velocity, we find position of a particle after 
substituting the initial conditions.
Furthermore, we applied the principle of work-energy theorem and principle of conservation of 
energy for a particle.
Now let us try to define and analyse the motion of any extended object which, in general, is a 
“system (or group) of particles”. When a system of particles experiences some external forces, each 
particle need not experience a net force. If we apply the Newton ’s laws of motion particlewise, the sets 
of equations will be many. This complicates the calculation of acceleration, velocity, etc. In other
*f2
F32
F3
mi
'V
...(ii)
—(ii)
1)]
F2i
^12
= mj a,+ m2 a2+... + mn an
As we have grouped the internal forces as action-reaction pairs, using Newton’s third law, the 
net internal force turns out to be zero. This helps us in reducing the RHS of the force equation to a great 
extent which is left with the external forces only. Hence, the net force acting on a group of particles is 
equal to the sum of external forces acting on it. Then, the force equation for a system of particles can 
be given as following:
(>net)l - ?\ + >12+ >13+ ••• + P\n
Applying Newton’s law on mx, we have
(>net)l = >
where ax = acceleration of m1
Substituting (Fnet)[ from eq. (i) in eq. (ii), we 
have the following force equation
For /Wj : 5j + /r]2 + J?13 + ... + ...(0
Similarly, by application of Newton’s second law on other particles, we have the following force 
equations:
For m
2 G.R.B. Understanding Physics MECHANICS (Part-B)
words, it is practically impossible or extremely difficult to analyse the motion of a system of particles by 
using Newton’s laws particlewise. Hence, we need to develop a method in which the entire system can 
be made equivalent to a particle and then conveniently apply the laws of motion, work-energy theorem, 
etc.
>2 : F2 + F2x + F23+ ... + F2n - m2 a2
13 
A
§11.2 Definition of Centre of Mass
Let us consider a system of n particles and try to 
apply the laws of motion. Let us assume that external 
forces Fx,F2,...,Fn act on the particles of masses 
mx,m2,..., mn respectively. If we assume any interaction 
between the particles, we have to consider the forces of 
interaction, which can also be called internal forces. Let 
the internal forces acting on mx due to other particles be Fi 
Fx2,Fxi,...,FXn respectively. Then, the net force acting 
on mx is the sum of external and internal forces acting on 
it which can be given as : Fn
...(i) System of n-particles; the particles interact 
with each other. The net force acting on the 
system is equal to the external force because 
F|nt = 0. The acceleration of CM is given as:
a " Em,
Fig. 11.1
Fn2; . T 
A''/Fn3
For mn : Fn+ FnX + F„2+ ... + Fnn_x = mn an
Summing up all “n” force equations, we have the net force Fnet acting on the system which can 
be given as :
>net = (>l + >2 + - + >«) + I(>I2 + >21) + (>13 + >31) + ••• + (>„-l,n + >n,
jQm2
/ ' vF23
F31-*- 
>'/m3 
/ >3n
System of Particles 3
Sol.
Then
Ans.
Student Task
y
m2
Ans.
0 x
> A system of two oscillating particles of masses m^ and m2 inter­
connected by an ideal spring is released in gravity. If the accelera- 
tion of mt is a, find the acceleration of m2.
Substituting 
we have
Substituting 
we have
iiHiinn/ininirtiiln/it/iii
Fig. 11.3
This gives a.
Hence, m{ accelerates towards left with a magnitude of Iplane between first and second inclined 
plane. (Assume 0 = 2)
y 
h
'77777777777777777777777777777777777
7777777777777^77777777777777
k B C
77777^77777777777777777/77^777^77/7777
my-Ii
77777777/7777777
N smooth ball of mass m is kept on a plank of mass M which is falling freely 
from a height h. If the collision between all surfaces are elastic, find the 
maximum height attained by the ball when :
(i) M = m
(ii) M » m
//-numbers of identical smooth balls are kept maintaining 
small distance between them. The ball at the left end is 
pushed with a velocity vq. If the coefficient of restitution 
of collision between all balls is e, find the speed of the 
A th ball.
Collisions
Three identical smooth balls A, B and C are placed on a horizontal 
surface. The ball A is pushed with a velocity v0. The coefficient of 
restitution of collision between A and B is | and between B and C is 
1. Find the velocities of the bodies after all collisions.
A particle is projected from the edge of a step of height h and 
width d such that it collides again and again rising a constant height 
h' relative to the point of collision. Find the :
(i) coefficient of restitution.
(ii) speed of the particle so that it can strike the edge of each step.
A pendulum bob is released from rest from a height h when the 
string of length I is horizontal. If it does not rebound just after the collision, 
find the:
(i) energy loss during the collision.
(ii) maximum angle that the string swings with vertical.
A particle of mass m collides elastically with the pan of mass M{= 2m) of a 
spring balance. Find the :
(i) velocities of the bodies just after the collision.
(ii) maximum height raised by the particle m relative to the point of impact.
kJ h' I
H Ixr: /.a \
h' \
♦ —v0 |m|
23.
Cliff
—«-v
B
I s
- v k ----- 2v 
2m —► 'TftTOW' m —► 
"777777777777777777777777777777777777777777777/
A particle of mass m} is connected to another particle of mass m2 by 
an inextensible string of length /. If mt is projected on a smooth horizontal 
plane with a velocity v0, find the :
(i) velocities of the particles just after the string is taut.
(ii) loss of energy during collision.
A bead of mass m collides with a smooth hemisphere of mass M with 
a vertically downward velocity v0. In consequence, the bead moves 
horizontally just after the collision. Find the :
(i) speeds of the bead and hemisphere just after the collision.
(ii) coefficient of restitution.
(iii) energy loss during collision.
Two blocks of mass mt = m and m2 - 2m are connected by an inextensible string A. 
The system (mj + m2) is hanging from a rigid support by an elastic string B of stiffness 
k. If the string A is cut, mx will bounce up and just touch the point of suspension P. Find 
the natural length of the string.
A bullet of mass m strikes a hanging sand bag of mass M inelastically, with a 
horizontal velocity v0. Find the :
(i) fraction of KE of the bullet delivered to the sand bag during the collision.
(ii) maximum angle swung by the sand bag.
(iii) change in tension in the string during collision.
Two blocks of mass m and 2m are moving towards a massive cliff with velocities
2v and v respectively. The cliff moves with a velocity v. If 
the coefficient of restitution of collision at the surface of 
the cliff is e = find the :
(i) velocity of the block m just after colliding with the 
cliff.
(ii) energy loss during collision in between the mass m and cliff.
(iii) work done by the cliff during collision.
(iv) maximum compression of the spring of stiffness k which is fitted with the block m.
1—
|- - R/2^V0 
R / \
G.R.B. Understanding Physics MECHANICS (Part-B)
A hammer of mass m falls freely from a height h onto an iron nail of mass M. If 
the collision is perfectly inelastic, find the :
(i) fraction loss of energy of the hammer during collision.
(ii) maximum distance of penetration of the iron nail, assuming R as the resistance 
of the wood.
(iii) time of penetration of the nail with reference to (ii).
t
h
__ L
miP-
24.
25.
26.
u
27.
28.
'////✓///
29.
30. m,
31.
32. m
/Zyzzz
✓ZZZ/ZZ/ZZZ/Z/Z/
Bl
Th
Collisions
A toy gun fires n balls per second each of mass m with muzzle velocity w. The 
balls collide elastically with a disc which makes the disc to float in air. If the 
mass of the disc is M, find the height h.
In the above problem, if we substitute the toy gun by a vertical hose pipe of 
cross-section a which ejects water with a velocity v, find h. Assume that the 
water falls dead just after collision and p = density of water.
There are w-persons each of mass m standing on a trolley 
car of mass M. If they jump one by one with a velocity
u relative to the trolley car, the car receives a net impulse
A/j. If the persons jump simultaneously out of the trolley [ 
car in same direction with same speed relative to the 
trolley, the car receives the net momentum AP2. Compare 
AFj and AP2.
A particle is projected with a velocity v0 at an angle 0 with horizontal. It collides 
with a vertical wall and rebounds to strike the point of projection. Find the value 
of R if the coefficient of restitution is e.
Two balls A, B are kept on a smooth horizontal surface at a small distance of 
separation. If we project the ball A with a velocity v0, find the final velocities 
of the balls after all possible elastic collisions if:
(i) W| m2
Two blocks of masses and m2 are connected by an inextensible string 
which passes over a smooth pulley. If the block is kept on a horizontal 
surface and the block m2 is dropped from rest through a vertical distance h, 
describe the motion of the blocks.
A ball is dropped from rest from a height h onto a smooth horizontal surface. If 
it collides elastically, find the force exerted by ground averaged over the time 
period of to and fro motion of the ball.
In the above problem, if the coefficient of restitution is e, the ball will stop after repeated collisions, 
find the total:
(i) time of motion.
(ii) distance covered.
(iii) momentum delivered by the ball, till it stops bouncing.
A ball of mass m is pushed with a horizontal velocity v0 from one 
end of a sledge of mass M and length /. If the ball stops after its 
first collision with the sledge :
(a) find the speeds of the ball and sledge after the second collision 
of the ball with the sledge.
(b) what is the distance traversed by the CM of the system till the ball experiences (n +1) 
collisions with the sledge?
m 
A
• ».
M
zzzzzzzzzzzzzzzz/zzzzzzzzzzzzzzzzzzzzz/
A .& a.,it 
■t' rM? v* .i
■ rt - - w w -ft
v0
?zzzZzzzz/?zz2///z
-------- I —
34. M
length /. The sledge is kept at a distance d\ = d
35.
ANSWERS
4. (b) 5. (c)
3. (b)2. (a)
3. (b)
9. (F)8. (T)7. (T)5. (F) 6. (T)3.. (F) 4. (F)2. (T)
H. (T)
2. (c)
10. (b)
6. (b)
14. (b)
5. (d)
13. (c)
7. (a)
15. (d)
8. (c)
16. (d)
4. (c)
12. (a,b)
1.30
33.
A body after falling freely through a distance h collides with an inclined plane of 
angle of inclination 0 . If the coefficient of restitution of collision is e, find the :
(i) distance between two consecutive impacts.
(ii) energy loss during the first impact.
M + m
a rigid vertical wall. If the string is cut, find the :
(a) distance moved by the CM of the system (AY+ m) till the block collides with the free end of 
the sledge.
(b) compression of the spring when the block strikes it after colliding elastically with the free 
end of the sledge. Ignore the friction between all contacting surfaces and assume I» x0.
°T
h
G.R.B. Understanding Physics MECHANICS (Part-B)
A ball of mass m is projected horizontally with a minimum 
speed v0 so that it leaves the top of a fixed smooth 
hemispherical surface of radius of curvature R.
(i) Find v0.
(ii) If the ball collides repeatedly with the smooth 
horizontal surface, find the total time of vertical motion of the ball and the corresponding 
horizontaland vertical distances covered by the ball.
A block of mass m compresses the spring by x0 by a string 
that interconnects the block and a sledge of mass M and
6 7777> '////////////////////////////////////'///■ '/W
I --------------------►
Multiple Choice Questions
1. (d) 2. (c) 3. (a)
9. (d) 10. (b) 11. (b)
17. (c)
Assertion-Reason Type Questions
1. (d) 2. (a) 3. (d)
Match the Columns
1. a - r. b-s, c-p, d-q 
Comprehensions 
Passage-1
1. (b)
Passage-2
1. (b) 2. (c)
True or False
1. (F)
10. (F)
from —
Collisions 131
Fill in the Blanks
1. change in momentum 2. negligible 3. 5. -4W-
6. 7. 6£
Problems
(>) V]1.
(ii) v =
(iii) -Yma.x —
5.
7.
9. (i) v =
(iii) E =
11. VA
13.
(ii) P = sin
15.
= ^0
4
k(m} + m2) 
m}m2
^l»rel 
m\ + m2
m
M + m
m2n,3 
(m2 + m2)k
P = sin”1 (e2” sin 0), with horizontal
3v0
= 16^’Vc
(i) mgh(] -sin4 0)
y (1 -sin4 0), where 0 = y
(i) >/2gA(e2sin2(|) + ccs2(|))
(ii) 2ej^(tan)
\ S
(iii) 4esin0(l + etan2
+ sin20 v0
/
R and y = + e R 
1-e2
(io
* = -L
2g
(ii) , = !+* >2R 
1 — e
V2
Rotational Kinematics
StudyC^Points
§13.1 Introduction
In particle kinematics, we found displacement, velocity and acceleration of a particle if one of 
them is given. In the kinematics of rigid bodies, we will establish the concept of rotation as the angular 
motion of any point relative to another point of the rigid body. For this, the concepts of angular motion, 
angular velocity and angular acceleration of a rigid body will be explained. Different types of motion of 
rigid body like pure translation, fixed axis (pure) rotation and combined motion (translation and rotation) 
are explained. I explained the concepts of co and a of a rigid body by the ground frame method and 
relative motion method. Then the concept of instantaneous axis of rotation is explained. The concept of 
rolling is explained as a special case of combined translation and rotation of a rigid body.
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
13.9
INTRODUCTION
DEFINITION OF A RIGID BODY
Rotation (Angular Motion)
Angular (Rotational) Quantities
Types of Motion of a Rigid body 
Kinematical Equations for Rotation
Physical Interpretation of Angular Motion (Rotation) 
Calculation of Velocity of a Rotating Body 
Calculation of Acceleration
13.10 Concept of Rolling
13.11 Instantaneous axis of Rotation
13.12 Radius of Curvature of the Path Traced by a Point of Rigid Body
13.13 Rotation About a Point
13.14 Summary
CHAPTER 13
134
Sol.
Fig. 13.1but not
G.R.B. Understanding Physics MECHANICS (Part-B)
After reading this chapter, you will understand the practical applications of rotational kinematics, 
e.g., the transfer of energy via gear-system, belt-drive of wheels, ball-bearing system, power transfer 
from the engine to the wheels. The relation between the linear motion of particles and angular motion of 
rigid bodies is applied in rotating parts (shafts, gears, cranes, ball-bearing etc.) in various rotating 
machines. In fact, this chapter will form the basis of understanding the dynamics of rigid bodies.
§ 13.2 Definition of a Rigid Body
Generally we call an object “rigid” when we cannot deform (squeeze) and twist it. This means, 
the shape and size of a rigid body do not change ideally. In consequence, the distance between any two 
points of a rigid body does not change. However, in practice, any object can be compressed or elongated 
by certain extent. In this way, nothing is perfectly rigid. When the deformation of an object is insignificant 
compared to its dimensions, practically we call it a “rigid body”. Here, we use the word “deformation” 
to mean compression, elongation and shearing (twisting or bending).
Recapitulating the above facts,
> In the above example, will the relative acceleration between the two points be zero? Ans. No
§13.3 Rotation (Angular Motion)
Confusion between rotation, revolution and circular motion : When we say that an object is 
rotating, what does it mean basically? Does it mean that the points of the object (like CM etc.) are 
rotating? When you observe a rotating fan, you find its CM stationary. Does it mean that the fan is not
= v^, we can write = 0,
There is no relative motion between any two points of a rigid body along the line of separation, 
but there may be a relative motion (v^ * 0) perpendicular to the line joining the points.
Student Taskf
dr 
dt ,
- component of relative velocity vAB between A and B along (or parallel to) the
We can define a rigid body as a body which undergoes no deformation ideally or practically 
there is only a slight deformation compared to its size, under the action of a force or torque. 
Hence, the distance between any two points of a rigid body remains ideally constant or almost 
constant in practice.
Ex. 1. If the distance between any two points inside a rotating rigid body remains constant, do you 
think that the relative velocity between the points will be zero? Explain.
As the relative distance r, that is, distance of separation between any
two points A and B, say, remains constant,
^ = 0
dt
dr
Since, —dt
where i
“’ll ’ - • ■
line of their separation. However, the component of vAB perpendicular to AB, that is, 
may not be zero when r*AB (or r^A ) changes in direction.
Rotational Kinematics 13 5
rotating? Absolutely not! Then you may try to define the rotation of a fan as the circular motion of the 
points of the fan. Yes we do agree that the points of the fan are moving in circular paths relative to the 
axle, say. But when you watch a spinning ball leaving the hands of a bowler, neither there is an axle 
(axis) nor the particles of the ball move in circular paths (relative to you). Then how do you define the 
rotation of the spinning ball? By this time, you might have been more cautious in defining the term 
“rotation”. Now, you may be tempted to explain “rotation of an object” by using the idea of “revolution 
of a point” about any other reference point. Suppose your argument is that, “when a particle revolves 
(moves around a point in a “closed path”), the particle (point) is said to be rotating. For instance, 
revolution of electrons around the nucleus, revolution of the planets around the sun etc. Well; when we 
throw a spinning disc, the CM of the disc moves in a parabola relative to the point of projection which 
is not a closed path. Then, can you say that the disc is not rotating? Of course it is rotating! Then your 
idea of “revolution of a point” fails to define the rotation of the spinning disc. Now let me try to analyse 
the mistake.
Whether a point revolves (moves in a closed loop) or moves in a general curve such as a parabola 
etc., we call it curvilinear motion of a particle. The word “curved path” is basically a line in which the 
particle moves. When a particle (or point) moves, it passes through different points of space. If we join 
all these points, we get a curve, that is what we call path of the particle. In a rigid body, different points 
move in different types of paths. For instance, the CM of a body rolling on a horizontal surface moves 
in a straight line whereas other particles move in curves. Then, how can the motion of a particle define 
the rotation of an object? In fact, linear motion of an object is defined as the motion of the point of its 
centre of mass, but the rotational motion of an object can be defined by the motion of something other 
than the points.Then, what is that which can define rotation? Probably the nearest geometrical symbol 
to “POINT”!Oh. yes, it may be a “straight line”!
Realising the fact that, “rotation of a point has no meaning", we will see how a straight line will 
help us in defining rotational motion of an object as described below.
Angular motion of a straight line : For 
the sake of simplicity, let us take a plate like (laminar) 
rigid body. Draw a straight line connecting any two 
points A and B, say, of the plate. Then produce the 
line AB towards a fixed reference line XX'. Let the
line AB make an angle 0, with XX. We call this 
angle “angular position” of the straight line AB. When 
the plate moves from position 1 to position 2, we 
can see that the angular position of the line AB 
changes from 0| to 02. That is what we call 
“angular motion” of the straight line AB. More 
popularly “angular motion” is known as “rotational 
motion or rotation”.
From the above discussions we conclude that;
X1 X
The straight line AB changes Its angle of 
orientation with X'X line from 01 to 02 as the 
plate moves from position 1 to position 2
Fig. 13.2
Whenever the angle of orientation of a straight line drawn inside a rigid body changes with 
respect to a fixed reference line, we say that the straight line and hence the rigid body undergoes 
mgular motion or rotational motion or rotation.
Then,
where 4> = p - 6.
Hence,
This gives;
where 0 = angle of orientation of any straight line of the planar rigid body measured relative to 
a fixed reference line.
Since, rotation of a rigid body has a sense (such as clockwise, anticlockwise etc.), angular 
velocity of a rigid body is a vector quantity.
All lines drawn inside a planar rigid body rotate through same angle which can be called “angle 
of rotation or angular displacement” of the rigid body.
X' X
Each straight line (AB and CD) drawn 
on the plate moves with same angular 
displacement and angular velocity 
Fig. 13.3
de ^p 
dt dt ’
which signifies that these two lines change their angular position with equal rate.
It means, when a rigid body rotates, each line inside the rigid body changes its angular position at 
the same rate at a given instant. In other words, at a given instant, the angular velocities of all lines 
drawn in the plane of a planar rigid body are constant, that is what we call “angular velocity” of the rigid 
body which can be given as :
de co = —, 
dt
Awhere — = U dt
This gives;
The above equation tells us that both lines rotate with same angular displacement. Hence, we 
conclude that;
vZb
Angular velocity : Since, between the straight lines can be given as :
(J) = p — 0
Since, the body is rigid, = constant.
^0. 
dt
X C x 
W “CD A
c/p = de 
dt dt
59 = 5P
1.
2.
3.
4.
d coa = — 
dt
Dividing both numerator and denominator by de, we have 
d® de a =---------
dQ dt
(ado)a =-----
dQ
X' x
When a body rotates, the angle 9 made by the line 
AB increases. Hence, the sense of rotation is same 
as the “anticlockwise” sense of measuring 0
Fig. 13.4
■z
Substituting — = ©, we have 
dt
Rotational Kinematics
Direction of © : Let us now find the 
direction of co, that is, sense of rotation of a 
rigid body.
For this, we need to keep an eye on the 
motion of the straight line AB drawn inside the 
rigid body. If during a small time interval measured 
from an instant /, the angle of orientation 0 of 
the line AB increases, the sense of rotation of the 
rigid body is same as the direction or sense of 
increasing 0. That means, direction of increment 
of 0, that is, "de" and co are same. In other words, 
if 0 is measured in anticlockwise sense and 0 
increases with time, then the direction of co is 
anti-clockwise. Similarly, we can argue that, if 
the body rotates so as to decrease the angular 
position 0, the sense of rotation is clockwise.
From the above discussion, let us mention the following ideas :
The angular velocity co of a planar rigid body is defined as the rate of change in the 
angle of orientation of any straight line drawn inside the rigid body which can be given
deas co = —, where 0 = angular position of the reference straight line drawn inside the 
rigid body relative to a fixed reference line.
If the body rotates so as to increase 0, the sense of rotation (direction of co) is same as 
the sense of measuring 0 and vice-versa.
Angular velocity of a rigid body does not depend on any coordinate system or reference 
frame.
Angular velocity is an internal property of rotating rigid body.
Angular acceleration : Once we find the magnitude and direction of an angular velocity, we 
need to observe whether the magnitude of angular velocity is changing or not. If a planar rigid body is 
not allowed to change the plane of rotation, it can only change the magnitude of its angular velocity. 
When the rigid body is speeding up or slowing down, we say that it is accelerating. In other words, 
when the angular speed of a rigid body changes, it is said to be accelerating. The rate at which the 
angular velocity of the rigid body changes is defined as the angular acceleration of the rigid body which 
can be given as:
138 G.R.B. Understanding Physics MECHANICS (Part-B)
Substituting
1.
5.
2.
3.
4.
(b) co increases; a and oo 
have same direction
dQ co - —, 
dt
§ 13.5 Types of Motion of a Rigid Body
Plane motion : In this section we will describe different types of motion of the rigid bodies 
encounted in our day to day lives. In all types of motion we will include the examples of motion of rigid 
bodies when all particles (or points) of the rigid bodies move in one plane (or parallel planes). That is 
what we call plane motion.
(i) Pure translation : During the motion of a rigid body 
when the angle of orientation 9 of any straight line drawn inside 
the rigid body does not change during any time interval, we say
that the body is not rotating because, co = — = 0. In
dt
consequence, all points of the rigid body have equal linear 
displacements, linear velocities and linear accelerations. Then, 
the rigid body is said to be translating perfectly (without rotation). 
In fact, pure translation is a special case of rotation when co = 0.
you can also express a in terms of 0, which is given by :
d2Q 
a = —v dt2
/^e /^e
t = tj t = t2
Pure translation of a rigid body 
because the straight line AB does not 
change Its angle of orientation 
Fig. 13.6
Direction of angular acceleration : To find 
the direction of a, first of all we have to find whether 
the angular speed is increasing or decreasing. If co 
increases, — will be positive; then a and co will 
eft
have same direction. If co decreases, ~r will be “ decreases;a andrigid body respectively.
