NCERT Solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Physics
Chapter : 7
Chapter Name : System of particles and Rotational Motion

Q7.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder. (iii) ring, and (iv) cube,
each of uniform mass density. Does the centre of mass of a body necessarily lie inside the
body?

Answer. Geometric centre; No
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated.
For the given geometric shapes having a uniform mass density, the C.M.
lies at their respective geometric centres. The centre of mass of a body need not necessarily lie
within it. For example, the C,M, of bodies such as a ring, a hollow sphere, etc., lies outside the
body.

Page : 178 , Block Name : ExerciseQ7.2 In the HCI molecule, the separation betw

een the nuclei of the two atoms is about 1.27 A (1 A = 1 10 m). Find the approximate
−10

location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive
as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer. Let us choose the nucleus of the hydrogen atom as the origin for measuring distance
Mass of hydrogen atom,m = 1 unit (say) Since chlorine
1

atom is 355 times as massive as hydrogen atom, mass of chlorine atom, m 355 units
2

−10
x1 = 0 and x2 = 1.27A = 1.27 × 10 m

Distance of centre of mass of HCl molecule from the origin is given by
−10
m1 x1 +m2 x2 1×0+35.5×1.27×10
X = = m
m1 +m2 1+35.5

35.5×1.27 −10 −10
= × 10 m = 1.235 × 10 m = 1.235A
36.5

Page : 178 , Block Name : Exercise

Q7.3 child sits stationary at one end of a long trolley moving uniformly with a speed V on a
smooth horizontal oor. If the child gets up and runs about on the trolley in any manner, what
is the speed of the CM of the (trolley + child) system?

Answer. No change The child is running arbitrarily on a trolley moving with velocity v.
However, the running of the child will produce no effect on the velocity of the centre of mass
of the trolley. This is because the force due to the boy's motion is purely internal. Internal
forces produce no effect on the moti«, cf the bodies on which they act. Since no external force
is involved in the boy—trolley system, the boy's motion Hill produce no change in the velocity
of the centre of mass of the trolley.

Page : 178 , Block Name : Exercise

Q7.4 Show that the use of the triangle contained between the vectors a and b is one half of the
magnitude of a x b.

Answer.
→ → →
Let a be represented OP and b be represented

→
by OQ. Let ∠P OQ = θ, Fig.

Complete the ll gm OPRQ. Join P Q .

Draw QN ⊥ OP

QN QN
In ΔOQN , sin θ = =
OQ b

QN = b sin θ

→ →
Now, by definition, |a × b| = ab sin θ = (OP )(QN )

2(OP )(QN )
= = 2 × area of ΔOP Q
2

∴ area of ΔOP Q =
1
→
∣ ′ ∣
∣a × b ∣
→
2

which was to be proved.

Page : 178 , Block Name : Exercise

Q7.5 Show that a. (b x c) is equal in magnitude to the volume of the parallelepiped formed on
the three vectors, a, b and c.

Answer. A parallelepiped with origin O and sides a, b, and c shown is in the following gure.

Volume of the given parallelepiped = abc

→ →
OC = a

→ →
OB = b

→
OC = c →
Let n
^ be a unit vector perpendicular to both b and c. Hence, n
^ and a have the same

direction.

→ → ^
∴ b × c = bc sin θn
∘
= bc sin 90 n
^

= bc^
n

→ →
a ⋅ (b × c ) →
= a ⋅ (bcn)
^

→ →
a ⋅ (b × c ) →
= a ⋅ (bcn)
^

= abc cos θn
^
∘
= abc cos 0

= abc

= Volume of the parallelepiped

Page : 178 , Block Name : Exercise

Q7.6 Find the components along the x. y, z-axes of the angular momentum I of a particle,
whose position vector is r with components x, y, z and momentum is p with components
p , p and p Show that if the particle moves only in the x-y plane the angular momentum
x y z

has only a z- component.

Answer.
lx = yρz − zρy

ly = 2px − xpz

lz = xpy − ypx

^
→ ^
Linear momentum of the particle, p = px i + py j + pz k

→ ^ ^ ^
Position vector of the particle, r = x i + y j + zk

Angular momentum, l = r × p
→ → →
^ ^ ^ ^ ^ ^
= (x i + y j + zk) × (pr i + py j + pr k)

^ ^
Comparing the coefficients of i , and k, we get: ⎫
⎪
⎪
⎪
⎪
lx = ypx − zpy
⎬
ly = xpz − zpx
⎪
⎪
⎪
⎭
⎪
lz = xpy − ypx

The particle moves in the x-v plane Hence, the z-component of the position vector nd
linear momentum vector becomes zero i.e.,
lx = 0 ⎫
⎪
ly = 0 ⎬
⎭
⎪
le = xpy − ypn

Therefore, when the particle is con ned to move in the x-V plane, the direction of angular
momentum is along the z-direction.

Page : 178 , Block Name : Exercise

Q7.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel
lines separated by distance d. Show that the vector angular momentum of the two particle
system Is the same whatever be the point about Which the angular momentum is taken.

Answer. Let at a certain instant two particles be at points P and Q, as shown in the following
gure,

Angular momentum of the system about point P:
→
LP = mv × 0 + mv × d

= mnd

Angular momentum of the system about point Q:
→
L0 = mv × d + mv × 0

= md

QR = y

∴ PR = d − γ

Angular momentum of the system about point R:
→
LR = mv × (d − y) + mv × y

= mvd − mvy + mvy

= mnd

Comparing equations (i),(ii) and (iii), we get:
→ →
Lp = LQ = LR →
We infer from equation (W) that the angular momentum of a system does not depend on
the point about which it is taken.

Page : 178 , Block Name : Exercise

Q7.8
A non-uniform bar of weight W is suspended at rest by two

strings of negligible weight as shown in Fig. The angles made by the strings

∘ ∘
with the vertical are 36.9 and 53.2 respectively. The bar is 2m long.

Calculate the distance d of the centre of gravity of the bar from its left end.

Answer.

Length of the bar, l = 2m

T1 and T2 are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have
∘
T1 sin 36.9 = T2 sin 53.1
∘
T1 