139
(ii) Pure rotation : We leamt that angular velocity of all
instant. That is what we call “angular velocity” of the rigid body. 
However, the angular velocities of the points of the rigid body 
about any fixed point inside or outside the body may or may not 
be constant. Now we will need “that point P, say”; about which 
all points of a rotating body must move with same angular velocity. 
If this condition is satisfied we can say that the rigid body is 
rotating perfectly about P.
If a rigid body rotates perfectly about a fixed point P, 
angular velocities of all points (three points A, B and C, say;) 
about P are equal to the angular velocity co of the rigid body,
®ap = ®bp ~ ®cp = m
Rotational Kinematics
(a) Rectilinear translation : If the points of a rigid body executing 
perfect translation move in straight lines, the motion is called rectilinear 
translation.
In rectilinear translation, all 
points, (A, B, C and D etc.) 
move in straight lines 
Fig. 13.7
(a) Axis P passes through CM 
"C”; centroidal rotation
Three points A, B and C of a rigid 
body move with same angular velocity 
about P. Then, the body is said to have 
pure rotation about P 
Fig. 13.9
(a) Centroidal rotation : When the point P (axis) passes through the CM of the rigid body, we 
call it centroidal rotation.
(b) Non-centroidal rotation : When the point P (axis) passes through any other point of the 
rigid body, we call it non-centroidal rotation.
(b) Curvilinear translation : If each point moves in parallel curves, 
the motion of the rigid body is said to be curvilinear translation 
which should not be confused with the rotation of a rigid body 
because the angle of orientation of any straight line AB, say, 
does not change (co = 0).
In curvilinear translation, the points 
lines drawn inside a rotating rigid body is constant at a given A, B, C, D and E move in parallel curves 
Fig. 13.8
°’AP
(b) Axis P does not pass through 
CM; non-centroidal rotation
Fig. 13.10
Pure rotation is basically rotation about a fixed point (axis). All particles of the rigid body must 
move in concentric circles about the axis with equal angular velocities. However, the linear velocities of 
different points will be different.
(Hi) Combined motion : In this type of motion, the linear and angular velocities of different 
points of the rigid body are different relative to a fixed point. In other words, no two points of a rigid
110
V
1.
2.
3.
Combined motion of a cricket ball in 
air; unconstrained combined motion 
Fig. 13.12
The hanging sandbag Is free to 
rotate about any axis passing 
through Its point of suspension P 
Fig. 13.13
(i)^
In pure translation, there is no angular and linear velocity between any two points of the 
rigid body. It means that velocity of each point of the body is constant at any instant. 
However, angular velocities of different points of the rigid body relative to an outside 
fixed point will be different which must not be confused with the angular velocity of 
rigid bodies; wbody = 0 and v = c .
In pure rotation, if all points move in a plane all points of the rigid body must have same 
angular velocity (to = c) relative to “the fixed point” (axis) for planar rigid body; and the 
linear velocities v = rco are not equal at a given instant. In this case all points move in the 
concentric circles relative to the fixed point.
In combined planar motion, different points of a rigid body possess different linear and 
angular velocities relative to a fixed point outside the body. Generally, the centre of mass 
of the rotating bodies moves in combined motion. You should remember that pure rotation 
and pure translational motion-are the special cases of combined motion.
140 G.R.B. Understanding Physics MECHANICS (Part-B)
body undergoing combined motion have same linear and angular velocities relative to a fixed point. Most 
of the motions of extended bodies fall under this category. However, in a combined motion, it may be 
possible to imagine a “point” relative to which the rigid body can have pure rotation. We will define it as 
“instantaneous axis of rotation” in next section.
(a) Constrained motion : There are many 
examples of rigid bodies executing combined motion under 
the action of “constraint forces”. For instance, “rolling” 
of a body on a horizontal surface, motion of a bar in 
vertical plane under the action of vertical and horizontal 
surfaces, etc.
(b) Unconstrained motion : When a body has a 
combined motion in space or under a field force without touching 
any surface, we call it an unconstrained motion. The motion of 
a spinning cricket ball in air is a familiar example of unconstrained 
motion.
(iv) Gyroscopic motion : When the plane of rotation 
changes, even though magnitude of angular velocity remains 
constant, its direction will change. This type of motion is called gyroscopic motion. The change in 
direction of angular velocity will give rise to an angular acceleration which is different from that described 
in previous section (due to change in magnitude of angular velocity).
(v) Rotation about a point: Sometimes we see the sandbags 
hanging from rigid support such that you can rotate them about any 
axis you like, passing through the point of suspension P of the 
sandbag. That is what we call rotation about a fixed point.
Recapitulating different types of rotation we have following 
important points:
6 'ZZZZZZZ ZZZZZ^ZZ ZZZZZZZ
(I) Rolling of a body (ii) motion of a bar in 
vertical plane under the action of gravity and 
reaction (constraint) forces, are the suitable 
examples of constrained combined motion 
Fig. 13.11
(ii) \
141
4.
Fig. 13.14
•••(•)
and ...(ii)
ewhere
Average angular velocity : This gives;
Sol.
0 mdt,
where
This gives; Ans.
Then the average angular velocity over a time interval 
Ar = t2 - /| can be given as:
= f2/3 Jo
= 0
t '
= f 2 co dt
J'i
0 = $a>dt
“av
§ 13.6 Kinematical Equations for Rotation
Recasting the general kinematical equations for a rotating rigid body, we have
dS 
0) = — 
dt
dm mdm d2ea = — =----- = ——
dt de dt2
Angular displacement: If co is given as the function of time, the total angular displacement 
can be given as :
I”1*
Z2 ~Z1
co = 2t - 3r2
0 = Jo2/3(2r-3/2)c/r = rad
0
®av —
Ex. 2. When a disc rotates so as to vary its angular velocity with time as co = 2t - It2, find the total 
angular displacement of any point on the disc relative to its centre when the disc will stop 
momentarily.
2 
When the disc will stop, co = 0. Putting co = 0 in the equation co = 2r - 3/2, we have t = —.
2 ... 3
The angular displacement 0 during a time interval of — s measured from starting is given as :
Rotational Kinematics
If the plane of rotation (or axis of rotation) changes due to the change in 
the direction of angular velocity of the rigid body about the axis of rotation, , 
the angular acceleration can be given as, a' = coco', where co' is the 
angular velocity of the axis of rotation of the body about a fixed axis. The 
direction of a' will be perpendicular to the rotating axis while lying in the 
plane of its rotation.
G.R.B. Understanding Physics MECHANICS (Part-B)
Sol.
This gives;
we have
This gives; Ans.
Student Task
Ans. (i)
M-I :
> In the above example, find the :
(Z) total angular displacement.
(ii') average angular velocity over the time of motion of the disc.
142
Student Task
> In the foregoing example, find the average angular velocity during first (Z) two seconds 
2(ZZ) 3 second.
2^/2
3K
To find 0 we have two methods, when a = /(0).
Express co = f(f) using the expression P dt
Ju)o Vo) Jo
Then using the equation 0 = find 0.
M-LI : Substituting a = -K J® in cot/co = ac70, obtain co as the function of 0
_, dQ , . r dQThen use co = — to obtain / = -----
dt J co(0)
Ans. (i) -2 rad/s (ii) | rad/s
Sometimes a is given as a function of co . In this case, how to find the angular displacement and 
average angular velocity ? Let us see this in the followingexample.
Ex. 3. A disc rotates so that angular acceleration a varies with its angular velocity as a = -Kyjw, 
where K is a positive constant. Find the total time of motion of the disc. Assume the initial 
angular velocity of the disc as co0.
The “total time of motion” signifies that the disc will stop after sometime, /, say, as it is rotating 
with an angular deceleration. We need to find that time *7”.
Substituting a = -/fVco in the equation (ii), we have
dt
^ = -Kd,
VCO
J?'
.... °0(") y
When the disc decelerates from coo to 0 during time /,
Juo Vco
 2^ 
I — — 
K
which gives;
where 0 = co0r +
This gives;
where
®av =then,
and t -Substituting coav =
we have
where co = co0 + at 
Then, we have
Rotational Kinematics 143
Constant accelerated motion : When the body rotates with constant angular acceleration, we 
have a = constant.
= C co eft, 
Jo
CO + COq
2~
“av
Then during a time/, the change in angular velocity is Aco(= In the above example, find the :
(/) number of rotations of the fan before it stops rotating.
(ii) average angular speed of the fan over the total time of rotation.
(iii) Since, p,| = |or|,
where \a, | = |ra| and pr| = |rw21,
a. = -K in co = co0 + a/,
G.R.B. Understanding Physics MECHANICS (Part-B)
Let us tabulate the expressions for rotation with constant angular acceleration
2
... “0 ®°Ans. (1) —— (n) ~
co = Va, coc = 2^JK,
1 
/ = —7=4k
(ii) Substituting co =
»av
Ex. 4. When you switch off a ceiling fan of radius R rotating with an angular velocity co0, suppose it 
retards uniformly to rest with an angular deceleration K. Find the : 
(/) total time of rotation of the fan.
(ii) angular displacement of the fan, over the total time of its motion.
(iii) time after which any point of the fan has equal magnitudes of tangential and radial 
acceleration, assuming co0 = 2\I~K ■
(i) As the fan comes to rest, co = 0; Then substitute co = 0 and a = -K in co = co0 + at to 
obtain
co + co0 
2~
co = )
Now let us stand at A and observe B. We find that B seems to move 
with a velocity ~vBA perpendicular to AB. It means that B turns about A with 
an angular velocity,
VBA
B turns around A with an 
angular velocity In the above example, draw the snap shots about 
C, D and the geometrical centre of the plate.
The above analysis tells us that the motion of a point 
relative to any other point inside a rigid body is circular.
When any rigid body, a flat plate, say, moves from y 
position 1 to position 2, we can see that the straight line AB 
rotates through an angle 0 and simultaneously translates by a 
distance SB. Hence, the displacement of A relative to ground (or any fixed 
point) is given as :
Whatever may be the point of reference inside a rotating rigid body, the angular velocity 
of other points relative to the reference point is same, which is called angular velocity of 
the rigid body. At a given instant, the magnitude and direction of angular velocity remain 
the same relative to any reference point inside (or outside) the rigid body. It means that 
the angular velocity does not depend on the reference point or coordinate system.
In general, a rotating body does not require any axis (pivot, hinge, etc.). Sometimes we 
provide the axis (for instance shaft in the ceiling fan, hinge at the door) to equilibrate the 
rigid bodies against gravity and other external forces.
------------- 'B
Fig. 13.17
G.R.B. Understanding Physics MECHANICS (Part-B)
Imagine yourself standing on a square plate rotating in a horizontal plane. If D 
the plane of the plate is lying in a horizontal plane (plane of the paper, say;), 
draw the snap shots of velocities of different points of the plate and sense of 
rotation of the plate, relative to the corners A and B of the plate. A
Since there is no relative motion along the line joining any two points, each 
point seems to move perpendicular to the line 
joining the other points. This means, each point 
moves in a circular path relative to a.iy other point 
of the rigid body. Following this logic, when you 
stand at the comer A you can see that other points 
move around you in concentric circular paths with 
same angular velocity in clockwise sense.
Similarly, when standing at the other comer B of 
the plate, you can observe the points of the plate 
moving in concentric circular paths around you with same angular velocity in clockwise sense. 
From the above discussion we find that,
$A ~ $AB + $B
Differentiating both sides with respect to time we have
Rotational Kinematics 147
dS,
rAB
Hence, VA ~ ®AB x rAB + VB
.Cc*
Sol.
i
I e
bU
a
Student Taskf
Bi 
r;
Ai
Pure rotation about C + translation of rod
Fig. 13.22
Please note that both translation and rotation of the rod happen simultaneously during the 
motion of the rod. Hence, we call this motion a combined motion.
Ir
A1
, d^tt - 
and ------ = i u
di= 1 IB
An
> In the above example, find the position of the end points A and B of the rod, if the initial 
position of CM of the rod is assumed as origin.
Ans. xA = xc +- sin 9 , yA = - — cosO ; xB = xc - — sin 9 , yB = - — cos 9
2 2 2 2
This tells us that the combined motion of a rigid body (or any line BA of the rigid body) can be 
thought as the translation of any reference point B and circular motion of the other point A 
relative to B. Hence, the choice of the points (axes) is arbitrary.
Let us execute the above idea in the following example.
Ex. 6. In all possible ways, explain the combined motion of the rod from the 
position 1 to position 2, as a combination of pure translation and 
rotation of the rod.
In this example, we take the CM at C. We observe that the CM 
moves through a distance xc and the rod rotates about the CM by an 
angle of 30° simultaneously. Hence, the motion of the rod is a 
combination of a pure translation of the rod by a distance 
xc and pure rotation of the rod about C by an angle 9 = 30° 
in anti-clockwise sense. Alternately, we can first rotate the 
rod through an angle 0 = 30° in anti-clockwise sense about 
C and then translate the rod by a distance xcto obtain same 
final position.
Fig. 13.20 
B -.B'z 
■ 
—
A?
Pure translation of CM “C” of the rod 
+ pure rotation of the rod about C 
Fig. 13.21
d S j dS.IB dSB , d S j 
dt dt dl dt
—> —> —> ♦
Then, v ( = vAB + vB , where vAB = aAB x
IB
= V-" —
2 .
0J
\\ 1 \\l 
60°A\ -xc —
y
— XA
i
-------- X0 ,B
O
...(iii)
and ...(iv)
Yb I
X
The relation between coordinates of the 
points relative to origin can be given as : 
xB = xA + / sine and yB = yA -1 cos0
Fig. 13.23
II
148 G.R.B. Understanding Physics MECHANICS (Part-B)
§ 13.8 Calculation of Velocity of a Rotating Body
In this section we will find the velocities of a rotating body. This may mean the linear velocities 
of any point of the rigid body and angular velocity of the rigid body. As discussed earlier, to find the 
angular velocity co of the rigid body we have to draw a straight line inside the rigid body. Then we find 
the variation of the angle of orientation “ 0 ” of the straight line relative to a fixed reference line (x, y or 
z-axes, say;). Once we have 0 = f(t), its t.me derivative will give us the angular velocity co. Since we 
need two points to draw a straight line, we will need velocities of two points to find the angular velocity. 
In this way, three quantities are involved; two linear velocities of two points and angular velocity of the 
line joining the points, that is, angular velocity of a rigid body. If you choose two points A and B, we can 
have vA, vB and co. Then we need any two quantities to find the third; either vA and vB to find co, or 
vA and co to find vB, or vB and co to find vA. For this, we have two methods as described below.
Method-1 (Ground frame or absolute motion)
In this procedure, we express the coordinates of any 
two points (A and B, say;) of the rigid body relative to a fixed 
point (origin O, say;). If the angle of inclination (orientation) 
of the straight line AB withy-axis is 0 and the distance between 
A and B is /, we can write the coordinates of the points as :
xfi = x^ + Zsin 0 ...(i)
and yB = yA - I cos0 \ ...(ii)
However, you can also take 0 relative to x-axis and 
build the last two equations (i) and (ii). Try to take 0 in anti­
clockwise (or increasing 0 as the body rotates) to get a
mathematical advantage of writing — = +co. Now we have the positions of the points in terms of 0.
dt 
Then differentiating the positions given by equations (i) and (ii) with time, we have
dxB dxA dB
dt dt dt
dyB dyA , . . dQ
dt dt dt
When x and y-coordinates increase with time, write = vx and = vy. If the coordinates 
dx dy
decrease with time, write — = -vx and — = -vy. Similarly, when 0 increases with time, write 
. dt dt
— = +co and if 0 decreases with time, write — = -co. Assuming that xA,xB,yA,yB and 0 increase 
dt dt
with time, from the equations (iii) and (iv), we have vB* = vA* + co I cos 0 and vB? = vA^ + coZ sin 0.
Hence* we can find vA when co and vB are given or vB when co and vA are given or co when 
vA and vB are given, in any direction.
149
B
Sol.
Fig. 13.24
y
B
0
A...(ii)and
(given)
Ans.
...(iii)
Ans.
Ans. (i) -(/ - tan 0 j) (jj) Circle
2
> In the above example, find the :
(/) velocity of the mid-point of the rod.
(ii) nature of path traced by the mid-point of the rod.
de— = co 
dt
H— X —*4
Fig. 13.25
Method-2 (Relative motion)
In this approach, we need not take the help of a fixed reference point (origin of a coordinate 
system). Just we consider two points A and B, say; inside the body as we did previously. Let the points 
—> —>
move with velocities vA and vB respectively. Now, resolve the velocities parallel (along) and 
perpendicular to the line joining the points. Let (vx)]|» (v^)|| and (vx)_l»(vb)a be the parallel and
v ]
I cos 0 J
Using eq. (ii) and eq. (iii), we have
vB = -v tan 0
Negative vB signifies that B moves down (so as to decrease y).
Student Task^
t 
y J_ 
o
(vertical) as 0, we have x = / sin 0 
Differentiating with time we have 
c/x . -de— = i cose—
dt dt
dy - ■ ade— = -I sin 0 —
dt dt
. . . dxAs A moves, x increases; hence we put — = vA = v
Substituting = vA in eq. (i), we have 
v co =-------
ICOS0
(ii) The positive sign of co signifies an anti-clockwise rotation (in the increasing order of “0”). 
Since, 0 increases with time, we have
Rotational Kinematics
Let us use the foregoing method in the following example.
Ex. 7. A rod of length I is moving in a vertical plane (x-y) when the lowest i 
point A of the rod is moved with a velocity v. Find the : 
(/) angular velocity of the rod (ii) velocity of the end B.
(i) Assuming the coordinates of A and B asxandy respectively relative 
to the origin O and the angle of orientation of the rod with 7-axis
and y = I cos 0.
This gives us;
Sol.
• Fig. 13.28
/(VB)||
I vabl I 
CD = --------—
/
\(Va)1
In the combined motion of the rod, 
(va)|| = (vab)is vab
.. a IvabJ IvabI Hence, co = —= —y—
Fig. 13.26 .
This equation will give you one velocity (either vA or vB ) if 
the other is given.
Then we substitute vA and vB in the equation of angular 
velocity given as:
where vABl = vAB as = 0 The body rotates In anti-clockwise 
sense with co =
Fig. 13.27
I
Let us use the above idea in the following example.
Ex. 8. The ends A and B of a rod of length I have velocities of magnitudes v 
and 2v respectively. If the inclination of vA relative to the rod is 0, 
find the :
(a) inclination p of vB relative to the rod.
(b) angular velocity of the rod.
(a) For the rigid bodies v/tn = vfin, 
where vX|| = VCos0 and vsN = 2vcosP
150 G.R.B. Understanding Physics MECHANICS (Part-B)
perpendicular components of the velocities of A and B respectively. Since, the points cannot move 
relative to each other along the line joining them, we have
151
P = cos Ans.Then,
(b) As we know,
where
Fig. 13.29
This gives;
sinP =Substituting
VBA.
we have co =
Student Task^f
B
0
A
Sol. kVA
...(ii)and
rBC
Substituting co4C = coBC
COS0
~2~
> Referring to the previous example, using relative motion method, find the 
velocity of B and angular velocity of mid-point of the rod.
Hint:
B
Fig. 13.30
✓
Fig. 13.32
I (vax)± i
I
Fig. 13.31
Rotational Kinematics
This gives : v cos 0 = 2v cos P
Equate the velocities of A and B along the rod to obtain \ 
vB cos 0 = v sin 0. Then substitute
I l= VB sin 0 + vA cos0 in the equation co =
Ex. 9. If the ends of a rod of length I have velocities ~vA and v*B, find the velocity of mid-point of the 
rod. —>
Let us assume that, the mid-point of the rod has velocity vc. Then, we have
rAC *c 7:
czz
A
/VA.
A
VBA±
CO = ——, 
/
VBA = vBAl =l ~vBl -C+V^) I
= 2vsin P + vsin 0
V
co = (2 sin p + sin 0)y
^3 + sin2 0
~2 '
(^3 + sin2 0 + sin 0)v \
I '
vB = 2v 
\sA//
VA = VAC+ VC 
VB = VBC+- VC
Adding eqs. (i) and (ii), we have
vA + ZB =vac + Zbc+2vc 
Substituting vA(- = q>ac x rAQ and vBq =Zbc) + 2 vc
152
Fig. 13.33
Student Task^f
Ans.
C
\vcP
VP - VPC + vc '
G.R.B. Understanding Physics MECHANICS (Part-B)
—> —>
Since, C is the mid-point of the rod, rAC = -rBC . Then, we have
where | | = rPCa>
Then the magnitude of vP can be given as :
I vp l=l vpc+ vc I
If the angle between ?pc and is 0, using parallelogram law of vector addition, we have
> In the foregoing example, derive an expression for the velocity of centre of mass of the system 
if the ends A and B of the rod are attached with point masses mt and m2 respectively. 
Assume the mass of the rod as M.
This tells us that when any two points A and B, (say;) of a rigid body 
have velocities and Vg respectively the mid-point C (but not 
necessarily the centre of mass) of the line joining the points will have a
{2m} + M)va + (2w2 + M) vB 
2(M + ffi| + m2)
 = tan
A
Sol. CD
B
Fig. 13.35
Vac
VA =
where
Fig. 13.36
2v
CD = —
/
0 = 90° - p 
Let us summarize the above ideas as following.
vp = yvPC +vc+ ^vpcvc cos 9 
and the direction of vP can be given as :
30° —*-v
where vc = v, vAC = - • co and 0 = 90° + 30° = 120°
Then, we have
2v
Ex. 10. A uniform rod of length I is spinning with an angular velocity cd = — 
while its centre of mass moves with a velocity v. Find the velocity of the end 
A of the rod.
The velocity of A is,
+ V2 +2yc, 
0 can be given as :
I /2co2 
v2 +-----+ lv(D cos 120° ,
4
If the angle between and vPC is 0, the velocity of P, that is, vP is given as :
Vp ~ yVp^ + Vq + 2vp^v^ cos 0,
where vPC = rco(r = distance between P and C)
and co = angular velocity of the- body;
0 can be calculated from observing the orientation of the line PC with the line of motion of 
centre of mass of the body. We can also adopt component method to find vP. In this method, 
resolve vc and vPC along and perpendicular to the rod or line PC, assuming the x and y-axes 
parallel and perpendicular to the rod (or line PC) respectively. Note that vPC is perpendicular to 
PC.
I
2
154
and = v VAC2 v
in the expression
 = tan 1
B
Fig. 13.37= tanwe have Ans.
Student Task^f
> Find the speed of other end B of the rod in the above example.
= vcos 30°,
I VA l =Then, 73v
Then, 
Substituting
G.R.B. Understanding Physics MECHANICS (Part-B)
Ans.
I (2v_
I
n
2
V/J
A/"
vsin 0 
vAC cos 0 + vc ’ 
vsin 120° 
vcos120° + v
Ans. V3v
vc± = vsin 30° and vAC± = vAC = ^co.
2
= v
You can set the axes parallel and perpendicular to the rod.
After finding the velocities, let us find the ways of calculating the (i) acceleration of any point of 
a rotating rigid body (ii) angular acceleration of a rigid body.
§ 13.9 Calculation of Acceleration
Relative motion method : Let us consider a flat plate rotating in the plane of this page. In the 
previous sections you learnt that any point moves around (relative to) the other points of the plate in 
' circular paths. Hence, an arbitrary point P revolves in a circle about the easiest point, that is, centre of 
mass C. If the acceleration of the centre of mass is given as ac, how to find the acceleration of an 
arbitrary point P, say, let us discuss.
When we have the acceleration a>c of the centre of mass, acceleration of the point P is given as:
ap ~ apc+ uc ;
where vc
x2 v I — +2)
I 
Vac — — (0 — — 
2 2
In the above example, you can also find the velocity of any point of the rod by component 
method. For instance, we want to find vA. For this, resolve vAC and vc parallel and 
perpendicular to the rod. Since, vAC is perpendicular to the rod; its component parallel to the 
rod is zero; v^c = 0
Then, =^cli +^C±+ ^C±.
va = v
0 = 120°, vc = v
155Rotational Kinematics
C
>
'PC,p
will be zero. In that case, we have
...(H) .
Fig. 13.39
Sol. ,y
aP_ (= a - fcd2)x
.2? aP
V a
path traced 
/ by C
aP = apc +ac;
—> —> —> —>
apc = apc,+apcr’apct =ra and aPCr
Fig. 13.38
where
aPC,
The acceleration of a point Pina rigid body 
can be given as :
Then how. to find aPC, let us see. Since, P moves in a 
circle of radius r relative to C, the radial comp- ’ent of aPC, 
that is, aPC which is called centripetal acceleration of P relative 
to the centre .C of circle, can be given as ;
aPC = r(^
—>
Then what about the tangential component of aPC, 
—>
that is, aPC/ (or component of aPC perpendicular to the 
radius r)? As you know, aPC/ is given as :
aPC, = ra,
ac
P6
and , you can
aPC,
where the angular acceleration of the rigid
= ra2
The acceleration of P can be given as :
aP = aPC,+ aPCr + ao 
*. ”* 2 • jaPCi = -ru.j, aPC = -rco i and
ac = ai.
ac
C aPC(
_-^apc, aPy(=ra)
Fig. 13.40
body.
If the rigid body spins with a constant angular velocity, a
aPCi = 0. However, for variable co, we have
aPC = aPC, + aPCr
Substituting a"PC from eq. (ii) in eq. (i), we have
aP = aPC, + aPCr+ ao
where aPCt = ra and aPCr = rco2
Please bear in the mind that co and a are the angular velocity and angular acceleration of rigid 
body respectively.
After knowing the magnitude and directions of the accelerations aPC), ciPCr 
add them in suitable methods as described in the following example.
Ex. 11. A uniform disc of radius r spins with angular velocity co and angular 
acceleration a. If the centre of mass of the disc has linear acceleration a. 
find the- magnitude and direction of acceleration of the point P.
156
This gives:
2\2 Ans.
 ;
The acceleration of any point in a plane rigid body motion can be found by “relative motion 
concept”. Taking the centre of mass “C” as a reference point, the acceleration of any particle 
(or point) P of the rigid body can be given as the sum of tangential and radial accelerations of P 
relative to centre of the circle C.
G.R.B. Understanding Physics MECHANICS (Part-B)
—> 2 * *
aP = (a - rco )/ - ra j
> In the foregoing example, find the accelerations of Q, R and S.
Ans. aQ = (a - ra)/ + rco2 j 
ap =(a + rvi )i + raj 
as = (a + ra)/ - rw2y
aPC - aPC, + aPCp
2 > ; a = —. 
dt
a and co must not be confused with the angular acceleration and angular velocity of any point 
of the rigid body relative to a point outside of the rigid body, respectively. The key idea to 
remember in this case is that angular velocity and angular acceleration of any point relative to 
the other points in a rotating rigid body for its planar motion are constant, at a given time. After 
finding ^pc just add it with to find ^P . Since, “CM” is unaffected by the internal forces 
as a whole, it moves in a most simplified way. Hence, “ ac ” can be calculated easily.
Ground frame absolute motion method: Sometimes we 
observe that, any other pointexcept centre of mass moves in a more 
simplified way. For instance, in constrained vertical motion of a 
uniform bar under the action of the constrained forces of the surfaces 
in contact and gravity, the end points A and B move in straight lines 
whereas the CM of the rod moves in a curve (circle). In that case, 
how do you find the acceleration of some other points of the rod 
when the point A moves with uniform velocity v? In previous example 
we have found co by using both ground frame (absolute motion) and ang|e Of inclination 0 with vertical 
relative motion methods. Fig. 13.41
Hence, the magnitude of ap = - ro>2)2 + r2a2
and the direction of aP is given as :
ra 
a - ra>2
The rod undergoes a combined 
motion with angular velocity m, 
angular acceleration a at a given
157Rotational Kinematics
where
This gives;
Then, aB ~ aBA, + aBAr
and aBAr = la> , you can find aB by following the component
aB = /(a sin 0 + co2 cos 0)
de 
dt
de_
dt '
Now you have co and a. 
Since, vx = v, we have 
dvAa A ~ —— = 0A dt
where aA
. d® .de where — = a and — = co
dt dt
This gives;
v
As calculated previously ®
If you want to find a, just differentiate co with time. Then, we have
 = vs
Vp = VpC+ vc,
Vp (= v - Rco) 
□-►vs 
For clockwise rotation, 
vs = v - Rco for pure rolling 
Fig. 13.44
vp (= v + Rco) 
□—vs 
For anti-clockwise rotation, 
vs = v + Rco for pure rolling 
Fig. 13.45
§ 13.10 Concept of Rolling
Definition : Take a uniform disc of radius R, (say;). Place 
it vertically on a smooth horizontal surface. Spin the disc and 
simultaneously push it. In consequence, the disc will rotate with 
an angular velocity co and its centre of mass C moves with a 
velocity v. Using relative motion method, the velocity of lowest 
point P of the disc can be given as :
• + °pyJ-= aP,
at any angular position 0. Then substitute co and a
In absolute motion method, express the positions (xP and yP ) of the point P, say; (for which 
you are going to calculate the acceleration), in terms of increasing 0- (so that we can write 
dB 
dt h
Once we have
P
iniifinm
S
The disc undergoes combined motion 
on a smooth horizontal surface
Fig. 13.43where vPC = -Rd>i and vc = vi
Then, we have vP=(y- Ra)i
The above expression tells us that for different v and co , the point P 
will move with different speeds. For suitable velocities v and co, let us 
assume that P moves with a velocity which is equal to the velocity of the 
horizontal surface S; then, ~vP = ~vs . It means, relative velocity between P C 
and S is zero.
Then you can say that, “P does not slide on the surface S ”. In other 
words, the body is said to be “rolling without sliding on the horizontal surface”.
For this purpose, we have
where vP = (v - Pco)/"
Then,
When we assume anti-clockwise rotation, reverse the sign of co to t
G.R.B. Understanding Physics MECHANICS (Part-B)
Now substitute the obtained values of a and co.
xP = /(0)
V,,= /,(9)^ 
dt
tad tn _ du' 
~de ~ ~dt\
} —> —>
of acceleration components. Finally adding aP and aP> , find aP
yP = (0),
,,zen de 
vPy = I (9) “ • at
159Rotational Kinematics
However, in general, vectorially we can write,
vs = vP,
1.
Substituting
we have v/>c = -vc cx rPC
VC ~ rPC x ®PC ’
For instance, if yc = yj
rPC
p.,
This gives;
where a>PC = co (angular velocity of rolling body) and rPC = position vector of P relative 
to C.
Hence, for pure rolling on a fixed surface
VPC = “ x rPC
Summarizing the above discussions we note the following points :
VC = rpc x w 
/ and © = -ak.
we have v = 7?co
We can write, v = 7? co as the condition of pure rolling if 
the following conditions are met :
(a) The horizontal surface is fixed
(b) The CM moves due right.
(c) The disc spins clockwise.
4. Therefore, it is always advisable to remember the general 
condition for pure rolling which is given as :
vP = ’
where y*p - vc ; = co x
vc
vPC = *c for pure rolling
In pure rolling, there is no relative sliding between the points of contact. If the point “P ” 
of the rolling body touches the surface “S ” on which it rolls, the relative velocity between
P and S is zero for pure rolling; = 0.
2. If the surface is stationary, vs = 0. Then, we have vP = 0 for pure rolling.
vp = vpc+ = 0*
160
Sol.
This gives;
Since,
substituting
Cv Xand
we have
Then using the formula
where
Ans.we can find
Student Taskf
v 3v 2v
Ans. (i)--;(ii) — P *77^
2v
> A uniform disc ofradius R rolls perfectly over two horizontal planks 
A and B moving with velocities v and -2v respectively. Find the :
(i) velocity of CM of the disc.
(ii) angular velocity of the disc.
—*>VPC 
------- ►v
Fig. 13.48
Fig. 13.47
Vq = -V/
vPC = 3vi Q
where vPC = &PCx rPC
, 3v
Substituting u)PC = co# (assumed), vc = -vi and vq = 2vz, obtain co = —.R
If you obtain a negative co, just alter your assumption, it means that the actual direction of 0 
(or sense of rotation) is opposite to the assumed direction.
vp = 2 v
Vp = VPC 4* Vq
vP = 2vi
vpc — CO pc X rpc ’
= ~RJ.
3v •
co pc (— co) — k
R
’cX R /ViV, 
hnimmn/nmi
Fig. 13.49
—> —> —► —>
You can directly use the formula vP(= vPC + vc) = vq ,
GR.B. Understanding Physics MECHANICS (Part-B)
Let us execute the above ideas in the following examples.
Ex. 12. A uniform disc is spun with an angular velocity co and simultaneously 
projected with a linear velocity v towards left on a plank, while, the 
plank moves towards right with a constant velocity 2v. If the disc rolls
■ without sliding on the plank just after its spinning, find the magnitude 
and direction of co .
For pure rolling of the disc on the plank, vP = vq,
where \>q = 2vi
o
Fig. 13.50
Sol.
y
c 
o X
/
I
_R cos0 
yc (= y)
(x. y) 
pl
r
i
i 
i 
"xr
Rotational Kinematics I6l
By this time you would have understood the basic meaning of rolling. Let us now analyse the 
rolling motion of an object by finding the nature of motion of any point inside a rolling body. In this 
section, I will give you the kinematical ideas of finding velocities, accelerations of any point in a rolling 
body.
Finding velocity and acceleration of a point of a rolling body : In earlier sections we leamt 
two methods, i.e., ground frame (absolute motion) and relative frame (relative motion), to find the 
velocities and acceleration of a point of a rigid body in its combined motion. Little ago I explained that 
rolling is not a special type of motion, it is just a special case of combined translation and rotation, that 
is, combined motion without relative sliding. It means, we are not going to apply any new technique to 
find the velocity and acceleration of a point in a rolling body; it is just a reapplication of theprevious two 
methods (absolute motion method and relative motion method) executed in the following examples.
Ex. 13. A uniform disc of radius R rolls without sliding on a horizontal
surface. The velocity of centre of mass of the disc is v. At 
t = 0, the point P of the edge of the disc is situated at the 
origin O. At time t, find the :
(a) condition of rolling.
(Z>) coordinates of P as the function of time.
(c) nature of path traced by P.
(d) velocity of P.
(e) distance covered by the point P relative to ground (origin O).
(a) As the disc rolls without sliding, the centre of mass moves towards right moving a distance 
xc = vt during a time t. Simultaneously the peripheral point P revolves in circular path 
about C describing an angle Q.
Hence, the distance covered by P relative to C is given as 
the length of the arc PP' which is equal to RQ. Since, 
there is no relative sliding between the curved surface of 
the disc and horizontal surface, the lengths of the rolled 
portion of each surface must be equal.
You can see that during a time t, the length of the horizontal 
surface over which the disc rolls is OP' and the length 
of the portion (arc) of the disc rolled on the horizontal 
surface is PP'. For pure rolling, we have 
OP' = arc PP',
where OP' = xc (distance moved by the CM) = vt
and the length of arc PP' = RQ (because the point P describes an angle 0 relative to C)
i
 = v
we have
| vP |= y/[v(l - cos 0)]2 + [vsin 0]2This gives;
C 
el
Alternate method :
Using relative motion method, we have
I
I 
I
x^p = V
The magnitude of vp can be given as
; e
*
where vpc = Rat and v( 
Since,
, vt1 - cos —
R
n ■ 9
= 2vsin —
2
"=0 
R
T . Vt ~ 
i + sin —jR J
(c) Substituting xc = RG, the coordinates can be given as the function of 0, 
x = /?(0 - sin 0) and y = 2?(1 - cos 0) 
This tells us that the path traced by the point P is a “cycloid”.
_ . vi x = vt - 2? sin —
R
D(. vty = /? 1 - cos — 
k R
Vp - vPC+ vc, 
’c=v
G.R.B. Understanding Physics MECHANICS (Part-B)
In fact, we got this by using ground frame (absolute motion) method.
(b) The coordinates of P can be given as :
x = xc - R sin 0 and y(= yc) = R - R cos 0
vt0 = —, xr = vt 
R c
-Pl
Fig. 13.52
-» dx - dy «(d) The velocity of P is given as : vP = — i + — j,
vP = (v - v cos 0)/ + v sin 0/
] vc = v
I
I
I
0
2
vt
R
vt
R
163
Substituting
Dwe have,
Ans.
Student Taskf
we have 4" VPq — 0
It means that if vP = 0, the sum of ?PC and v^ will be zero.
§ 13.11 Instantaneous Axis of Rotation
Apart from a pivoted rigid body such as a rotating fan, doors, etc. 
generally we do not find an axis (axle) fitted with a rotating rigid body, 
such as a spinning ball in air, spinning planets in space, etc. We call this 
motion as “general motion”. Then we should not expect a permanent 
stationary point inside the rigid bodies. However, we may expect a point, P 
(say;), inside the rigid bodies which should be stationary at least for a 
moment.
Substituting vp = 0 for the stationary point P, in the formula,
vp = vpc+ vc In the above example, find the :
(/) ratio of average speed and the magnitude of average velocity of the point P over a time 
period of rotation of the disc.
(n) acceleration of the lowest point of the disc relative to (a) ground (Z>) centre of the disc.
4 v2 a
Ans.(i)- (ii) (a) 0; (b) — ?
After evaluating the integral, we have
D = 8/?
The point P Is momentarily 
at rest obeying the relation 
Vpc = -Vc 
Fig. 13.53
Rotational Kinematics
You can also obtain the above expression by using the formula
VP = yvPC + VC + ^VPCVC cos •J* ’
6 = —
R
> g
where I v>| = 2vsin-
where = 180° - 0
(e) The distance covered by the point during one period of rotation, that is, T = can be 
found by the kinematical formula,
o - /Ji I dl.
164
Sol.
G.R.B. Understanding Physics MECHANICS (Part-B)
This tells us that we have to choose a point P such that it moves with respect to the centre of 
mass of the rigid body with a velocity vPC which is equal in magnitude and oppositely directed to the 
velocity vc of centre of mass of the rigid body, at the given instant.
In other words,
////////////////^^
P
Any point seem to move 
perpendicular to the line 
joining It and the lowest point 
P, having a speed v > ra>
Fig. 13.54
For the instantaneous stationary point P of a rigid body, we have vPC = -vc . Hence, the point 
P moves with a velocity relative to centre of mass which is equal and opposite to the velocity of 
CM of the rigid body.
That is the necessary condition to have a stationary point (point of zero velocity) at an instant. 
Since, any point moves relative to the others with same angular velocity at a given instant, we can say 
that all points of the rigid body move relative to the stationary point P with same angular velocity. In 
other words, at the given instant all points of the rigid body have same angular velocity and angular 
acceleration about the “instantaneous stationary point”. Hence, the rigid body is said to be rotating 
perfectly about the instantaneous stationary point (at that instant), that is, what we call “instantaneous 
axis of rotation” (IAR).
Instantaneous axis of rotation has nothing to do with the centre of mass. It strictly depends on 
the linear and angular velocity of the body. Sometimes instantaneous axis of rotation lies inside the body, 
sometimes it lies outside the rotating body. When IAR lies inside the rigid body, we can easily see it. If 
it lies outside the body, how can you express and experience it? Well. For this, we need to imagine an 
extended rigid body. In other words we can say that we will have to mentally extend the given rigid body 
so as to get a point of zero velocity, that is, “stationary point at that instant” inside the fake (imaginary) 
area, but lying outside the original (or actual) area of the rigid body. Then, we can define the IAR as an 
instantaneously stationary point of an actual (or mentally extended) rigid body rotating with a combined 
motion.
The IAR by definition is instantaneous point of zero velocity. This means that the IAR changes 
from time to time. Then, does it mean that IAR moves? We will answer it in the next section, where we 
will try to locate the IAR. Before that, let us discuss the position of IAR in a rolling body in the following 
example.
Ex. 14. Locate IAR for a body rolling on a fixed horizontal surface. How do you observe IAR, explain 
with physical interpretation.
For a body rolling without sliding on a fixed horizontal surface, always the velocity of the 
lowest “points” of the body does not move as per the definition 
of pure rolling. Since, we define the stationary (motionless) 
point inside (or outside) the extended rigid body as IAR, we 
can say that at any instant the “lowest point” (point of contact) 
of the rolling body with the horizontal surface is the IAR. This 
means, the body rotates perfectly about the point P at a given 
instant In consequence, velocity of any point Q, that is, 
must be perpendicular to PQ. The points further away from P 
move more rapidly and the points nearer to P move less rapidly 
leaving the point P motionless. That is why, when we take a 
snap shot by a camera, we can see a blurred image of all farther 
points leaving a clear mark at P.
Rotational Kinematics 165
Student Taskf
From the above discussions we learn that;
> In the foregoing example, is there any acceleration of the lowest point P of the rolling body 
relative to its centre of mass?
This signifies that the speed of the instantaneous axis of rotation is zero. Hence, as the body 
rolls, instantaneous axis passes throughthe lowest point of the body at any instant. However, 
IAR passes through different points of the body at different instants. Since, the lowest points 
do not slide (move) and the peripheral points of the rolling body move, you should not think that 
IAR moves (rather it changes from one stationary point to the other stationary point of the 
rolling body).
The IAR of the body is the point P 
of the given (shaded) body or extended body 
which remains at rest at that instant 
Fig. 13.56
Finding instantaneous axis of rotation
When v and 
it has magnitude equal to the product of the distance PQ between the Fjg. 13.55 
point and the axis and the angular speed cd of the rigid body;
Vg = PQ • co
In general, we can write v = rco,
where v = vq, r = PQ and co = instantaneous angular speed of the rigid body.
We should note that the points of the rigid body do not move in circular paths about IAR. 
Then differentiating both sides of the equation v = ra> with time, can you write 
oq = r a ?
v2
Ans. Yes, — upward, where v = vc and R = PC
166
A v
Sol.
VA=V
we have If the diagram of the foregoing example is modified as shown in Fig. 13.59 
find the IAR.
v r = — >
co
VPA
Fig. 13.58
, vwhere co = —
21
Then, we have r = 21
It means, the IAR is situated at a downward distance 21 from point A and hence lies outside the
real rod.
.(0 y
r
I
I
U)AP = angular velocity of A relative to P = angular velocity of the rigid body = 
= v4 = v because vP = 0, we have
VP = vpa + VA ’ 
= 0,
where vPA = -rcof and vA = v
This gives;
v
r = — 
co
Which gives us the position of IAR relative to the given point/.
Ex. 15. Find the instantaneous axis of rotation of a rod of length I when its end A 
moves with a velocity vi and the rod rotates with an angular velocity
co =---- k •
21
Let us choose the point P as IAR in the extended rod.
Substituting vP = 0 in the expression,
G.R.B. Understanding Physics MECHANICS (Part-B) 
co. Substituting
167
,va
vA
--,P
>P ! B
B
I0
Sol.
X
.VA
Vp = 0,
we have ®bp = ®ap(= “).
because = a)Bp = co^, as described earlier.
The above analysis gives us the idea that,
The point P of intersection of perpendiculars 
drawn at A and B to their lines of motion
(vA and vB) gives IAR
Fig. 13.60
The IAR P Iles inside the given body 
Fig. 13.61 (b)
Y 
PI _____ P—
^//Zw/Zvatw/TTT, W/lrr/U 
A
Fig. 13.63
al-
vB
Rotational Kinematics
the points A and B move perpendicular to the lines AP and 
BP respectively about P. Hence, AP and BP are 
perpendicular to the velocities vx and vB respectively. 
This means that ~vA 1 AP and vB 1 BP. Using the 
kinematics of relative motion, we have vA = (AP)caAp 
and vB = (BP)®bp. Then the angular velocity of the rigid 
body is given as :
The instantaneous axis of rotation is the point of intersection P of the normals AX and A Y drawn 
from A and B to their lines of motion. The IAR passes through a point perpendicular to the plane 
of rotation. P may lie inside the real or imaginary (extended) portion of the given body. The 
points of the real or extended rigid body cannot move in circular path about an IAR because 
IAR changes from time to time.
’ihnffrfHmfffrffrn/frifrn/rrr
Fig. 13.62
/ A'''
( vB''
The IAR P lies outside the given (real) 
body and inside the extended 
(hypothetical) body
Fig. 13.61 (a)
Ex. 16. J rod of length I leaning against a vertical wall is pulled at its lowest point 
A with a constant velocity v (say). In consequence, the rod rotates in the 
vertical plane. Locate the position of IAR of the rod when it makes an 
angle 0 with vertical and find the angular velocity of the rod.
From the geometry of the arrangement we observe that the ends A and B 
slide on the horizontal and vertical surfaces with velocities vA and vB 
respectively, where vA = v (given) and vB is assumed. As per the 
procedure, let us drop perpendiculars AX and 5/ from A and B respectively 
onto their lines of motion (vA and v#). We observe that the perpendiculars 
intersect at P. Then, P is the IAR. Ans.
Since,
AP BP
G.R.B. Understanding Physics MECHANICS (Part-B)168
where ®ap
PA - I cos 0
VA =
we have Ans.
Student Task
Since,
we have
This gives;
Substituting 
and
= 22.
PA
X x -1
> In the foregoing example, as the rod changes its angular position, will the perpendiculars BX 
and AY meet at same point? If no, does it mean that IAR moves which is contrary to its 
definition?
Ans. 1AR will be different at different instants, but velocity of each IAR is zero.
/ A\
Hii ra
\ip/
Since, A and B move perpendicular to 
the lines AP and BP joining the particles
v co =-------
I cos 6
Two given points move along parallel lines
When vA is parallel to : When the two given 
points A and B of a rigid body move in same direction with 
velocities vA and vB , at any instant, the perpendiculars 
dropped from these points with their lines of motion will meet 
at infinity. Does it mean that the IAR is situated at infinity? Let 
me explain it as following.
As two particles move unidirectionally at the given 
instant, is parallel to . Let us drop the perpendiculars and the IAR at the 0,ven instant>we have 
from A and B to and v^ respectively. We find that the “ " TT = x-/
perpendiculars pass through the line AB joining the points A F,fl-13 64
and B. This tells us that the IAR lies “somewhere” on the line AB. Let us choose the point P in the 
extended rigid body as our IAR, on the line AB at a distance of x from A, (say;). Since, P is the 
instantaneous centre of circular paths of the points A and B, we have
VA □ VBwAp = — and wBP = —— 
X X - I
I
X = ----------
l-^
VA
169
vA
Since, ®AP
VB
after equating and coflp.
we have
®AP
and
•—(i)
A' -^vAA' -» 
-tva
where ®AP =■
[ I 
/ x / 
vir- 
y_\^ -.11-----VB\% B /
-> VB V
“BP
^A=_X_ 
vB /-X
Fig. 13.67(H)
2!a=_x_ 
vB x-/
Fig. 13.67(1)
This gives; - — = and ultimately we will get correct answer only after equaling the 
x I - x
magnitudes alongwith the directions of angular velocities.
The above two cases may be considered in 
geometrical point of view as following :
When we have two points A and Bm/s2.
a* F - ?_ _ , ...., , f3
Em, a,
Em,
After being familiar with the expression FneI = Em, a, , let us now divide both sides by total 
mass of the system, that is, Zm,. This gives;
-> -> 
Fnet = ai 
'Lml Em,
The RHS term is the ratio of “total force” and “total mass” which gives us an “acceleration” a , 
say. Then, we can write ;
Om2
Fig. 11.4
^et = (Sw,)a
Here, we have three terms Fnet, Em,- and a . Let us look at each term. First term F\et signifies 
the net(total) force acting on the system, second term Em,- gives us the total mass of the system, but 
what about the third term a ? Well; it is an acceleration, but acceleration at which point? Furthermore, 
—>
you can also ask, where does the force Fnet act? Let us try to answer these questions.
—>
As the force Fnet is the sum of the external forces acting at different point masses of the 
—>
system, it makes no sense to tell that Fnet acts on any particle of'the system. Then, it is essential to 
imagine a “special point” at which the net force acts so as to move it with an acceleration a which is 
equal to net force divided by total mass of the system. This hypothetical point moves with an accelera- 
tion a as ifit carries the entire mass (Em, ) of the system experiencing the net force Fnet acting on the 
system. Hence, it is natural to imagine that the entire mass of the system is centered at that “special 
point” which can be logically called “centre of mass”. As this is just a “hypothetical point” by definition, 
it is not essential that centre of mass should carry some mass.
Ex. 2. Find the acceleration of centre of mass of the system of two particles of masses
mf and m2 connected by a light inextensible string that passes over a smooth 
fixed pulley.
p -► -►
—— = a , where a = 
Em,
This can also be given as :
System of Particles 5
yI
0 Q
X
—>
F = -(t»ig + m2g - 2T)j, mig
where
,.(i)
•••(ii)°CM =
aCM
t7j = -a2
Then substituting and a2 in the expression
aCM
Ans.aCM
Ans. Zero
Thus, when we say, momentum and kinetic energy of centre of mass, it may be misleading in 
strict sense. If you want to use these terms, momentum of centre of mass (CM), say, it may roughly
Here the system is (w, + m2).
The net force acting on the system is,
-> m, - m7 
a^ = —-------
mt + m2
mt - m2 
mt + m2
I
I
I
I
I
I
’ m2g
Fig. 11.5
i
i
Student Task^
> If we take the pulley into account, assuming mass of the pulley as M,find the acceleration of 
pulley + particle system.
Method-2 :
Applying Newton’s second law on each particle we find their accelerations as,
mi + m2
Ans.--------(*, - --------
(/W] + mf)(M + W| + mf)
> If we take earth + particles as the system, how does the answer change?
m, a. + m-> a7
= ——---- —-, we have
/K| + m2
2-»
g
Sol. Method-1 :
T = as derived earlier
Wj + m2 
. 7t (w, - rnf?
This gives F =------!
mi + m2 ' mt + m2
Then the acceleration of CM of the system is given as :
F 
mi + m2
Substituting F from eq. (i) in eq. (ii), we have
.................... -
g
6.
and the other is,
Sol.
Fig. 11.6
T
y
2
r
’mg
Fig. 11.7
IT cos0 
I 
I 
9J 
Sq
a(-> F 
aCM - Z— Lmt
Basically we have two equations, one is,
F 
aCM= — 
Lrrij
G.R.B. Understanding Physics MECHANICS (Part-B)
I 
mean the product of mass of the system and velocity of CM. Similarly KE of CM roughly means —
2(mass of the system) x (velocity of CM) . In this way, using the powerful concept of centre of mass, 
we will derive the expressions for KE, momentum and other parameters of the system of particles.
§11.3 Characteristics of Centre of Mass
(i) Acceleration of CM : After realising the requirement of the concept of centre of mass, let 
us discuss all the factors which define the centre of mass. As you know, in kinematics, a 
point is characterised by its position, velocity and acceleration. Since, we find the accelera­
tion from force equation, then velocity from acceleration and position from velocity, let us 
define the centre of mass by using the idea of force as following:
Centre of mass is a point which moves with an acceleration equal to the net force acting on 
the system divided by the mass of the system. Symbolically
The forces acting on the chain are tension 7’\ and weight mg I.
Let us write the force equation of the CM which moves in a horizontal circle 
of radius r. Let the CM has acceleration a.
£Fr = mar, where Fr = T sin 0
and ar = rco2. This gives T sinO
TsinQ = mr®2 •••(*) '' —tyrjC
Y.Fy = may. Since, the CM moves in a horizontal circle,
ay = 0
__ —>
-> Dn,a.-
^=-7—
Please note that both equations give same thing, but the approach may be different. This 
will be clear in the following example.
Ex. 3. A flexible chain of mass m and radius R is rotating about a vertical axis 
passing through the point P by an inextensible light string. Find the radius 
of the circular path traced by the CM of the chain if the string makes a 
constant angle 0 with vertical.
7
...(ii)
r = Ans.
Student Task
> In the foregoing example, find the tension in the string. Ans. mg sec 0
Substituting
vc =
v
2v2v v2v v
O2m
(iii)(')
-» 
Avc =
System of Particles
Substituting
(ii) Velocity of CM : The change in velocity of centre of mass from t = r, to t = t2 can be 
given as:
(ii)
Fig. 11.8
mo- 2m-o
mo
2m
O
mo-
Avc =
Then, we have
TLFy = T cos 0 - mg and ay = 0, we have 
T cos 0 = mg
Eliminating T from eqs. (i) and (ii), we have
gtan0
f tl
j, a, dt = A Vj
All forces T and mg are imagined to act at the CM. If you take the radius of the ring or radius 
of the circular path traced by any other point of the chain (such as Q), you will get wrong 
answer. We have nothing to do with other points except centre of mass. That means, in F = ma, 
a is the acceleration of CM and F is the net force acting on the body (CM).
Zm, Vj
"Lm,
Ex. 4. Find the velocity of centre of mass of the system of two moving particles of mass m and 2m, as 
shown in the figures (/), (n) and (iii).
Zm,
Since, A vc = vc- uc and Zm;(A v() = Zm, vz- Zm, u,
Comparing both sides we have,
Avc = f‘2acdt
—>
-» _ Em, at_ jiave
c Zmz
Zm, r‘2 a, dt 
------- !------- , where.
Zm,
8- G.R.B. Understanding Physics MECHANICS (Part-B)
Referring to the equation vcSol.
we have
(>) vc =
(ii) vc =
(iii)
Student Task
This gives 5c =
sc =
y
/
m x
A —21-------
Fig. 11.10
—>
Sm, Vj 
"Lm,
4v « 
= — i
3
—>
Sw, Vj 
Y-nij
m(2vi) + 2m(yi) 
m + 2m
B 
2m
sc = fvcdt, where vc
Ew, £'2 v,dt
Xmt
2v - -
= yG + j)
c 
3m?
o 1 
‘+T3J
Zm, Sj 
Zmj
Ex.(say;), moving in 
parallel lines at a given instant, we just join the points 
(A and B) and extend the line AB. Then join the tips 
A' and B' of the velocity vectors vA and vB 
respectively and extend the line joining the tips of vA 
and vB. Let the extended lines meet at P. In this 
process we obtain two similar triangles APA' and 
BPB', as shown in the Fig. 13.67 (i) and 13.67 (ii). Using the properties of triangles, we have
AA' x 
BB' x-l
/A\ An
\ p /
Rotational Kinematics
When v\ is antiparallel to : If the points move in opposite 
directions, let us choose the IAR at P at a distance x from A (say;),
_ VA . „ VB- — and Wap = ----- »x I - x
aAP ~ ®BP ’
- — k and aBP = k 
x I - X
or ^-k
l — x
Here, we cannot equate the angular velocities by writing. The angular velocities In the previous example if the plane 1 moves with a velocity vi, find the :
(a) location of I AR.
(b) angular velocity1 of the cylinder.
Ex. 17. A cylinder is rolling without sliding over two horizontal planks 
{surfaces} 1 and 2. If the velocities of the surfaces A and B are -vi 
and 2vi respectively, find the :
(a) position of I AR.
{b} angular velocity of the cylinder.
(a) Since, there is no relative sliding between the cylinder and the 
planks 1 and 2, the points A and B of the disc will move with 
velocities equal to the velocities of the respective surfaces.
= —vi and vB = 2vi respectively. Joining A and B and A' 
and B' we find the point P as IAR. Then we have the similar 
triangles PAA' and pbb' ■ Using the propert’es of similar 
triangles we have
3v « co = — )
2R /
> BB' BPco =---- = —
AA' AP
where AA' = vA = v, BB' = vB = 2v, AP = 2R - x 
2v x 
v 2R- x 
AR x = — 
3 
ARHence, IAR is located at a distance of — below the top of the disc. 
co = ^-, 
AB
A
VA — V
Fig. 13.69
and BP = x
• ~• y X —yIn case (i) {vA is parallel to vB} and---- =------- . In case (ii) (yA is antiparallel to vB),
BB' I - x 
AA' J |
BB' ' I I
Using eqs. (i) and (ii), we can find x.
v~ , ,,1
Fig. 13.68
B B' 
*vb
171
vP
-A
I
CD
P =
-QoSol.
Pp =
where
Since, aPCn
substituting
we have
0 (where pc = oo).
Substituting
->
VP
Rotational Kinematics
(c) linear velocity of centre of the cylinder.
Ans. (a) 2R below the lowest position of the cylinder
v . 3v
2R ) (C) T
luiuiinimiHiiluiiiiiniu
Fig. 13.71
(ap)n^
3j=P
By this time you leamt the procedures to find the velocity and acceleration of a point of a rigid 
body in combined motion. As each particle moves in a curve we can find the radius of curvature of the 
path traced by the particles (points) of the rigid body.
§ 13.12 Radius of Curvature of the Path Traced by a Point 
of Rigid Body
Suppose you want to find the radius of curvature “ p ” of the path 
traced by the point P of the rigid body. For this, we recast the formula,
vp
(ap)n
where (aP)n is the component of net (total) acceleration of the
point P perpendicular to its velocity vp . We can find vP by using 
following equations.
Then, we have (^c)n
The radius of curvature of the 
path (dotted line) traced by 
the point P is given by : 
vp2 
px(U
Fig. 13.70
VP ~ VPC+ VC’ 
where vPC = ox^,c. aP can be found as discussed earlier.
Ex. 18. Find the radius ofcurvature of the path traced by a peripheral point ofa perfectly 
rolling disc of radius R, at its highest position P.
The radius of curvature of the p .th traced by P is,
(Op).’
aPn -I aPC„+aC„ I
_ VPC _ I VP~ VC |2 
rPC rPC
= 2vi, vc = vi for pure rolling and rpc = R , 
v2
(aPC)n =
Since, the point “C” moves in a straight line, the radius of curvature of a straight line is infinite. 
.1.
Pc
(b) 2R ) (c) 2
v2 v2
Vp — 2Vjtfp — Qpc + ^c„ — 0
172
in the formula Pp =
we have Ans.
Student Task^
Ans. VL?
coCP
“CP
coCp
Then, we have
During the time dt, the horizontal projection of PC rotates through d§ (but not d$) about the 
• —• axis yy'. Since, d§ > de and direction of d§ is along positive y-axis whereas de is directed
This gives;
where vPC = v and PC = I (say;),
where vPcL = vPC 
because
G.R.B. Understanding Physics MECHANICS (Part-B) 
v2P
(ap\ ’ 
Pp = 4R
rCP
Student should not be confused in thinking that p = R as the point P moves in circular path 
relative to C. However, pPC = R but pp = 4R. Some students may be tempted to write 
Pp = 2R arguing that the distance of P from O (instantaneous axis of rotation) is equal to 2R. 
This is wrong because P does not move in circular path about the IAR.
§ 13.13 Rotation About a Point
Angular velocity about a point: When you weld one point of a 
rigid body with a pin, it will be free to rotate about an axis passing through 
the pin P Since, the distance PC between the pin (point of suspension if 
you have hinged the body at the pin) and the centre of mass C of the 
body, is constant, C can trace any curve (as shown as paths 1,2, 3 and 4) 
maintaining a constant distance from P. In this process we can observe 
that the angular velocity of C relative to P can be given as :
VPCL
PC '
v 
wcp - -
VP(\ - 0
_ VPC 
PC ’
> In the foregoing example, find the radius of curvature of the path traced by peripheral point 
at Q.
C can move in many paths 
(1, 2, 3 and 4 etc.) relative to P 
Fig. 13.72
I
I / 1 z
1/ 
p,X
: v, . 
1
® CP 4 **CP
Fig. 13.73
Angular velocity about an axis : As the body rotates about the point P, the line PC move 
through an angle d£> (say;), during time dt. Then,
dB
173
C
Substituting d§ = and dQ =----
we have
Since,
P
coCp is a component of coc/x
P' >C
Sol. 8
Then,
co
Kinematics:
Ex. 19. In a conical pendulum find the angular velocity of the bob B relative to P if 
the string makes an angle Q with vertical.
Angular velocity of any point about (relative to) any reference stationary point must not be 
confused with the angular velocity of the point about (relative to) any axis passing through the 
reference point.
However, we will study in the next chapter that, the angular momentum L or torque t about 
an axis may be a component of L and r about a point on the axis.
we have
Hence, we conclude that;
P 
 | > |d01 we call 
“d0" a component of d$ 
Fig. 13.74
wcprx
1
COSp ’ 
®CP = Cp' COS P
V
CO = - 
/ ’
where v can be calculated as follows :
Referring FBD, the forces acting on the bob are T and mg
Force equation: T sin 9 = mar
T cos 0 = mg
v2
ar = —.r
Let ®BP = co and vBP(= vB~) = v
>c!
I / 
w I /
y/d6
XWCP zf__.
p\
pV
A
I \
I __ \
,c;
Fig. 13.76
Rotational Kinematics
perpendicular to PC, we call dQ a component of dfy . In other words, 
dQ . _ d(f) dQ ->—is a component of —. If you write —= a>CP and -j- = coCp dQ
CC___ 11
° "r P'C
■7/0^
Since, - is directed alongy-axis, you can call angular velocity
of PC along y-axis or angular velocity of C relative to p'. “Cp is a component of ®cp
Recapitulating the above facts, Fi9'13,75
CC
PC ’
®cp' _ PC
(i>Cp ~ P'C
PC
P'C
I
1 XI /
P \
IV\\
I—
Fig. 13.77
174
T
T sink ar
v =
sin 0we have Ans.CD =
Student Task^
It is very important to note that;
w'
but not
P co' = CO +Hence, CD = CD + CO]
This means, co is a component of cd' .
CM
iar»;
z
> In the foregoing example, we can see that co varies in direction as it rotates forminga cone 
of revolution. Then, what is the angular velocity of the bob relative to (about) y-axis? Is it 
the component of co along y-axis?
2. It is always advisable to take the CM (or IAR) as 
the reference point because the CM moves in a 
most simplified way as there is no net effect of 
internal forces at CM and the IAR moves with 
zero velocity.
Simplest path of CM and stationary IAR 
Fig. 13.81
§ 13.14 Summary
1. A rigid body does not require any axis to 
rotate in an unconstrained rotation. In this 
case you can mention the angular velocity 
in any way you like keeping its sense 
constant as shown in the figures (i), (ii) 
and(iii).
cd = co'sin 6 
cd' = cd sin 0 •
mg
Fig. 13.78
where r = I sin 0
Using the force and kinematical equations, we have
' S
I COS0
G.R.B. Understanding Physics MECHANICS (Part-B)
|T cos0 
I
0i
•©.■©.O>
In all the ways (i), (II) and (iii), angular 
velocity a of a rigid body can be given, keeping 
the direction (sense of rotation) constant 
Fig. 13.80
CD
Ans. —No sin 0
g/sin2 0
COS0
pre
re\
i \
i \
i x
p1*--------- _\B
Fig. 13.79
v
Then substituting v in the expression cd = — ,
175
C
t = ‘i t = t2
IAR has zero velocity (vp = vQ = 0) 
Fig. 13.83
03AB = “body * “AC 
Fig. 13.82
Rotational Kinematics
3. Angular velocity between any two points of a rigid body at any 
instant is constant which is defined as the angular velocity of the * 
rigid body, whereas at any instant the angular velocities of different 
points of the rigid body relative to any point outside the rigid body 
are different.
4. The velocity of IAR is zero. Hence, we can use it to 
find the velocity of any point of a rotating rigid body. 
However, IAR accelerates and hence cannot be 
usually used as an inertial reference frame to find 
the acceleration of a point of a rigid body.
5. The points of a rigid body cannot move in circularlAR change their positions from P to Q 
path about IAR because by nature IAR cannot be bl^ *ny instant>th® corresponding 
any particular point of the rigid body which is needed 
for centre of circular path. It changes with time.
6. The IAR is stationary as per its definition. However, as the body moves, the locus of all 
IAR may be a curve. This should not give the idea that IAR moves. It may be correct if 
you take many IAR at different instants of times. However, at any instant, IAR is stationary.
—> —> —> ~~►
7. (0)2 = ^2I» but V|2 = “v2b
8. (0|2=©j-W2 is vague, but i^2 = is valid.
9. When no net force acts on a rotating rigid body, except CM, all particles (points) must 
have acceleration.
10. The path of any point A of a rotating rigid body relative to any other point B of the rigid 
body is circular in relative sense for all times but the radius of curvature of the path traced 
by any point (relative to ground) may not be equal to the distance of separation between 
the points A and B.
176
11.
12.
15.
22.
23.
24.
13.
14.
In a rotating frame, we have two sets of accelerations; coriolis acceleration and centrifugal 
acceleration. Explain. Do these accelerations depend on the nature of motion of the particles? 
Can the spin and orbital angular velocity of earth change if we change the reference frame?
Can the angular velocity of the hands of a clock be changed upon the change in relative motion of 
an observer relative to the clock?
If we reverse the direction of rotation of a reference frame, what will happen to centrifugal, 
centripetal and coriolis accelerations?
A body rotates about instant axis of rotation; comment.
Instant axis of rotation does not move and does not accelerate; comment.
What is the difference between pure rotation and pure translation?
Can we call pure rotation “a fixed axis rotation”?
Can we call a body rotating perfectly (pure rotation) about the instantaneous axis of rotation?
Centrifugal and centripetal accelerations counterbalance each other producing zero acceleration; 
comment.
If a rigid body has zero angular acceleration but non-zero angular velocity, can you call it an 
equilibrium?
If a rigid body possesses a non-zero angular acceleration and zero linear acceleration of the CM 
of the rigid body, can you call it a body in equilibrium? Explain.
How does a particle (point) appear to move relative to other points of a rotating body in case of 
fixed axis rotation and unconstrained rotation?
G.R.B. Understanding Physics MECHANICS (Part-B)
Assignments
4.
5.
6.
7.
8.
9.
10.
16.
17.
18.
19.
20.
21.
1.
2.
3.
Discussion Type Questions
Can angular motion be attributed to a particle? Explain.
What is the meaning of angular motion of a rigid body?
What is the basic difference between angular velocity of a rigid body and angular velocity of a 
point of a rigid body relative to a fixed point?
Is angular velocity of a rigid body dependent on a reference frame?
What is relative angular velocity? Does it depend on the reference frame? Explain.
There is no relative motion between any two points of a rotating body, comment.
Why do small angles behave differently from large angles?
Angular velocity and angular acceleration are pseudo vectors. What does it mean?
How can you relate the angular velocities of a pair of gears coupled with each other?
In pure rolling a = Ra, where a = acceleration of CM of the rolling body and a = angular 
acceleration of the rolling body. Comment.
In pure rolling, vp =0 and ap = 0; comment.
25.
1.
(b)(a) VjV2
(d) none of these
2.
(b) vc(a) VB
(d) aB = 0
3.
4.
5.
y
o X
(d) aPo = 0 and fig = 0
26.
27.
28.
Two particles of masses mt and m2 are connected with two ends of a massless 
—rigid rod of length/. If the velocities of particles are V[ and v2 respectively, the 
velocity of centre of the rod is :
2v2 ,
(c) aP0 =---- cos 0sin0
h
Rotational Kinematics 177
The particles (or points) of a rotating rigid body move in circular paths about the I AR (instantaneous 
axis of rotation); comment.
If the points of a rigid body move in the curves, the body is said to be rotating. Explain.
Does a rigid body require an axis to rotate? Explain.
Can any point of a body roll' ;g on an inclined plane have a vertical velocity?
Multiple Choice Questions
+ m2v2
+ m2
(c)ii5.
2
The end A of a uniform rod is pulled with a constant velocity v. In 
consequence, the rod rotates in vertical plane. Then :
4
(c) aA = 0
The lowest point P of a rolling disc has acceleration :
(a) (b) Ra^
K
(c) zero (d) none of these
A rod of length I rotates in the vertical plane under the constraint of horizontal 
and vertical surfaces. The path of the point O on it is :
(a) a circle (b) an ellipse
(c) a hyperbola (d) spiral
A point P moves with a constant velocity v parallel to x-axis. Then :
/ s v -> , V2 3 a
a) Wp0 =-cos20 and —y =—cos 0 
po h dt2 h
2
(b) (Opo )radial * 0 ar*d ^centripetal ~
(b) W&4 = coBC
7.
B •—►v
(b)(a)
3v
P
8.
0)
9.
v---*-x
i
B10.
p11.
B3
a ' 
—-v
>Q 
2(b) 1 :2
(d) not defined
I
i
G.R.B. Understanding Physics MECHANICS (Part-B)
A disc has angular velocity co and the speed of the CM of the disc is v. Then :
= wk
P
ZZZZz^’/^W'ZZZ
178
6.
S' 
21
A
2 4
 (b) a = -Ra
(c) (ap)i =0 (d) aP=Q
A rod rotates about a perpendicular (vertical) axis in horizontal plane with an angular velocity co. 
If the axis passes through the centre of the rod and the axis is moved with a velocity v, the 
instantaneous axis of rotation of the rod is situated :
(a) in +y axis
(b) in -y axis
(c) in +x axis
(d) on the straight line passing through the rod
—* 2v •
In the Q.-9, assuming co = ——k, the ratio of speeds of A and B is :
(a) 1:1 (b) 2 : 1
(c)3:l (d) V3:l
Three rods 1, 2 and 3 are smoothly pinned at A and B. If the rods 1 and 2 
are pivoted at P and Q respectively, the ratio of angular speeds of the rods 
is:
(a) 1 : 1
(c) 2 : 1
A
a
(a) coB?)
(c) (i)BC = V »
(d) aBA =--k
K
The end A of a string wrapped over a disc is pulled with a constant velocity3vi. The centre of 
the disc moves with a constant velocity vi. Then : s'
2v 
CD = -
(a) ax =— sinO-^anda^ = ^-cos0'L
(b) ax = -sin0and vy =—cos0i
1.
2.
3.
4.
1.
S
(b)
(c) aR
Rotational Kinematics
12.
VPQ
179
A rod of length I rotates in vertical plane having angular velocity co and angular acceleration a as 
shown in the figure. At the given instant, let the rod make an angle 0 with horizontal. Then, at the 
given position:
I ■ n la.___
2"'" y 2........'
I ■ « .
2’' y 2 ’
(c) vx = ^sin0 7
Ho
180
2.
(p) secO
\C
(q) tanO
(r) sec3 0
(s) 1
a>AB = = co (say)
co can also be given as:
1.
I
2:1
where 9 = angle of inclination of line AB,
and co^a = angular velocity of A relative to B (or vice-versa)',
v r
de
Comprehensions
Passage-1 (Concept of rotation)
Rotation of rigid bodies is defined as the angular motion of any straight line drawn inside the rigid 
body. The rate at which the angular position of a straight line changes with time is called angular 
velocity of rigid bodies;
where vAB1 = component of vAB perpendicular to AB(= I).
The angular acceleration a can be given as :
da) dma = — = co— 
dt dQ
Using the above ideas answer the following questions.
Two identical rods 1 and 2 each of length I are smoothly pinned at P. 
The rod 1 is smoothly hinged at 0 and the free end of the other rod 
2 is pulled with constant velocity v.
What is the ratio of angular speeds of the rods 1 and 2 ?
(a) 1:1 (b)l:2
(c) 1:^2 (d)
VABX 
(0 = —— 
/
G.R.B. Understanding Physics MECHANICS (Part-B)
(d) Average speed of a peripheral particle (s) aPC
over one period of rotation of the disc
A rod moves in the vertical plane against horizontal and vertical surfaces. The lowest point P of 
the rod is pulled with a constant velocity v. At any angular position 0 of the rod, match the 
following:
(a) —
V
(b) -f
v
(O
V̂
(d)
V
j_______ \y—»v
zzzzzzzzzzzzzzzzz/zzzz
181
3.
= (PO)a'po,
Q
2
1.
(b) ZjCO| 4-Z2C£>2
(d) none of these
2.
(b) —/2’2
(c) ^Z,2©2 + l2a2
The acceleration of Q is :
(a) >/(/]+/2)2cd4+(/2a2+/2a2)
True Or False
Relative velocity between any two points is zero in a rotating body.
(V^2 + to2)(/| +/2)
2
vl®2 ^2® I
2
/] CD | + Z2CO2
~2
Passage-2 (Finding velocity and acceleration of a point in a rotating rigid body)
The velocity of a point P of a rotating body can be given as
vp = '>o+ vo J where “po = (PO)a
The acceleration of a point Pis given as aP = aP0+a0\ aP0 =at 
and aP0 = rco2.
Following the above theories, answer the following questions.
Two rods 1 and 2 are hinged at P smoothly. The rod 1 is pivoted smoothly 
at O. The angular velocities and angular accelerations of the rods are 
(co,,cd2) and (a|,a2) respectively when the rods are perpendicular to 
each other.
The velocity of Q is :
(a) +l2 )o>2
+ aPOr, where a POi
. a2 dp
4
(d) none of these
The relative velocity between the centres of the rods is :
7A2(X>2 + ^2(o2
2
Rotational Kinematics
2. The ratio of speeds of the points P and R of the rods 1 and 2 is : 
(a) 1 : 1 (b) 1 : 2 sin 0
(c) 2 : sin 9 (d) none of these
The ratio of angular accelerations of the rods 1 and 2 is : 
(a) 1 : 1 (b) 1 :2
(c) 2 : 1 (d) none of these
G.R.B. Understanding Physics MECHANICS (Part-B)182
3.
4.
5.
1.
4.
 J—3v
v
Problems
A disc spins with1.
The path of the peripheral point of a rolling disc on a fixed horizontal surface is 
The path of the mid-point C of the rod sliding along the horizontal and vertical surfaces in vertical 
plane is 
6.
7.
8.
2.
3.
( 2v
I R>)
Find the velocities and accelerations of the points 1,2, 3 and 4 of the 
disc.
2v
W= -R
A body rolls on two horizontal plates 1 and 2 moving with velocities 3v and - v respectively. The 
distance of instantaneous axis of rotation from the lowest plate is equal to
1
CD|2 - of any point P situated at a distance r = 2 cm from the 
axis of rotation of a wheel of radius r = IO cm moves with non-uniform circular motion obeying 
the relation v
16.
17.
P
18.where
19.
20.
I 
I 
ie
d^-R
K /U)
_______________________ V
zzzzzzJzzzzvzzzzzzzzzzzzzzzzzzzZww/
n 
h—-v I I
/ R \
zzzzz/zZZzzzzzzzzzzzzzzzzzzzzzzzzzzz/zzzz
2v
CO = —
G.R.B. Understanding Physics MECHANICS (Part-B)
A rod is resting on a step in a vertical plane. If the end P of the rod of length I 
is pulled with a velocity v, find the :
(i) angular speed of the rod.
(ii) angular acceleration of the rod .
(iii) linear velocity of the points Q and R.
(iv) accelerations of the points Q and R.
A rod pivoted smoothly at O is resting on a block of height h. If the block 
moves with a constant velocity v, find the :
(i) angular velocity of the rod.
(ii) angular acceleration of the rod.
(iii) magnitude of acceleration of the free end of the rod.
A disc of radius r is kept on a hemisphere of radius R. If the disc rolls 
without sliding on the hemisphere, find the :
(i) angular velocity of the centre C of the disc relative to O and P.
(ii) acceleration of C relative to the point O and P.
(iii) angular velocity of the disc relative to O and P.
A sphere of radius r is kept on a wedge having a hemispherical 
surface of radius of curvature R. The sphere rolls on the hemi­
spherical surface with a velocity vc(= 3v) when the wedge moves 
with a velocity v. Find the :
(i) velocities of the points M and Q of the sphere. zz/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz^x
(ii) accelerations of C relative to the (a) point P of the wedge (b) ground
In the Q.-l 5, if the acceleration of the wedge is a, find the (i) accelerations of M, C and Q relative 
to 0 and P.
A disc of radius r pivoted with a rod at C rolls on a cylindrical surface of 
radius of curvature R. The rod is pivoted at the point O. If the rod rotates 
with an angular velocity co , find the :
(i) velocity of the centre C of the disc.
(ii) angular velocity of the disc.
(iii) acceleration of C relative to O and P.
2 v A cylinder of radius r rotates with a constant angular velocity co = — T 
r '
v = velocity of the plank on which the cylinder is rolling. Find
the (i) vc (ii) vP (iii) vQ (iv) aPQ
In the Q.-l 8, if the plank moves with an acceleration a while it has 
velocity v, find the velocities and accelerations of P, Q and C.
A cylinder rolls on the planks A and B without relative sliding. If the planks move with velocities
'^777777777777777777
hie
11
0
///)//J//////7/}/7////////////J
R~IM
A r )
Rotational Kinematics
a
21.
22.
23.
24.
25.
26.
r 
bC
/ 
I 
V R
________ B_
( C-R-4Q
2 y \ >A
7777/7///////////////////////////
-2vi,v‘i respectively and the plank A has acceleration
a (=ai), find the :
(i) instantaneous axis of rotation of the cylinder,
(ii) vc and vq.
(iii) ac and oq.
Two rods of lengths b and c are pinned at Q. The rod OQ is pivoted
at 0. If the end P of the rod is moved with velocity v, find the :
(i) angular velocities of the rods.
(ii) velocity of the point Q.
A rod PQ having length I is rotating in a vertical plane. Find the :
(i) instantaneous axis of rotation.
(ii) velocity of Q.
(iii) angular velocity of the rod, if the end P of the rod is moved
with constant velocity v.
A rod of length I rotates in a vertical plane against the horizontal and vertical 
surfaces, with an angular velocity to and angular acceleration a . Find the :
(i) velocities of A and B.
(ii) accelerations of A and B.
(iii) velocity and acceleration of the mid-point C of the rod.
A small ring Q which is free to slide along with a bigger ring of radius R, is 
moved with a constant velocity v. Fmd the :
(i) angular velocity of the bigger ring.
(ii) linear velocity of the center C of the ring.
(iii) acceleration of the centre C of the ring, after a time t.
An insect P is sliding with a velocity vr relative to a rotating disc along the 
straight groove. If the disc rotates with an angular velocity co and angular 
acceleration a , find the :
(i) velocity of the insect.
(ii) acceleration of the insect.
A sphere of radius r rolls without sliding on the horizontal surface such that the centre of mass C 
of the sphere moves in a circular path by the connecting rod 
of length R. If the rod rotates about the vertical axis with an 
angular velocity co , find the:
(i) angular velocity of C relative to 0 and P.
(ii) angular velocity of the sphere.
(iii) angular acceleration of the sphere.
(iv) linear and angular accelerations of C relative to 0 and P.
G.R.B. Understanding Physics MECHANICS (Part-B)186
ANSWERS
4. (b) 5. (a,c,d) 6. (c) 7. (a,d) 8. (a,b,c)
4. (a)
2. (b) 3. (a)
2. (b) 3. (a)
3. (T) 4. (F) 5. (F) 6. (F) 7. (T)
2x/2/? 2. cycloidal 3. circular 4.1.
1.
3. (i) cycloidal
5.
7. (i) vc=-^-
9.
11.
R
2
1 + 11 
r>
(iii) a2t2
(ii) vP = 2vsin^
Za
(ii) o) = -^- 
' ' R + r
(ii) 1:6
at = (a + Ra)i - Rd)2j
a2 = (a - Ra>2)/ - Ra j 
a-} = {a- Ra)i + Rti>2j
= (a + Ru2 )i + Ra j, where co = at and a = Ra
vr
R + r
(i) 4.5 rad/s
(1) COi =—, (02 =-
h =”> a2 =7n
Multiple Choice Questions
1. (c) 2. (a) 3. (a)
9. (b) 10. (d) 11. (b)
Assertion-Reason Type Questions
1. (c) 2. (c) 3. (a)
Match the Columns
1. a - q, s; b - p; c - s; d - r
2. a - q; b - r; c - p; d - s 
Comprehensions 
Passage-1
1. (a)
Passage-2
E (c)
True or False
1. (F) 2. (F)
Fill in the Blanks
Problems
i’l = 3v V2 = >/5v vj = v ' = -——co
r 22
..... ,D . 21 (R-r) w -(111) aCo ={R~ r)ta T, aCP =------- ------ J
—> A A A —> A
vP = (v - 2rco)z, vq = (v - rco)z + rcoj and vc = (v - rco)z
a 2 * 2 * *aP = (-2ra + a)i -rta j, oq =(a-ra- Rd) )i + Raj
ac =(a- ra)i
vcosp
dsin(9 + P)
 vcos 9
QP csin(9 + P) 
vcosp 
sin(9 + p)
(i) v^ = co/cos 9, vB = co/sin9
(ii) aA = (w2cos9 + asin9j|, aB
(iii) vc =y(cos9z?-sin9/j, (ac)x =|(acos9-
and (ac)? = ^(co2 cos 9 +a sin 9)
(i) v = yjv2 +r2co2 +2vrrcocos 9
(ii) ax = (racos9-co2rsin9)z
ay = —(co2r cos 9 + 2covr + ra sin 9) j
2
(iii)
CHAPTER
PointsStudy
Torque on a Particle and 
Angular Momentum of a Particle
14.1
14.2
1 4.3
1 4.4
1 4.5
14.6
14.7
14.8
14.9
Introduction
Torque About a Point
Torque Due to Many Forces Acting at a Point
Torque About an Axis
Angular Momentum About a Point
—>
Physical Significance of L
Angular Momentum About an Axis
Relation between t and L
Conservation of Angular Momentum
§ 14.1 Introduction
In this chapter we talk will about the turning or rotational effect of a force, that is, torque on a 
particle. Using the idea of torque we will develop the concept of angular momentum. The concept of 
angular momentum helps us in understanding the meaning of “turning of a particle”. Then we will 
explain the principle of conservation of angular momentum as an independent principle. The relation 
between torque and angular momentum will be verified as Newton’s second law of rotation (turning of 
a particle about a point of reference). After understanding the concepts of torque and angular momentum, 
we can use them for system of particles in the next chapter. The physical significance of torque and 
angular momentum has been explained by the examples of physical pendulum, conical pendulum, particles 
moving in a straight line, planetary motion etc.
§ 14.2 Torque About a Point
So far we have seen that a force causes the motion of an object. An object may move in a straight 
line or a curved path. In both cases we will see how a force is responsible for “turning” an object around 
a fixed (inertial) point.
Let us consider a stationary particle P connected by a rigid light rod pivoted at one of its end O. 
The particle can turn around a fixed point only when it possesses a velocity perpendicular to the rod.
189
F
I
0
Fi
r>
F
i 
i 
i
ty 
7 p
When a force acts at a particle, its per­
pendicular component (component 
perpendicular to r ) Fx is responsible 
for turning theparticle about the iner­
tial reference point P
Fig. 14.1
-*
The turning effect de­
pends on r, F and 0
Fig. 14.2
t = r x F
i
The ease of rotation of the door de­
pends on the magnitude I F k direction 
—>
0 of force F and the distance r of the
Torque on a Particle and Angular Momentum of a Particle 
This is possible only when the particle will experience a net 
force perpendicular to the rod.
If we apply a force F on the particle, the component of 
force F perpendicular to the rod, that is, FL is responsible for 
turning the particle around 0, whereas the component of F 
parallel to the rod, that is, Fj| will partly nullify the reaction in 
the rod. In consequence, the rod will rotate. If the rod is not 
there, will pull the particle out of the origin O. When we 
apply more force, more promptly the particle turns about (9.The 
turning effect is associated with an angular velocity given as
V1 wpo = —. r
where v_|_ is caused by the force Ft (= F sin 0) 
—>
9 = angle between r and F
Hence the turning effect of the force is associated with the quantities r, F 
and sin 0.
The turning effect of the force can be termed as “Twist” or “ Torque” 
denoted by the Greek letter 7 (tau).
t = f(r,Fjf) = /(r.Fsin 0)
Let us take a simple example.
While opening a door we need to push it with some force 
F, say. The component of the force F perpendicular to the 
plane of the door actually helps us in turning the door. At any 
point, if we push the dpor harder (more force), more quickly 
(easily) the door will open. When we push the door at farther 
points from the hinge, it will turn it more easily. Furthermore, 
when we push the door perpendicular to its plane, it will rotate 
most easily for a given magnitude and point of application of 
force.
Practically it is verified that the “turning effect” or “twist’ 
or “torque” of the force F is directly proportional to both r and 
Fi(= Fsin 0). Then, we have
roc rF sin 0 point of application of the force F from
For the sake of simplicity the constant of proportionality the axis of rotation
can be taken as unity. Fig. 14.3
Then, r = rFsin0
Since the stationary door (or rod) tends to rotate by the turning effect of the force F in the 
direction of r x F, we can attribute a direction to the “torque”. Then, vectorially torque can be given as
190
y o
y
X
X
FP
9
The expression
P
= n
Fig. 14.6
P
o'
'V
/ line of application of force F
Fig. 14.5
OS
X
G.R.B. Understanding Physics MECHANICS (Part-B)
We can also call r as the moment of force (denoted as M) because it is multiplied with r . It 
must not be confused with work, even though both work and torque are dimensionally same; work is a 
scalar whereas torque is a vector. The unit of work is joule whereas the unit of torque is newton-metre.
When we reverse the direction of force at any point, the turning effect (torque) reverses its 
—>
direction. In fig 14.4(a), the torque of F about O is anticlockwise as it tends to turn the stationary 
particle P in anticlockwise sense.
'f± ] = r x Z7! (because r x = 0),
—>
where FL is the component of the force F per­
pendicular to the line joining the reference point and the point 
of application of the force.
This gives us the idea that the component of the force 
—> —■►
perpendicular to r (but not perpendicular to v ), that is, F± is
—>
F tends to turn the point P about 0 
in clockwise sense; T is clockwise 
or inward ®
Fig. 14.4 (b)
-4
P
Oz
t = rFsin Q(-k) = rF sin 0
t = r x F can be written as
t = rF sin 0 }
= (r sin 0)F ) = pF )
where p = perpendicular distance between the point of reference P and the line of application of 
force F\ p is also called moment arm. When moment arm is zero, any amount of force cannot produce 
a turning effect. It means, if we choose the reference point on the line of action of force, the torque of 
the force will be zero about the reference point.
Alternately, we can write the torque as
t = rF sin 0 )
sJ3 F ---
>V\ °"
F± causes turning effect which is given
as torque t
—>
F tends to turn the point P about
O in anticlockwise sense; 7 is 
anticlockwise or outward ®
Fig. 14.4 (a)
T = rF sin Qk = rF sin 0 )
Similarly, in fig 14.4(b) the torque of the force is clock­
wise,
(b) —>
Fig. 14.7
where
F
wrench
nut
/-----------►
SoL /
'pX
Ex. 1. When you apply a force F at the end ofa wrench how much torque 
can it generate to open the nut ? Assume that the force, wrench 
and nut lie in the plane of the paper.
x = | r x F | = rFj. = pF
F± = F sin 0 and p = r sin 0;
—> —>
0 = angle between r and F
When we reverse the direction of force F , the torque reverses its direction. This means that, 
the point of application of force tends to turn in opposite sense if the direction of force is 
reversed.
1 / F
7 ?=rFi)
I
!■ 0, F -> oo Ans.
Student Taskf
> In the above example, find the maximum ratio of maximum torques of the applied force for 
any two points of the handle of the wrench for any given orientation of the force.
If we move the force F towards the nut, t will be lesser. It means, it will be harder for us to 
rotate the nut. That tells us that the rotational effect will be lesser. It will vanish when the force 
passes through the point P.
I 
Ans. — 
ro
T
'oI
zo
Fig. 14.12
umiuii______
P|—x—|
Fig. 14.10
P ■*—x 
Fig. 14.11
> In the above example, we are applying a torque of the force F. Why is the rod not turning? 
Ans. The applied torque is counteracted by the torque due to gravity (weight of the load £), about O.
x - Fx = c
1 rSince 7x7 = 0 as tension passes through O, it cannot produce a torque about O. Then, we 
have
°net = a'+an
a = — = 0 (because v = 0) 
n I
a, = g cos 0
Net torque is equal to torque 
due to the net force 
Fig. 14.14
When many forces act at a point (on a particle), the net torque, that is, the sum of torques of all 
forces about an inertial reference point is equal to the torque of the net force acting at the point 
(on the particle) about the chosen reference point.
= r x Z Fj
Since, Z 7) = net force acting at p = Fne{ >
Ex. 3. Verify the formula xnet = r x Fnet, where rnet = Z r, x F,- for a swinging 
pendulum bob when the string of the pendulum makes an angle 9 with 
horizontal. —> —>
There are two forces, i.e., T and mg acting on the particle.
The sum of the torques of these two forces about 0 is
Torque on a Particle and Angular Momentum of a Particle
§ 14.3 Torque Due to Many Forces Acting at a Point
—> —> —>
When many forces F],F2 — act at a point P having position 
vector r relative to our reference point O. the net torque about 0 is given 
as the sum of torques of all forces about O. Symbolically,
Tnet = Z Xj = r x F] + r x F2 + ... + r x Fn
= r x (7] + F2 + ... + Fn)
194
Substituting
and
we have T
ki
r =
we have
and
r = xxi + xyj + xzk
This can also be written as
rx
Fx
r2
Fz
Fnel
§ 14.4 Torque About an Axis
Referring to the previous section, the torque of a force F = Fxi + FyJ + Fzk about a point 0 is 
given as
xnet = r x
The components of the —>
torque of F about 0 are
■t = r x F 
= rx + ry + Tz 
Fig. 14.17
G.R.B. Understanding Physics MECHANICS (Part-B)
Thus we have aneI = g cos 9
Then the net force acting on the particle is
F - man) A pendulum bob is swinging about z-axis
(a) A small bead situated at a horizontal distance b from the observer 0 is 
falling freely.
A swinging pendulum 
Fig. 14.21 (b)
mg
Falling body
Fig. 14.21 (a)
Student Task^
> Referring to the above example, find the torque of the forces about O and compare it with the 
torque about y-axis.
G.R.B. Understanding Physics MECHANICS (Part-B)
Jy
X 
p
rx = mayi, xy - marj and tz = -
—> A —> —►
Ans. x0 = marj; x0 = (xP)y
4 2 + v y
^y-k
r
Ft = ma, = ma (because a, = a as given)
_ mv2
Fr = mar =------
r
respectively, as shown in the figure.
Then, the net torque about P is given as the sum of torques 
produced by Ft and Fr about P which can be given as
+ Ftr t
_ c rnv2
= -Fryk + Ftrj , where Fr = and F, = ma 
mv2 y -
=---------- k + marj
r
m, 2-. 2= — {ar j - v yk) 
r
m [~~~Hence, the magnitude of the torque is xP = —fa^r
Ex. 6. We have three cases.
Sol. (a) The gravitational torque is the net torque about O which is given as
P
Fig. 14.22 (a)Torque remains constant both in magnitude and direction 0
-♦
(b) As proved earlier, we have I
>P
I
rotates with r about the
where
and
T 
mg
x0 = r x mg = mgr sin Or
r 
Ip
§ 14.5 Angular Momentum About a Point
Even though the force and torque are different quantities, let us try to discover some similarity
(c) The net torque about O is equal to gravitational torque because the 
tension cannot produce any torque about O as it passes through O. 
Then,
—>
In the first quadrant the direction of x0 remains 
constant, whereas its magnitude changes
i
i 
i
mg
Fig. 14.22 (c)
r = R sin 0
gR sin2 0 
COS0
Torque on a Particle and Angular Momentum of a Particle
(c) A conical pendulum revolving about y-axis executing uniform circular 
motion in xz plane.
In each case, find the torque of the net force (or net torque) about O.
In this case, the magnitude of the torque x0 is constant, whereas its direction changes from 
f A 
= To 
T
mg
Fig. 14.22 (b)
llllllllllll
oY
!e
I
I
__ L
mg
x0 = -mgl cos Ok
time to time as the unit vector along the tangent that is r
Conical pendulum 
Fig. 14.21 (c)
I
( 2mv 
j>-axis. We can get the same answer by taking the torque of the net force F =-----I r
v2
x0 = rxmg
= rmg sin Ok
= mgbk (because r sin 0 = b )
F v
->
pp. m
Now substituting F =----- in x = r x F,
we have
which can also be written as
= 0 Obecause = v x m v
L = r x P(= r x m v),
of
Fig. 14.24 (a)L(=7«P) 
is outward or anticlockwise
Fig. 14.24 (b) L (-7 x P) 
is Inward or clockwise
a
—♦ —>
Torque ( = r x F) about 
the point O changes the angular 
momentum of the particle about 0
Fig. 14.23
r x P. Let
198 G.R.B. Understanding Physics MECHANICS (Part-B)
between them. As we know, force can change the linear momentum. Following this logic, let us enquire 
if a torque can change something !
Retrospecting to the previous section, the net torque on the particle about O is
x = r x F,
where F is the net force acting on the particle P. According o, 
to Newton’s second law of translation, the net force pushes the 
/ \ -> 
F
= — .In other words, we can 
k mJ
express the force F as the rate of change of linear momentum;
c/f
-> dP x = r x----
________ dt
7 = l(7xP) 
dt
particle with an acceleration a
d r-----x P 
dt
The above expression tells us that the net torque can also change something, that is, 
—> —
us now take this new term “ r x P ” and name it carefully.
—>
As we know, a “force” changes the linear momentum P likewise, the angular analogue of force, 
that is, torque changes the quantity “ ~r x P ”. Hence this term can be termed as “angular momentum" 
—>
denoted by L . Hence we can define the angular momentum of a particle possessing a linear momentum 
—>
P(= m v) about a point as
where r = position vector of theparticle relative to the reference point.
The cross product of r and m v signifies that L is a vector quantity pointed in the direction 
r x P, given by right hand thumb rule.
The angular momentum L of a particle 
about any inertial reference point O is 
defined as the vector product of the 
position vector r and linear momentum 
—> 
P of the particle relative to the reference 
point. O.
Symbolically, L = r x P
dP
dt
199Torque on a Particle and Angular Momentum of a Particle
The magnitude of L is
0
-•0 r. =pu p
Then,
—♦o
L = rxP = rxp1we have \L\=pP
y
o
Fig. 14.26Sol.
| L | = rP sin 0, 
where P sin 9 = P±
L = r * P = rxx p, 
L = r±P^
Fig. 14.25 (b)
I L \=rPL
Alternately, substituting
r sin 0 = Tjl = p,
The angular momentum has a magnitude equal to the product of, “distance between the particle 
and the reference poinf’and “the component of momentum of the particle relative to the refer­
ence point perpendicular to the line joining the particle and the reference point”. Alternately, the 
magnitude of angular momentum is equal to the product of “linear momentum” and the “per­
pendicular distance between the reference point and the line of motion of the particle”.
Summarizing the above facts,
The angular momentum of a particle depends upon the choice of reference point. If the velocity 
of the particle relative to the reference point is zero, the angular momentum of the particle is 
zero about that reference point. Furthermore if the particle moves along the line joining the 
reference point, angular momentum is zero.
Ex. 7. A particle of mass m starts moving in a circle in xy plane 
when it is at x - 2R and y = 0. If the particle moves 
with constant speed v, find its angular momentum about 
the origin O after a time t measured from beginning of its 
motion.
Let the particle acquire a position r after describing an
v 
angle if) = (20) = —/, during a time t.
R
The angular momentum of the particle (at that instant) about O is
Fig. 14.25(a)
For different points of reference, r and P will be different. Hence the angular momentum depends 
on the reference point. Since the magnitude of linearmomentum p is multiplied with the distance p, we 
can call the angular momentum “moment of (linear) momentum”.
16 P•*HII
200
^mvi
v± = v cos 6
O
XSubstituting
we have
Fig. 14.27
>
I
Ans.
Sol.
/
P
0 Fig. 14.28
Ans. (i) No; (ii) Yes
Ex. 8. Prove that angular momentum of a particle moving with constant velocity is constant relative 
to an inertial reference frame.
From the basic formula, we have
The angular momentum about any point on the perimeter of a circle varies in magnitude, 
whereas both magnitude and direction of angular momentum remain constant relative to the 
centre of the circle.
> In the above example will you get same answer at the given instant when you consider the 
reference point O moving with a (J) velocity v' (ii) acceleration a?
where
and
Then, we have
T
2mvR
p
!
2
Lo = 2mvR cos2 0 
vt 
2R'
Student Task^
Referring to the above example, plot the variation of L as a function of time.
r - 2R cos 0
L = r x P
= rP sin 0 }
= (r sin 9)P
Since r sin 0 = p = constant, we have
~L = pP\
As p. p and the direction (clockwise or inward) of angular momentum is constant, we can say —>
that, the angular momentum L remains constant.
Student Task^
, „ 2 VtLn = 2mvRcos —k ° 2R
C.R.B. Understanding Physics MECHANICS (Part-B) 
lo = /?Mv± y
^4.
201
Xv±
P
-♦
Sol.
Substituting
xwe have Fig. 14.31
Angular momentum measures the state of turning (angular motion of r ) of the particle. For 
this, the particle need not move in a circular path. It may move in a straight line or in any curve. 
The direction of angular momentum gives the sense of turning and the magnitude of angular 
momentum gives the amount of turning (how quick the particle turns about the reference 
point).
with an angular velocity a>p0 
Fig. 14.29 (a)
v
*
V±
Fig. 14.30 
------- - 
v
the component of its velocity perpendicular to r, that is, Vj,. On the 
other hand, the particle approaches to (or) departs from the point O 
with a velocity ^|.
The above discussion confirms that the angular momentum (L = m r x v±) containing the 
—> —>
terms r and v± informs us that the particle of mass m turns about a point situated at a distance r from 
the particle with a velocity v± (but not v).
O'
A particle P having non-zero angular 
momentum must turn around the 
point of reference O 
Fig. 14.29 (b)
s /
 A 
L = -mv^rk
Vjl = r CD,
L = (mr2)co,
S’"?
w
p/z___-y__
Torque on a Particle and Angular Momentum of a Particle
§ 14.6 Physical Significance of L
Sometimes the term “angular momentum of a particle moving 
in a straight line about a point” baffles the student. Generally we think p/
that when a particle moves in a straight line, how does it possess a I J A. 
term “angular momentum” without undergoing angular motion or cir- / / X
cular motion etc ? In other words, how can we use angular quantity / //-; 
like angular momentum when the particle moves in a straight line? Let /// 
us explain the idea behind this. *0
We can provide a logical interpretation of this by saying that The particle P turns around O 
when a particle moves in a straight line or any curve, its position ... . . v,, , with an angular velocity a>p0 = —vector rotates about a fixed (or inertial) point. Hence we can say that r
the particle is turning about that point.
We can see that in the given figures, the particle P turns 
about the fixed point 0 with an angular velocity given by
- vPOi = V1 
r r
because v0 = 0
It means, the turning of the particle about O is decided by
02
—>
mv
0, Fig. 14.32
Fig. 14.33
l
L = x z
mvz
y
Sol.
Vo
%
O X
Fig. 14.34
G.R.B. Understanding Physics MECHANICS (Part-B) 
where mr2 is known as “moment of inertia of the planet about the sun”. We will talk about it in 
chapter 15.
114.7 Angular Momentum About an Axis
By using cartesian coordinate system, let us substitute ~r = xi + yj + zk, 
n the formula
J 
y 
mvy
v - vxi + vyj + vzk
Lo = r x m v, we have
Lo = m(xi + yj + zk) x (vxi + vyj + v.k)
After evaluation, we have
m
\"Vx
\ v
—> A A *
= m(yv, - zv )i + m(vxz - xv.)j + m(xvy - yvx)k
The above expression tells us that there are three components of the 
—>
ingular momentum L of the particle about any point along three mutually
perpendicular axes, i.e., Lx,Ly and L, respectively;
--- ► A --- ► A ~*
Lx = m(yv: - zvy)i, Ly = m(yxz - xv,)j and L, = m(xvy - yvx)k
Then, we can write
mvx
This gives us the relation between angular momentum about a point and angular momentum 
about an axis.
Ex. 10. Derive an expression for angular momentum ofa projectile relative to the point of its projection, 
as the function of time measured from the instant of 
projection.
Assuming v0 and 60 as the speed and angle of projection 
respectively, the velocity of projectile after a time t measured 
from the instant of projection is given by
v = v0 cos 0OZ + (v0 sin 0O - gt)j
The-position of the projectile relative to the point of 
projection 0 is given as
Lq — Lx + Ly + I.
The angular momentum about an axis may be a component of angular momentum about a 
point. In other words, angular momentum about a point can be split into three rectangular 
components.
The above expression can be tabulated in a determinant form :
k
203
r = VqI cos O0i +
Ans. 9|
Sol.
p\j
P
Ex. 11. If a conical pendulum has a bob of mass m and string of length I, find the 
angular momentum of the pendulum bob relative to the point of its sus­
pension "0", when the string makes an angle 0 with vertical.
> In the foregoing example, for what value of the angular momentum of the particle at its 
highest position about 0 willhave maximum magnitude.
Lo = m r x v
Substituting r and v, we have
-i 1 
= COt ~r=
2
2
— cos Qok
As the bob moves in a circular path, the string rotates in space
—> 
forming a cone. This means that the position vector r rotates
about the vertical axis passing through the point of suspension. In 
_> r
consequence, the angular momentum £ rotates in a conical plane l rotates forming a cone as 
as shown in the Fig. 14.36 tbe bob (r>and ?) rotates
Fig. 14.36
The angular momentum of the bob about the point of suspension 0 at any 
instant (as shown in the figure) is
->
Lo = m
Torque on a Particle and Angular Momentum of a Particle
■a #/2 1 * 
vo/sm0o - —JJ
Then, the angular momentum of the projectile relative to the point 0 is
. ( gt2}-
v0/cos90f + I v0/ sin 90 —— \j x [v0 cos90/ + (v0 sin 0O - g/)y]
= mv0/(v0 sin 90 - g/) cos 90 - mv0 I vor sin 90 - -— cos90 k
.2
— cos 90£
Fig' 14.35
'Wry
ZZO\
£0 = rxP=rxmv
The direction of Lq is perpendicular to the plane containing 7 
and m v.
7* r* mvpgtIn the above example, Lx = Ly = 0 and Lo = L, =----- —
for the given coordinate system.
Student Task^f
204
L
Lmv
-> mv
7
Lo
where Lo = mvl
Then, Lr = mvl cos 0r
where v =
Hence, L = m Ans.
4? = A- + Ly'1.
where
and
P = mv is inward; L is 
perpendicular to both r and P
Lr = mvl cos 0r
Ly = mvl sin 0/
where r = I 
It means,
Fig. 14.37
Lo = rP - mvr,
Oblique viewof L = r » mv, 
where r± mv because the 
bob with the thread rotates 
forming a cone
G.R.B. Understanding Physics MECHANICS (Part-B)
—We can notice that r is perpendicular to the momentum P. Hence the magnitude of angular 
momentum Lo is given as
Similarly, the axial component of Lo is
Ly = Lo sin 0/ = mvl sin 9/
If we resolve Lo along the radial, we have 
—> A
Lr = Lo cos Or,
gl I
1------ sin 0 (see the box for its proof)
cos 0 J
I- 11 sin 9 (cos 0 r+ sin 0 j) 
cos 0 J
Lo = mvl,
In other words, the magnitude of Lo is constant whereas its direction (orientation) changes 
while rotating in space forming a cone (imagine !).
I / 
I /
[9\
1 r*
i X
___ i____ \
i
»------R-------
i .
i 
i
Fig. 14.38
205
T
T sin0
and
where R = I sin 9
sin 9This gives v =
Hence
Ans. wivZ sin 9j
Then, we have
4 T cos0 
I 
I
gl 
cos 9
where r x P = L
In other words, when the angular momentum of a particle changes about any reference point, 
it must experience a net torque about that point which is equal to the “rate of change of angular 
momentum” about the point of reference.
Student Task
> In the above example, find the angular momentum of the bob relative to the centre of the 
circular path traced by it.
mg
Fig. 14.39
.sin 9 / 
cos 9 J
= ml sin 0 -
N cos 9
Lo = mvl = m
Torque on a Particle and Angular Momentum of a Particle
It means, £ varies in direction and Ly is constant.
2. We should remember that Lq and Lr rotate forming a cone and circle respectively, with a 
frequency equal to the frequency of circular motion of the bob.
3. The speed of the bob can be found by solving force equations :
T cos 9 = mg
Tsin0 = ^,
R
-> dL 
T = ------
dt
The above expression tells us that a torque about a point changes the angular momentum about 
that point. The rate of change of angular momentum about an inertial reference point is equal to the “net 
torque” experienced by the particle about the point of reference.
§ 14.8 Relation between 7 and L
Torque about a point: Let us recast the expression, 
t = 4 (r x ?)> dt
206 G.R.B. Understanding Physics MECHANICS (Part-B)
Sol.
Fig. 14.40
and
Then
y
Using the expression
,(x, y)-»we have
mg
0 x
where
Then,
Ans.
Student Task^f
> Verify
The torque about an axis brings about a change in angular momentum about that axis and it is 
equal to the rate of change in angular momentum about that axis.
— x------
Fig. 14.41
-» dL
T = -----
dt
mg = -mgj
t = (xi + yj) x (-mgj)
= -mgxk
-» dL
T — ------ ,
dt
Ex. 12. A particle of mass m is projected at t = 0, with a velocity v0 at an angle 60 
—>
with horizontal, from the origin O. Using the formula r =----- , find the
dt
angular momentum of the particle after a time t about the origin O.
The torque produced by gravity about O is
T = r X mg, where 7 = xi + yj
dL— = -mgxk
dt
L = -mgkixdt,
x = vQt cos 90
£ = -mgk v0 cos 0O £ t dt
1 2= - — mgvot cos 90£
Substituting 7 = t^ + t^, + 7 and L = Lx + Ly + Lz in 
we have
- dLx -> dL dL,
t„ = —— ,t., = —— and t, = —— 
dt y dt z dt
dLx dL dL.
X •’ • dt dt dt
Comparing both sides, we have
t = —— for a swinging simple pendulum, 
dt
dLy
dt
Torque on a Particle and Angular Momentum of a Particle 207
Sol.
d’C
Since Ly = constant,
Then,
Then,
d§
Substituting
we have
Then, we have Zrco;
dLrSubstituting
t or 6
we have
/
Substituting = gR tan 0,
Lr
Fig. 14.43
Lr
Fig. 14.42
dtp 
dt
dLv
= 0 
dt
where — = rate of change in direction of Lr which is equal to the angular speed of the 
revolving bob.
dL0 _ dL^ 
dt ~ dt
dt
mv2l cos 0
R
dtp
dt
v2
Ex. 13. Verify the formula r =-----for the particle (bob) in a conical pendulum. Account for the
dt'
direction of torque experienced by the pendulum bob.
Retrospecting to the expression
—. , • —>
As Lr rotates in a horizontal circle, for a small change in its orientation d§, the change in Lr 
can be given as
dLr = Lrd$
dLr T d$ 
------ Lr —, 
dt dt
d§
— = co, 
dt
dLr
-----= A. co
dt
dLp .
dt
Lr = mvl cos 0 and
v v
co = — =-------
R / sin 0
Lo = Ly + Lr 
and differentiating with time, we have
dLr
dt
208 G.R.B. Understanding Physics MECHANICS (Part-B)
we have = rngR
which is equal to -—must point tangentially given by the unit vector | or §.
0= -mgRfyThen, we have -»
® T
Since
We have
Since
and
using To =
we have
and
In the above example, since
we have
Hence
Fig. 14.45
->
Tr
dL0 
dt
—>
This implies that Ly = constant
However, there is a torque which rotates in horizontal plane.
This torque t0(= changes the direction of Lo(= Lr) in the tangential direction, that is, 
direction of $ or r.
dt
dtp
dt
dLy 
dt
t0 = r x mg =
mg
Fig. 14.44
dLr ->
which tells us that the torque rotates (changes its orientation). In 
consequence, the angular momentum L changes its orientation (by 
rotating) forming a cone of revolution. The torque is tangential to the 
—>
circle produced by the tip of L .
x - (JJ dt
dmint nnint tanopntiallv oivpn hv thp unit vprtnr A nr a .
dt
Torque on a Particle and Angular Momentum of a Particle 209
Sol.
TO = V + TJV + T = V,
N
because
we have I m F----- mv
Then substituting
we have Fig. 14. 46
Then,
Substituting
we have
When there are many forces, each force has its own rotational effect (torque). The torque of 
any one force may not be able to change the total angular momentum of the particle. However, 
each torque tends to change the angular momentum of the particle. The net torque is respon­
sible for changing the angular momentum of the particle. In other words, torque of a force can 
change or tends to change the angular momentum of a particle.
where F = net force
Since
Ex. 14. How do you apply r = 
forces? For instance, when
plane, it retards and stops after some time moving in a straight line. Verify the relation 
for the example.
Since there are three forces,/ N and mg acting on the particle, the net torque about 0 is
F = f + N + mg = f
N + mg = mav = 0,
- dv = myk— 
dt
dL0 - 
—- = -pmgyk 
-> dt
_ dLo_ js verjfie(j
dt
dt
dv
— = “MS dt
x0 = -\xmgyk
The angular momentum of the particle about O is
Lo = r x m v = mvyk
x0 =Xf = r x f 
= -fyk
f = V^g,
-----for a particle moving in a straight line under the action of many 
dt
we project a particle, a small block, say, on a rough horizontal
-> dL
T = ------
dt
Hence t0
210
y
Xh
Ans. 2 mvhk
Since
Fig. 14.48dtwe have
Substitute
—> —>
When xnet = 0 about any point, £ remains constant aboutthat reference point
y
P
Sol.
Fig. 14.49
Student Task^S
Ex. 15. In a conical pendulum, can you conserve the angular momentum of the bob relative to any axis 
passing through the point of its suspension ? Explain.
> In the above example, is it possible to conserve the angular momentum about any point on 
the vertical axis passing through the point of suspension of the pendulum bob?
Ans. About the centre of the circle.
In conical pendulum, we have seen that xy = 0 (but xP * 0). Hence 
dLv -> -—= 0: in other words, Lv remains constant, that is, Lv = mvRj
dt y y J
(but LP # c ).
§ 14.9 Conservation of Angular Momentum
As we leamt, a net torque can change the angular momentum of a particle. If there is no net 
torque, the angular momentum of the particle remains constant. That is what we call the conservation of 
angular momentum.
0
iLxJ—*v
/77777/777T7777777777777777777777777777777
Fig. 14.47
Tg,. = r x mg = mgxk
-> _dL
Tgr dt ’
A L = jdL = f(
, [2hand t = —
V 8
Tgr
G.R.B. Understanding Physics MECHANICS (Part-B)
P 
---------------- o
> A particle (a stone, say) is dropped from rest from a height h 
from an observer "O" standing on a trolley car which is 
moving with a constant velocity v. Find the change in angular 
momentum of the particle relative to the observer O till the 
particle strikes the ground, by using the basic formula
-> _dL 
T " dt
Hint: At any displacement x(= vt)
211
2'2
Fig. 14.50
Sol.
F
P
Fig. 14.51(a)
(b) (7ZI] = m2)
1 ■— r,This gives
ViSince W| Fig. 14.51(b)w2
2
we have
Student Tasker
Ans. I : I
Sol. Fig. 14.52
Fig. 14.53
Torque on a Particle and Angular Momentum of a Particle
Ex. 16. In sun-planet system,
(a) Can you conserve the angular momentum of the planet ? 14 
If yes, about which point ?
(b) Find the ratio of angular speeds of the planet relative to 
the sun at the positions 1 and 2.
(a) Since the planet experiences a gravitational pull F which 
passes through the sun (more precisely, the CM passes through 
the CM of the sun-planet system), it cannot produce a torque 
about the sun. Hence the angular momentum of the planet 
about the sun remains constant; LP = c.
LP = mr}V\ = mr2v2
v2 H
r2
> Find the ratio of areal velocities at any two positions of the planet.
Ex. 17. A particle P(a small ring, say) is connected with a rigid support O by a massless spring. 
Discuss the validity of conservation of angular momentum of the 
particle about 0 when it is projected with a speed v0 on a smooth, 
(a) rigid horizontal elliptical wire loop. 
(Z>) horizontal surface.
(a) As the spring force Fsp passes through the point 0, it cannot
produce a torque about 0. Then the normal reaction N can only 
produce a torque which will change the angular momentum of 
the particle about 0. Ans.
(b) Since the spring force passes through O, it produces a zero torque 
about O. The net torque produced by normal reaction N and 
weight mg of the ring about O is zero. Hence Lo = c. Ans.
= — and n
w2
Strictly speaking the planet revolves about the CM of the sun-planet system. Since the centre of 
m
mass of the sun-planet system is very close to the centre of the sun (because — « ms
that the planet revolves around the sun practically. Hence L about the CM of sun-planet system is 
conserved as the gravitational forces experienced by each other cannot produce a moment as they pass 
through the CM of sun-planet system.
212 G.R.B. Understanding Physics MECHANICS (Part-B)
,21.
,2
,2
Fig. 14.54
Sol.
which gives Ans.
Substituting
,2
we have Ans.
r2
T T
Fig. 14.55
(Ly)i = (Ly)f
= mvr,
> II
, External agent
r0~
Ex. 18. A smooth bead of mass m moves with a speed v0 in a circular path 
of radius r0 on a table by the action of an inextensible massless 
string, if the string is pulled down slowly by an external agent, find 
the:
(a) speed of the bead when the length of the string on the table is 
equal to r
(b) work done by the external agent in pulling the string in (o).
> In the foregoing example, can you write the equation F(= kx) =-----for the particle?
r
Ans. No
v = », 
r
2 lr2
(a) As the tension T passes through O, it cannot produce a torque about _y-axis
Then,
Since (Ly)/ = mvoro and (fy)/ 
we have mvoro = mvr,
V=^L 
r
(b) Work done by T can be given by using work-energy theorem;
W = tsK = Kf - Kt
1 2 1 2
= — mvj - — mVj
= |w(v2 - v2)
1 2 1In the presence of the rigid frame, E = — kx + — mv‘
1 2 12. In the absence of the rigid frame, E = — kx + — mV
2 2
= c but Lq * constant.
I 
I
I
I
I
Or —
■ T 
i 
i
= c and Lo = constant.
213Torque on a Particle and Angular Momentum of a Particle
Student Task^
>
Sol.
R
mg
Fig. 14.57
Ans.
where r0 = R sin 0
This gives
Fig. 14.56
iy
In the foregoing example, find IV by using W = j"/7 ds •
If we project the particle (bob) in vertical plane, minimum speed of the bob to reach the top is 
v = y]2gR cos 9. In that case, the bob revolves about horizontal axis in vertical plane. Since 
gravitational torque about horizontal axis is non-zero as described earlier, we cannot conserve 
the angular momentum about that axis. However, mechanical energy remains conserved.
where (Ly)j = mvQr0 and (£>)/ = mvR as the particle just reaches 
the top of the hemispherical bowl with a velocity v. Then, we have
Vov =----- ,
R
v = vq sin 9
E = C : Conserving energy, we have AX' + MJ = 0
I ? ?—m(v - v0 ) + mgh = 0,
where /i = /?cos9 and v=v0sin9
Then, we have
— Iqz 
k-A v'
I ° ,\N
Since isK > 0, work done by the external agent is positive. In the above example, we cannot 
apply the principle of conservation of mechanical energy because an external force F(= T) is 
r mv2acting on the bead. You can also find IT by using W = IT dr, where T =-----= mrw .
J r
Ex. 19. A particle (a small sphere, say) is projected with a horizontal velocity 
v0 from a point A on the inside surface of a smooth hemispherical 
bowl of radius R. Consequently, the particle reaches the top of the 
bowlfollowing a helical path, and then moves down. Find the minimum 
value of Vq to do so.
Since, N passes through y-axis, it cannot produce a torque about it. As 
—>
mg is parallel toy-axis, torque due to it is zero abouty-axis (but not 
zero relative to P as described earlier).
L = C : Since = 0, we have (Lv)y = {Ly\,
214 G.R.B. Understanding Physics MECHANICS (Part-B)
w
Sol.
Fig. 14.59
= m
Then, we have
Student Task^f
> In the above example, analyse the motion of the bead and discuss the origin of the torque 
acting upon the bead.
Ans. The bead moves in a spiral path; the horizontal (transverse) component of the reaction 
force given by the rod generates a vertical torque.
t-r0-l
Fig. 14.58
. Jco where — = a 
dt
Ex. 20. A light rod carrying a smooth bead of mass m situated at a dis­
tance r0 from its axis of rotation is rotating with a constant angu- 
lar acceleration a about the vertical axis passing through the end 0 
Oof the rod. When the rod has angular velocity co, the bead slides 
with a velocity vr relative to the rod. Find the torque exerted on 
the rod by the external agent in rotating the system (rod + bead). 
The angular momentum of the bead relative to 0 is
L = mvjj,
Vj_ = ra>
L = mr co
Even though we maintain the angular speed constant (a = 0), — cannot be zero. It means, 
dt
there must be a torque acting on the bead about O, but the net torque about O is zero!
, dr
and ~dt=V^ = Vr
dL , i .— = m(r a + 2covrr)
where
Then, we have
Differentiating with time, we have
d In the foregoing example, what is responsible for lifting the particle ?
Ans. Vertical component of normal reaction
215
>4
torque of force F about 0 is equal to x = r x F?
Similarly, can you find the angular momentum ignoring the mass of a particle ?
15.
16.
17.
21.
Can you state that “a particle experiences a torque”?
Can you use the word “torque” when a particle moves in a straight line?
A particle rotates; a particle turns; whichis a better statement and why?
What is meant by torque? In which way torque is different from force?
Torque changes angular momentum. What does it mean?
Does angular momentum of a particle explain whether the particle is turning or not?
Angular momentum of a particle can be non-zero when the particle moves in a straight line. 
Explain.
Angular momentum of a particle depends on the reference frame, why?
The torque due to a force about a point depends on the reference frame. Explain.
How should we choose the reference points to conserve the angular momentum ? Explain.
If we say constant angular momentum, does it mean the constant angular velocity of a particle? 
What is the relation between angular momentum and angular velocity? Are their directions same? 
How do you relate torque of a force and angular momentum of a particle?
Can you apply a force F at Q where no mass is present to ensure that
In Q.-14, when you apply a force at Q where no mass is present but the rotational effect of the 
massless rod is manifested as the rod rotates. How does it happen? What is the net force acting 
on the particle P, if the length of the rod is / ?
—>
Can L and x have different directions?
Torque on a Particle and Angular Momentum of a Particle
Assignments
18.
19:
8.
9.
10.
11.
12.
13.
14. > Q P
—r rod
-> —> -> dL , -*
Can dL, x and co have different directions? Can we always use the expression — = / a ? dt
The torque and angular momentum about an axis and about a point are basically different. Explain. 
The sum of torques of many forces, that is, net torque, is equal to the torque due to the net force 
acting on a particle. Explain.
Torque causes rotation. How can this statement be valid to reconcile the fact that a freely falling 
body released from rest moves in a straight line experiencing a net (gravitational) torque?
How can you use angular quantities like, 9,co,a,£ and x for a particle?
Discussion Type Questions 
i.
3.
4.
5.
6.
7.
20,
G.R.B. Understanding Physics MECHANICS (Part-B)
1.
,2
(d) kx = where I and x are the length and elongation of the spring respectively
/V22.
V1
3.
4.
yi
5.
i
(a) Ly = C (b) Lq -C m
6.
(a) jlgl sin 9 (b) y/Zgl cos0 (c)
7.
Q
T
2g/ 
cos 9
A particle P of mass m is moving in a circular path by an 
inextensible string. The other end Q of the string is pulled by 
an external agent. If the external agent pulls the string down, 
the radius of the circular path decreases from /•] to r^, say. 
Then:
216
Multiple Choice Questions
A particle of mass m connected with an ideal spring of stiffness k is projected in 
a smooth horizontal plane. Then :
(a) L = C
(b) E = C
(c) p = C
(d) Tne
The torque of a force F acting at a point P about a point 0 is given as
ic —
y
Z/°~*X
Two anti-parallel forces F} and F, act on a smoothly pivoted rod. The value of
— so that the net torque of the forces about P is zero, is:
%
—> —>
For a conical pendulum, | LP~ Lo | is equal to :
(a) mvl (b) mv/sinO
(c) mvlcosQ (d) mv/tan0
P w
Xi
Fl
x0 = r x F, where r =rP-r0
T may change its magnitude, direction or both. r0 changes the angular 
momentum of the particle m about O. The angular momentum of the particle 
about O is given as
4.
7.
I
8.
of the particle of2.
mass m is 
p
3.
block is Plank
4.
I zzzzzzzzzzzzzzzz ZZZZZZZZZZZZZZZ. [
(b) mvl cos 0
(d) 2mv/cos0
1.
2.
h 
|
3.
4.
5.
6.
ta
A block has a velocity v on an accelerating plank. If the coefficient of 
(dL\ 
friction between the block and plank is P, the value of J
The angular momentum of an oscillating two particle system about its CM 
is 
I of the
Torque due to gravity about the CM of a system of particles is 
( 
dLA particle slides on a smooth surface. The value of — 
\ dt Jp
Referring to Q.-7, = (oP sin 0, where and a>P are the angular velocities of the bob abou'
y-axis and about the point P respectively.
Fill In The Blanks
i.
Torque on a Particle and Angular Momentum of a Particle 219
In Q.-3, the magnitude of change in angular momentum of the bob, when it moves from position 
1 to position 2, is equal to: 
(a) zero
(c) 2mv/sin0
True Or False
A particle does not tum about a point when the point lies on the line of motion.
Angular momentum of a particle moving with a constant velocity is constant about an inertial 
point.
The net torque due to two equal and opposite forces does not depend on the reference frame.
Torque is a rotational effect of force.
Torque about a point and torque about an axis are same.
Torque experienced by the bob in a conical pendulum5. Four particles of mass m, 2m, 3m and 4m are placed at the 
vertices of a rectangle ABCD as shown in the figure. The 
particles move from A to B, B to C, C to D and D to A 
respectively. In this process find the displacement of centre 
of mass of the system of four particles.
Ans. - —
4
> Find the velocity of CM of the system of three particles of mass m,2m / \— /A 
m v 3m
Fig. 11.9
(iii) Displacement of CM : Integrating velocity of centre of mass with time, the displacement 
of centre of mass is,
P t, —>
Substituting vt dt = s, , we have
J/i
-> m(2vi) + 2m(vj") y _ ---------------------------
m + 2m
(W)(2vf) + (2W)(-vf) 
----------------------------------------- — y 
m + 2m
System of Particles 9
s2
Sol.
S3
we have
where m. m4
This gives sc
Ans.
Sol.
&
SCM
where j| = -s2 (because the string is inextensible)
and
 mj S| + m2 s2 + my Sy 
m\ + m2 + my
The system has three particles 1, 2 and 3. The displacement of the system can 
be given as:
S4
Fig. 11.11
>3
1° °2
Fig. 11.12
■ / _ si
Sr1 —------------
En.,
-» mt S| + m2 s2 + my Sy + m4 s4 sc -
m\ + m2 + my + m^
Ex. 6. Two particles 1 and 2 are attached with an inextensible light string that passes over 
a smooth pulley. Another particle 3 (an insect, say) sits on the particle 2 such that 
the system (1 + 2 + 3) is in equilibrium. If the total mass of the system is M and the 
mass of the insect is m, find the displacement of the CM of the system when the 
insect crawls with an upward distance “y” relative to the string.
Referring to the formula sc 
particles,
f) + 2m(2l i) + 3m(-l j) + 4m(-2Z i) 
Sr — —------------------------------------------------------------------
m + 2m + 3m + 4m
/ ? 21 ,
= --i-------j
5 5
1° °2
Fig. 11.13
/7i| = m, m2 = 2m, my - 3m, m4 = 4m,
Si = I j, s2 = 21 i, Sy = -I j and s4 = —21 i,
s32
o"3
J3 = 532+ J2 • Then, we have
-* _ W, Sj+THz S2+ my(Sy2+S2)
JCM “ ---------------- - ------------------mi + m2 + my
-> -» -> 
mx $]+ (m2 + my)s2+ my Sy2
+ m2 + m3
y 
for the system of four
Student Task^
> In the foregoing example, let us move the particle m to the centre of the rectangle. Find the 
displacement of the CM of the system. Ans. (2i + j)
I ?
5
10 G.R.B. Understanding Physics MECHANICS (Part-B)
Since,
we have
Then, 5cm -
Ans.
Student Task
Since, , we have
y
x
Since the displacement of the CM of the system m + M is not zero, we conclude that the 
impulses of the impulsive tensions during the crawling of the insect are responsible for acceler­
ating (moving) the CM of the system up.
> In the foregoing example, (/) why should the CM move up? (n) What are the displacements of 
1, 2, (2+3)?
Ans. (i) The upward impulse changes the momentum of the system in upward direction.
m
\2
m3 s32
ml + m2 + ,n3
1
m
nt} ~ m2 + m3 and 51 + s2 = 0 ’
—> —>
nil $|+ (ni2 + w3)^2 = 0
m •
(ii) M
(iv) Position of CM : At last, we want to find the location of centre of mass. In fact, we are 
very fond of characterising the centre of mass by its location given by the position vector 
rc. Then we characterise the CM by its velocity and acceleration. Please note that rc 
comes from vc,vc comes from ac and ac comes from Fnet.
Since, ni2 = ni, nil + m2 + m2 = M and = y j we have,
-> m « 
5CM =fiyj M
30°
-----1
Fig. 11.14
m • m -
yjM M
60713 
- m
J* ri 
rC - --- -----Lm,
Ex. 7. Locate the centre of mass of a system of three particles 1, 2 and 3 
each of mass m situated at the vertices of a right angled triangle.
System of Particles 11
y
Since, rcSol.
W| = m2 = m3 = m,
4
30* x
This gives
Ans.
Let us summarize the above discussion.
Arc =rC =
respectively.
->
VC =
The centre of mass of a system of particles is characterised by its position, displacement, 
velocity and acceleration given as :
and ac
1 (\ I 7* 
*—3//4-*l Z/4
Fig. 11.15
substituting
-► A
and r3 = li, 
we have
Zto(- Vj 
Em,
1
0*
ri 
Em,
2 Xix.y 
i\-60’ 
3
_ to] rl + m2r2+m3r3 
fr — --------------------------------------
to] + m2 + to3
to
rC =~
rl = °> r2 = + ~lJ
-» 11 - -fil , 
rC ~ —1 + —J c 12 J2
Student Task^
> If the particles have different masses, can the centroid of the triangle and centre of mass of 
the system have same position? Explain.
Ans. It may be possible when masses are adjusted properly so that mi r} + m2r2 + m3 r3 = 0, 
—> —> —>
where r,, r2 and r3 are taken from the geometrical centre of the triangle.
£to,A r,
Em,
= StoLat Fnet
Em, Em, ?
§ 11.4 Calculation of Position of Centre of Mass (Finding CM)
Now, let us find the location of centre of mass of various systems of particles. In some cases the 
arrangement of the particles are discrete and in other cases, the particles are continuously arranged. 
Generally, a solid or liquid is practically composed of continuously arranged particles like molecules. 
But, when we observe the things under microscope, the molecules do not touch each other; these are 
just discrete particle distributions microscopically. However, this appears as a continuous mass distribu­
tion in macroscopic scale (gross matter). Furthermore, if you go deeper into the atomic scale, you will 
find the molecules which we consider as point masses, are composed of tinier atoms. Atoms can also be 
thought as a point mass when we compare a molecule, but it has many subatomic particles like neutron, 
proton and nucleus. This discussion emphasizes the fact that,
31 - Jj ?
+«(//)
4 4 7
to + to + to
3/ - V3
4
12
y
2m
m
m 3m
m
2m
z
There are eight particles;Sol.
y
2m
(/, A /)m
2m
+m(li + Ij + Ok) + m(0i + Ij + Ik) + m{li + Oj + Ik) + 3m(li + lj + Ik)
(0,1,1)
m
m
Fig. 11.16
G>
®/
(0, /, 0)
G.R.B. Understanding Physics MECHANICS (Part-B)
Microscopically matter is quantised. It means that matter has discrete (discontinuous) mass 
distribution at microscopic level, whereas macroscopically matter appears to be continuous.
Let us discuss the methods of finding the position of centre of mass of a system comprising (a) 
discrete mass distribution (b) continuous mass distribution.
(A) Discrete mass distribution
Ex. 8. Find the centre of mass of a system of eight particles placed at the 
corners of a cube of edge I.
m
(0,0.0)
_______ rnn
(/. o, I)
Fig. 11.17
Substituting the masses and position vectors of the particles, we have
/=8
- I3/m and
/=!
i=8 ,
J tw, r, = m(0i + Oj + 0k) + 2m(li + Oj + 0k) + 2m(0i + lj + 0k) + 2m(0i + Oj + Ik) 
/=!
J) 
3m
(0. 0. /) U- 
X 2m
z
—> r-
n = I -> 8. Applying the formula rc = ——, we have 
Em.
i=8
Ym' ri
.7* - /=l
'C -
/=!
J—► x 
2m
^^x 
/ (/,o, 0)
13
Then, rc
Ans.
2m
II
I
m mI
I I
dm
y
system
0 x
z
rC =
rC =
x□
Fig. 11.20
> Find the centre of mass of a system offour particles placed at the 
vertices of a regular tetrahedron of side I.
01 ■ -
--------x—►
System of Particles
- Iml^i + j + k)
 7ml(j + j + k) 
13m
2 
Ans. At a height of ~
line passing through the apex of the tetrahedron.
1 + *
I
urn
Fig. 11.18
density varies with x as X = Xo
7/- -
= — (' + J + k)
Student Taskjf
j r dm
j dm
(a) Linear mass distribution : If the object is linear, dm = Xdr. Then, we have
fhrdr
jkdr
Linear mass distribution can be straight or curved. Let us discuss the CM of straight and curved 
linear mass in the following examples.
(i) Straight line
Ex. 9. Find the position of centre of mass of a thread of uniform cross-section whose linear mass
mass distribution is given as : 
—♦
-♦ frtlrT1
rc=7^r
Fig. 11.19
/ from the base, on the vertical
(B) Continuous mass distribution
For continuous mass distribution it is difficult to distinguish 
zth particle from /h particle. In that case, we have to consider an 
elementary segment of mass "dm”. This behaves as a point mass.
Hence, we can write "dm” instead of " m, ”. Assuming the position 
vector of dm as r , we can substitute r, as r . Finally, substituting
“ f ” for “E”, we can find the formula for position of centre of The CM of a system of continuous 
mass for continuous mass distribution, which can be given as :
G.R.B.about its point of 
suspension is constant.
In a conical pendulum Ly = LP sin 0, where Ly and LP are the angular 
momenta of the bob about y-axis and about the point P respectively.
220
I.
x0
2.
3.
4. y
c
--------X
5.
6.
G.R.B. Understanding Physics MECHANICS (Part-B)
y*
n 
h
I
Problems
A particle of mass nt moves with a constant velocity v. Find the angular:
(i) velocity.
(ii) acceleration.
(iii) momentum, of the particle relative to the fixed point 0.
In problem-1, if the particle moves with an acceleration a from rest from the angular position 
0 = 0, at any time /, about O find the :
(i) angular velocity. (ii) angular acceleration.
(iii) angular momentum, (iv) torque.
A block of mass nt slides from rest along an inclined plane. If the coefficient 
of friction between the block and plane is p, find (i) co (ii) a (iii) L (iv) Tgr 
(V) TV (v>) T/ (v**) Tnct about O, as the function of time /.
A block of mass m moves along the x-axis so as to obey the relation 
x = t3 +r +1. Find the :
(i) angular momentum of the particle about C.
(ii) torque experienced by the particle about C.
A bead of mass m slides on a hemispherical surface with a velocity v at 
an angular position 9. If the coefficient of friction between the bead and 
hemispherical surface is p, find the :
(i) angular momentum of the bead about O.
(ii) — about 0. 
di
(iii) Tgf.TyandT/ about 0.
A pendulum bob of mass nt swings from 9 = 0 to an angular position 9. Find 
the:
(i) angular momentum of the bob about O.
(ii) torque about O.
(iii) ^gr’^iension and Tnel about Q.
221
7. M
v
8.
/
m
9.
vo
10.
11.
u
M
12.
13.
V
zo
6p 
4
ijx
[0
^//////^//////////////
Torque on a Particle and Angular Momentum of a Particle
A pendulum bob P swings about the point Mof the roof of a trolley car.
If the trolley car moves with a velocity v and acceleration a, and the 
pendulum bob has a velocity v relative to M, find the :
(i) angular momentum of the bob relative to (a) M (b) a fixed point v-~- 
coinciding with M.
(ii) torque about M and a point coinciding with M.
A conical pendulum of mass m swings with an angular velocity co relative to 
the point P of suspension.
(i) What is the angular momentum of the bob about (a) P (b) the axis of
revolution (c) 01 P
(ii) Find the net torque acting on m about P and O.
(iii) Find Tgr,rtension about P, axis OP and 0.
A spring of natural length /0 is connected with a smooth bead of mass m.
If the bead is projected on a smooth horizontal plane with a velocity v0 
perpendicular to the spring, the maximum elongation of the spring is /0, * zo ”
find the: j
(i) work done by the spring (a) upto the maximum elongation of the 
spring (b) at the time of maximum elongation.
(ii) spring constant.
Referring to problem-9, find the :
(i) velocity of the bead when the elongation of the spring is y.
(ii) rate of increase in length of the spring when the length of the spring will be equal to
A bead of mass m moves in a circular path of radius r0 with a 
speed v0 by a light inextensible string which passes through 
the smooth hole made on the table and connects another mass 
A/hanging below the smooth table.
(i) Find A/. II
(ii) What is the angular momentum of the bead m relative to Cl
(iii) Find the work done by the external agent in slowly pulling I
the mass M down through a distance y. -
In the problem-11, if we pull the mass M with a constant speed v, find the speed of bead m when 
the length of the string on the table will be y.
A particle P of mass m is connected to the peripheral point of a 
fixed disc of radius R by a light inextensible string of length /0.
(i) If the particle is pushed with a velocity v0 perpendicular to 
the string, what is the nature of motion of the particle?
(ii) Can you conserve the linear momentum, angular momentum 
and KE of the particle? Discuss.
222
14. /0
m
P
15.
ANSWERS
7. (a,b,c)5. (a,b) 6. (c)
4. (a) S. (b) 6. (d)
4. (d)2. (a) 3. (c)
8. (F)4. (T)3. (T) 5. (F) 6. (F) 7. (T)
3. -mh(pgi-aj) 4. zero
J (ii)(i)1. CD =
3.
5.
(ii) = [mg/?(cos9-psin9)-pnv2] )
Tnet = 0
G.R.B. Understanding Physics MECHANICS (Part-B)
A bob of mass m swings in a horizontal circle when the string makes an angle 0 F 
I
1. zero
Problems
A pendulum bob of mass m swings in a horizontal plane when the string 
makes an angle 0 with vertical. What minimum additional KE should we 
impart to the bob so that the string becomes horizontal?
(V) XN = Tgr (vi)
(i) L - mvR )
' ' di ' -..................................
(iii) Tgr = mgR cos 9 )
t^=0, Xf = pm(v2 + gflcosO)
with the vertical smooth tube. If we slowly pull the end P of the smooth string 
reducing the length of the swinging string from / to -p find the :
(i) angle made by the string with vertical.
(ii) work done by the external agent in pulling the string.
vcos2 9 
h
-* 2v2 a =—-cos' 
h2
(i) w = 0 (ii) a=0 (iii) 2 = 0 (iv) 2gr = mg2'2C0S6(*m9~^-C° — )
Multiple Choice Questions
1. (a,b) 2. (a,b,c,d) 3. (a,b,c,d)4. (a,c) 
Assertion-Reason Type Questions
1. (c) 2. (b) 3. (d)
Match the Columns
1. a-r, b-s, c-q, d-p 
Comprehensions 
Passage
1. (b)
True or False
1. (T) 2. (T)
Fill in the Blanks
2. mghsin Q
;3 Osin 0 (iii) L = mvh )
223
7.
29. (b) zero
(ii) L = mvor)11.
13.
15.
♦ ♦♦
Torque on a Particle and Angular Momentum of a Particle
(i) (a) mvl ) (b) mvl (1 + cos 0)J
(ii) ml(a cos 0 - g sin 0)^ , mgl sin 0^)
(1 + cos2 0)
(iii) mvl
(i) (a) ^sp = -^v0:
(i)
Sd)
(i) Spiral with constant speed.
(ii) No, No, Yes. Since the point 0 changes continuously, we cannot conserve Lq, but 
Lc = constant.
mgl
2cos0
(-1^1
I2 J
Dynamics of Rigid Bodies
1 5.4
the rigid bodies about their CM.
1 5.1
15.2
15.3
1 5.5
1 5.6
15.7
1 5.8
15.9
15.10
15.11
15.12
15.13
Study Q^Points
Introduction
Torque on a Group of Particles
Angular Momentum of a System of Particles
CHAPTER 15
§15.1 Introduction
In previous chapters we applied the concept of torque and angular momentum for a particle. 
Now the same will be applied for an object that is “a group of particles”. Then we will use the ideas of 
torque and angular momentum for a rigid body, for which we will deduce two equations. One for linear 
motion of CM of the rigid body given as oc = where M = total mass of the rigid body. The second 
equation defines the cause of rotation of rigid bodies as the change of angular momentum of the particles 
relative to the CM, which gives us the angular acceleration as a = where Ic = moment of inertia of
Jc
Relation between t and l
Conservation of Angular Momentum of a System of Particles
Newton’s Laws for a System of Particles
Angular Momentum of a Rigid Body about its CM
Calculation of Moment of Inertia of a Rigid Body about Centroidal Axis
Newton’s Laws for Motion of Rigid Bodies
General Procedure to Solve the Problems of motion of Rigid bodies
Need of Friction in Rolling
Alternate Solution for Fixed Axis Rotation (or, it Rolls on a Fixed Surface)
Newton’s Equations for Rigid Body Motion Relative to a Non-inertial (Accelerating) Frame
Dynamics of Rigid Bodies 225
forces
forces
t0 = ^x^+^xF2+... + r„xFn
= Zr, x Fj
~ Text
Then, xo ~ Text
Since the action-reaction pairs of the internal 
forces are equal and opposite and collinear, the 
net torque produced by the internal forces is 
zero. Hence the net torque experienced by a 
system of particles is equal to torque produced 
by the external forces.
Fig. 15.1
= Z riCx Fi 
i = \
The net torque experienced by a system of particles is equal to the sum of the torques of all 
external forces because the net torque due to the internal forces is zero.
Torque about the centre of mass: Let us now calculate the torque about the CM of the system 
and relate it with the torque calculated about the fixed point O. The net torque about the CM is
rj = r\Cx F\+ ^Cx F2+ - + Zcx Fn
— ZT/ext
These two equations form the basis of rigid body dynamics. We will solve many problems by 
using these two dynamical equations with other equations such as alliedUnderstanding Physics MECHANICS (Part-B)14
Directly using the formula rcSol.
xc =
Ans.
Student Task
Ans. When the thread is uniform.
In the formula rc
21
C
311
DA
> In the above example we found that the centre of mass and the geometrical centre (mid-point) 
of the thread are different. In which case both coexist?
fordr
fXdr ’
T 
I 1
41 
-2/-------
Fig.11.21
l+±
/
, we have
Jo'X0
-.1, 
9
where X = Xo
fXrdr
= , we cannot take X out of the integral and cancel in both numera­
tor and denominator before evaluating the integral when X = f(r) in case of non-uniform 
mass distribution. However, we can do the same when X = c for uniform mass distribution. In 
that case, the centre of mass and mid-point of a straight uniform object coincide. Another 
important thing we need to remember is that, in the above formula, we must treat “r” as 
“vector” even though we mention the formula in terms of scalars. In other words, r is an 
algebraic scalar. If the position lies in -ve x,y and z-directions (axes), we write it negative and 
vice-versa. Lastly, we note that m = M is valid for uniform objects only; however m = Jx A uniform rod of mass m and length I is bent to form a right p 
angled triangle as shown in the figure. Locate the centre of 
mass of the rod.
-» m{ rx + m2 r2 + m2 r3 + w4 r4 
rC ~ >W] + m2 + m2 + m^
m\ = 1/, m2 = (21)(2Z) = 41/, m2 = (31)(Z) = 31/, /n4 = (41)(2/) = 81/,
. -» I ?and r4 = — 1.ri = 7^’ r2 = + A .R d
distance of r =
zZ/d'
Rfe’2sin4.rf* ,
1
J-9/2 Y
After evaluating the integrations you can see that second term of the RHS will vanish giving us
0
27? sin -
“*■ 2 ■
rc =----- ------ 1
If the arc subtends an angle 0^= which is symmetrical about x-axis, we put lower limit of
0 0
4> = — — and upper limit of = - to obtain
„ fe/2- 2?f_e/2cos4>^4>
7^—' + 
J-e/2
(sintydty .
G.R.B. Understanding Physics MECHANICS (Part-B)
Let us draw x-axis such that it passes through the centre “0” of curvature of the circular arc 
and divides the circular arc into two equal halves.
Then considering an elementary segment of mass dm = XRd^ at an angular position relative 
to x-axis and substituting this in the equation
Substituting 0 = we have
-> 2R2 . 
rc ~ sin
This result signifies that,
_> J r dm 
rc=~TT~ \dm
-> p dfy 
rc=T?
Substituting x = R cos t|> and y = R sin cZ. R 
p4> 2+"
The CM of an arc of radius R subtending an angle 0 at the centre of its curvature lies at a 
0 
2Rsin-
—- from its centre of curvature on the radial line passing through the mid- 
0
point of the arc. As the uniform arc is symmetrical about x-axis, logically we can conclude that 
the ^-coordinate of rc must be zero. ’
17
> Locate the centre of mass of the circular arc as shown in the figure.
Fig. 11.25
dm = a dA
P
7
O
, we have
rC =
rc =
Substituting, dA = dx ■ dy and r
->
rc =
System of Particles
Student Task^
3R
Ans. — from the centre C towards right
A
This formula is undoubtedly simpler in the absence of a which can also be a function of distance 
for non-uniform distribution of mass.
(i) Flat sheets : Let us try to apply the above formula for flat surfaces. First of all we will use 
cartesian coordinate system. For this, let us imagine that the sheet lies inx-y plane. Then the elementary 
area dA at a point P having position vector r can be given as
dA = dx • dy
_► —> [ r —>
Substituting, dA = dx ■ dy and r = xi + yj in the formula rc = — I r dA,we haveA
where “
friendly way. So don’t panic to have this strange looking integral and wait for a while to understand this. 
For uniform surface mass distribution, o is a constant. In that case, we can write m = - pdA
r o dA is given as : rc = '
Fig. 11.26
” is the surface (double) integral. We will talk about the surface integral in a user
18
rc = dA
P“> A A
where rc = xci + ycj (x. y)
Then, we have i
x- X
y
y2
X0
y
r ' I ' i i
A
i
i
_X-
II
encouraged to use the basic (original) 
formula in more complex problems.
dy
y
o
Xc = 7 !!xdxdy
yc = 7 fjydxdy
Let me explain the physical significance of surface integral. 
The surface integral “ j ” can be written as double integral “ j j ”, 
It may seem mathematically bit different from line integral “ J ” 
what we call simple integral.
When we analyse the double integral j jx dx dy, we can see 
that j dxdy(= j In the above example, find xc by using the formula xc ~ ~ ^x^4'-
System of Particles
Ex. 12. Locate the centre of mass of a triangular lamina.
Sol. To find the y-coordinate of centre of mass we will have to take 
an elementary strip of length I at a constant y parallel to the base 
of the triangle which lies on x-axis. The area of the strip is given 
as
C
G.R.B. Understanding Physics MECHANICS (Part-B)20-
Then putting dA in the formula,
r d
XC =
d£ - rdr d
we have xc =
where
This gives XC
Similarly, putting dA in the formula yc = , we have
yc =
where A = area of the given sector.
and
Ex. 13. Let us apply the previous formulae to locate the CM of a semicircular plate.
and
Ans.
where
In a cylindrical coordinate system, the coordinates of the centre of mass of a uniform lamina 
can be given as
dr
ik
The area of the shaded element 
of the circular plate is dA = rdr d| 
in cylindrical coordinate system 
Fig. 11.32
.3
— (cosO - cosn)
2 
xc = —f Mr
2 
yc =-~2 ■Mr
H-4
A '
j $xrdr d§
A
x = r cos 
J j(r cos 4>)(r dr)d§
A
A
| j(r sin 4>)(rc7r)c/
A
XC ~ ~~ jr2dr Jcos (}> 
yc = Jsin c7
'R 2.r dr 
0
2 R
~ nR2 3 
47?
” 3%
Alternate method (1): Applying cartesian coordinate system we derive the formula 
yc = ± !ydA',
y - fp2 x2 and dA' = 2x dy
lo“!
r2
tiR2
Sol. Substituting r, = 0, r2 = R, = 0, 4>2 = n and A = —— in the derived formulae, we have
= o
r2dr JJ1 sin (J) d^
System of Particles 21
y
dy = -Substituting
dA'
/dA' =we have A.2 x
/
Then, we have
y
A,
d$- y
x
yc
and substituting
we have yc
where
and
This gives
0and
F///////////J//////////' /Ia
Fig. 11.33
ft! ♦'
Fig. 11.34
R2d$ 
2
2R fit . , ,, 47?^=^-Jos,n + rf+ = -
27? fir . .
*C = 3n C0S irf2
= ■— , where A =-----
3A 2
47?
yc = Y"
xdx
Jr2
2x2dx
Jr2 -x‘
-x2
-x2 ’
=^fo(^2-^2)
This gives
Alternate method (2): Let us take a thin sector OAB of radius 7? which subtends an elementary 
angle d§ at the centre of curvature 0.
This sector behaves as a triangle whose CM “C” is located at a distance
2©
OC = — from its apex O, as proved earlier. Then the mass dm of the elemen-
( 27? sin 
tary sector OAB can be imagined at a distance /I - —~—I-
Then, following the basic formula,
fydm [ydA'
fdm A
2R .y = -y sm ,
1 ff 27? . A= — jl — sin = 7t.
Since, the total mass of the elementary sectors is equal to mass of the semicircular plate, the CM 
of the semicircular arc produced by the CM of the differential sectors gives us the CM of the semicir­
cular plate yc.
where during
This gives dm = 2nvrdr
, 2r 
Since, the CM of the ring is situated at a distance y - — from
71
fy'dm
^dm
Substituting
2r
yc = — for the semicircular ring, 
71
r = -R,
3
4/?
yC=^
jy'dm fy'dm 
~ jdm ~ M ’
will be equal to the yc of the semicircular disc.
y ~ > dm = 2itcrdr ,
7t
fR('2r'\,^__ , x
Jo
yc = —
23
0
Fig. 11.37
Student Task
. For uniform surface M = aA, where A = area of the givenbe given as :
System of Particles
Recapitulating,
Curved surface
If the surface is curved, we can find the elementary area dA as shown as the shaded patch by 
using spherical coordinate system.
> Following the idea of thin coaxial rings and using the basic formula yc 
centre of mass of a thin hollow cone of height H.
Express y and dA' in terms of given parameters by using geometry, to find the position of CM 
of the system of particles.
When the thin curved surface is symmetrical to any axis, y-axis (say), first of all take a differ­
ential (elementary) ring at a height y. Then find its area dA. After that, its mass can be given as 
dm = a dA, where a = surface mass density. Since, the ring is symmetrical about 
y-axis, its CM is located at same height y on they-axis. Finally, the CM of the given surface can 
jydm 
~ M
surface. Substituting M - aA and dm = a dA', where A' = area of the differential ring, we 
have
H
Ans. — from the base on its axis
4 J? sin 
rc(= Xc) =
yc = jy^'
iy dm
If the circular sector of radius R (lamina) is uniform and subtends 
y 
an angle 9 at the centre of curvature and it is placed symmetrical 
with the x-axis, its CM will be situated on x-axis at a distance
0
2
30
G.R.B. Understanding Physics MECHANICS (Part-B)24.
Sol.
Since, 
where
In spherical coordinate system, the 
elementary area can be given as, 
dA = Rd0-r'd, where f = R sin 0 
Fig. 11.38
The shaded area of the ring can be 
given as, dA’ = (2?tr’)R d0, where 
r' = R sin 0 
Fig. 11.39
sin 0 cos 0 dQ J2* dfy
rc = — fr dA to obtain
A iS
rc = — J j r sin 0 dQ dty
However, in simpler problems such as hemispherical 
surface, we can easily find by ranging 0 from 0 to — 
and from 0 to2n , as described in the following example.
dA = (r'dMRdQ), 
r' = 2? sin 0, we have 
dA = R2 sin QdQdfy 
Now we can substitute dA in the formula
rn/2
Jo
wherey = R cos 0 and A = 2nR2 
This gives
Ex. 14. Find the CM of a thin hemispherical bowl of radius R.
Following the basic formula = — j J r sin 0 dQ d§ v/e understand that one component of 
A
r is directed along they-axis, that is what we roughly call
“axis of the hemisphere”. Since, the hemisphere is sym­
metrical about y-axis, logically we can conclude that its
CM is located at y-axis. Then we can write,
R2
rc = Pc = J y sin 0 c/G cZ4>,
R 
yc = —
= — f71/2 sin 2QdQ
2 Jo
_ R
~ 2
Alternate method : After considering an elementary ring of thickness RdQ and radius
r(= R sin 0) at a height y we can find its mass as dm = a dA', where dA' = area of the ring.
Substituting dA' = (2nr')R dQ = 2tc(/? sin 0)2? dQ = 2nR2 sin 0 dQ,
25
yc
rc =
Non-uniform distribution tion,
rc =This gives
Fig.11.40
? = v
The elementary volume can be found by the help of different coordinate systems. In cartesian 
coordinate system, dV -dxdydz. In cylindrical coordinate system, dV = Rdrdfydz and in spherical 
coordinate system, dV = R2dr sin 9 dQ d$.
_> J r dm 
j dm
\pdV
where = volume (or triple) integral and p = /(r)
/ Uniform distribution
If the object is “uniform”, we can take “ p ” out of the integral and cancel it in both numerator and 
denominator. Then, we have
yc
which will give the previous result
R
yc ~ 2
Volume-mass distribution: In volume-mass distribution, mass is distributed throughout the 
given volume (space) oi .er. The density of volume-mass distribution at any point is given as
dm 
P = —dV
This gives us the mass of the elementary segment if the density 
of the material is given as a constant or as a function of distance /-.After 
finding the elementary masst/zn = pdV, substitute this in the basic 
equation,
System of Panicles
we have dm = 2tiR2o sin 9 dQ
Since, the ring is symmetrical about y-axis, its centre of mass will be located at the y-axis at 
y = R cos 9.
As the hollow hemisphere is a combination of coaxial elementary rings of radii ranging from 
r = 0 to r = R and their CM lies ony-axis, we can imagine a continuous arrangement (array) of point 
masses “ JprdV
rr = , where p is a
fpdV
function of r
26
y
This gives yc =
we have
This gives Ans. Fig. 11.41
Student Task
= R2where r2
> Locate the CM of a solid cone of height H following the above method.
3H ,
Ans. — (from the apex on its axis of symmetry)
Let us now tabulate the important results of the previous discussion.
Take an elementary strip of radius r and thickness dy parallel to xz-plane at a distance y. 
The volume of the strip is given as
dV' = nr'
Then substituting dV' in the equation,
Tc=-y-> 
yc
and V = -nR2
3
o
G.R.B. Understanding Physics MECHANICS (Part-B)
If you are not familiar with the given ideas, we will consider some simpler cases such as uniform 
cone, hemisphere, paraboloid etc, where we can locate the CM. Avoiding the above rigorous mathemati­
cal expressions. However, the above expressions of “dP” are useful if the density “p” varies with the 
coordinates or the given object does not possess any symmetry about any axis.
Let us discuss a common procf lure to locate the CM of simple objects.
If we have an object symmetrical about y-axis (say), we will take a thin plate (elementary strip) 
at a height y parallel to xz-plane. Then find its volume. As it is uniform and symmetrical abouty-axis, the 
CM of the differential (elementary) strip lies at its geometrical centre, that is, at the same height on the 
y-axis. In consequence, the given object can be imagined as a continuous line mass distribution iny-axis. 
Now we can find the CM of the line distribution by using the formula,
fydm' fydm 
J dm M
-y1
Let us use this formula.
Lx. 15. Locate the CM of a uniform solid hemisphere.
Sol.
where dm' - pdV' and M = pV
Please note that dm' and dV' are the mass and volume of the elementary strip (but not that of 
the elementary segment of the elementary strip).
\ydv'
V
■2dy
27
C Mid-point
Line
(b) Circular arc
Geometrical centre
(b) Right angled triangle
Surface
(c) Circular sector
38
(d) Hollow hemisphere
(e) Hollow cone
Geometrical centre
Volume (b) Solid hemisphere
(c) Solid cone
(a) Any regular polygon 
such as equilateral 
triangle, square, 
rhombus, rectangle, etc
(a) Regular structures 
like cube, rectangle, 
cylinder
System of Particles
Table 11.1
R
H
rC = ~ from the base
H
rC -~T from the base 4
4y?sin - 
r - 1
rC ~------------
h__
,, yc
CM of some important uniform objects 
(a) Straight wire
rc = -R c 8
b h
xc = yyc = 3
2R . 0 
rC ~ “X-510 7 from O U 2
(ii)
Then,
Volume of the thin (elementary) stripsTable 11.3
y
*-r->
where
R for cone [Fig. (ii)].and r =
y
(ii)
[
'n
28
Table 11.2
dA = (2nr)( —>
second term gives the work done by the net force, that is, Fext(= IF,) on the CM even though any 
force may not act at the CM.
Work done by internal forces : Now let us consider the work done by the internal forces. 
Following the above derived equation substituting p. by f. (internal forces) the total work done by the 
internal forces can be given as :
Ex. 16. Derive an expression for work done by internal forces on the 
system of two interacting particles.
The work done by F{ is, dW\ = F] • dr{ = F- drx
~~~The work done by F2 is, dW2 = F2 - dr2 = - F- dr2
—> —> —> —>
The total work done is, dW = F- di\ - F- dr2
F, = F
/ r21
30 G.R.B. Understanding Physics MECHANICS (Part-B)
= F- drx - dr2
FSince, 21-
Ans.
1.2.
3.
for discrete particle system,
The total work done by the internal forces does not depend upon the reference frame 
which can be given as :
?2i
= F- di\2
Potential energy : Total energy of a system is basically defined as the sum of potential and 
kinetic energy.
We have many sources of potential energy. If the interaction is gravitational, the potential energy 
is gravitational; if the interaction between the particles is electrostatic, potential energy is termed as 
electrostatic potential energy, and so on.
For a system of particles we have, different formulae for potential energy. In the chapter “Gravi­
tation” we will derive the gravitational potential energy of interaction of a system of particles such as :
u„=
where x = relative distance of separation.
Ifx increases, work done is negative; when x decreases work done will be positive. Ifx 
remains constant no work will be done.
Total work done by internal forces on CM is zero because = 0
= ^21n
dr2l„
-> dr21 
dr2ix
Fig. 11.43(b)
- - F- d>2\
dr2\ - ^2i||+ —>
Since, F -L dr2ii , We have 
dW = -F-d,^n
Since, ^2iu and p are parallel, we have 
dW = -Fdr2X
Substituting Jr2l = dr, we have 
dW = -Fdr
Then, the total work done by the internal forces is,
W = jdW =~ jFdr ,
where F = magnitude of the internal force
and r = distance of separation between the particles. 
Let note the following points :
31
where
and
Sol.
yi yc yn
Ans.we have
Potential energy of interaction is equal to negative of work done by internal conservative forces, 
which does not depend on the reference frame.
Since, 
substituting
/77J7777f77777777777fT777777777777777
Fig. 11.44
U = ZUj
=
= g^yi 
^yj = ^m^yc, 
£mf= M), 
u = Mgyc
Ui = m,gy,
Summing-up the potential energy of all particles, we have the totaj po­
tential energy
y2?mn
System of Particles
Vj = gravitational potential at /th point due to all masses except m, 
mj = mass of ilh particle,
For continuous mass distribution, if gravitational potential K at any point of the system is given, 
the gravitational potential energy of the system is given as :
U =- [rdm
2 J
We have similar sets of formulae for potential (energy of interaction for discrete and continuous 
charge distribution) in electrostatics.
As we know, potential energy expression is valid only when the forces are conservative, no 
energy must be wasted in the form of radiant energy (heat, light and sound). For this, we can find the 
work done by all conservative forces using the expression as given in the previous section. Then using 
the expression IFcon = -MJ = (i/final - initial) and setting the value of C/jnjljal we can find the ex­
pression for C/final.
Sometimes we come across a system of particles such as water bottle, sand bag, any rigid or 
non-rigid objects falling under the earth’s gravitational field. In this case how to find the potential energy 
of the system? Let us see in the following example.
Ex. 17. Derive an expression for the gravitational potential energy of a system ofparticles due to the 
earth's gravity.
For any system of discrete distribution of particles, let us take the 1th 
particle of mass m, which is situated at a height y, (say), gravitational 
potential energy of m, is given as :
T1
„m2
Please note that U = Mgyc is strictly valid only when the size of the object (system of par­
ticles) is much smaller than the radius of the earth, so that we can assume a constant value of 
g (uniform gravitational field) for all particles.
G.R.B. Understanding Physics MECHANICS (Part-B)
Total kinetic energy, K =
where v, = vjC + vc ,
This gives
Since 'Em, vlC = 0 (Sum of momenta of all particles relative to the CM of the system is zero),
we have
Then, we have
The above expression tells us that;
32
Student Task
^3
Fig. 11.45
2
KE of a system of particles is equal to the KE of the system relative to CM plus KE of the CM 
(A/ 2
I ~ vc J • Hence, KE of a system is minimum relative to the frame attached with CM which is 
equal to K'. K' is called internal KE which does not depend on any reference frame. Thus, K' 
is an internal property of the system.
v, = velocity of , vlC = velocity of m, relative to CM and vc = velocity of CM.
K = - I vtc + vc I2
I 2 2 -* ->
= +VC +2viC-vc)
I 2 I 2
= + ~^mt)vc + ^miviC' vc
—> —► —> —►
substituting viC- vc = vc(Zmi viC) = 0 ,
l?l 0
K = S-,7i,v,c +-(Zw,)vc>
where 'Em, = M = total mass of the system and Z m,v2c = sum of kinetic energies of the 
particles relative to CM of the system, which is termed as internal kinetic energy or kinetic energy 
relative to CM frame (= K', say;)
> In the foregoing example, can you call "U" the potential energy of the given system of 
particles? If no, is it correct to call it potential energy of interaction between the system of the 
particle and the earth? Explain. Ans. No, yes.
After understanding the meaning of potential energy of a system of interacting particles, let us 
derive the expression of kinetic energy of a system of particles.
Kinetic energy of a system of particles : Total kinetic energy of a system of n particles (say) 
is equal to sum of kinetic energy of all particles relative to ground.
v ' 2
33System of Particles
Sol.
K' = K -
where VC
,2and
2
Then, we have
Ans.we have
Aliter :
Substituting VC
2
we have
Ans.
k
Sol.
3m 2m
Fig. 11.46
.2
2
Then, we have Ans.
->
v2-
2
-> 77^ + m2v2 
m{ + m2
I-* **l2|v,-v2|
1
^^2
V° k 2v0
m\ V1+ m2 v2
+ m2
/fyVl + m2v2 
m\ + m2
and v2
Jr 
where C - —x
2
 m} V| + m2 v2
7»l + m2
M = (nj| + w2)
1 2 1 2K = 2W>V1 + 2m2V1
Ex. 18. Find the KE of a system of two particles of masses m\ and m2 having velocities v( 
relative to the CM of the system.
The kinetic energy of the system relative to CM is,
Mv2c
2 ’
1 m{m2
2 m\ + m2
Ex. 19. Two particles of mass 3 m and 2m are moving towards each other with speeds v0 and 2v0 
respectively. If the particles are interconnected by a massless spring ofstiffness k and the spring 
is compressed by a length x, find the (a) total mechanical energy of 
the system (b) total internal energy of the system.
(a) Total mechanical energy is given as :
E = U + K, 
and /C = l(3m)v02+l(2/W)(2v0)2
E = — mvQ + —x2 
2 0 2
,,, 1 2 I 2 1 z x
K = 2mivi +2OT2V2 "2 +/”2
Expanding the square and simplifying the factors,
= m\m2 | vi - v2 |2 
2(77?] + 77?2) 
1 2 1 2
“ 2 mlvlC + ~m2v2C , 
—> —>
where vic = Vj- vc and v2C = v2- vc
7H|V] + 77?2V2 
777] + m2
K' = I"J1 vl-
34
(b)
This gives Ans.
Student Task
fTjnI - sum of work done by all internal forces,where
k
F
Fig. 11.47
Sol.
.2This gives Ans.
Ex. 20. Two blocks of masses m} and m2 interconnected by a light 
spring of stiffness k, are kept on a horizontal surface. The 
coefficient of friction between the blocks and horizontal 
surface is P- If a horizontal force F (> iim2g) acts on 
m2, the block m2 slides. In consequence the spring elon­
gates by a length x. Find the work done by (i) internal forces (zz) external force F(iii) all forces, 
on the system of the blocks and spring.
(i) For the system (m} + spring + mf), spring force is the internal force. The work done by 
the spring is given as :
For a system of particles, sum of the work done by all forces (internal and external) acting on 
each element or particle of the system is equal to sum of the change in KE of all particles of the 
system. This is what we call work-energy theorem for system of particles.
and
The above expression tells us that;
^ext = sum WOfk done by all external forces
AX = sum of change in KE of all particles of the system.
G.R.B. Understanding Physics MECHANICS (Part-B)
.,2
Wsp=-frar.
where F = kx and dr = dx
HIU H U t H H H H H f H n ! n ! in n n
Ir 
j
> In the above example, how does £inl change? Explain.
I 2Ans. Since, Eint = — kx + K , when the relative separation changes, x
and K' change. However, Einl remains constant.
Work-Energy Theorem : Now we have all expressions in our hand such as work done by 
internal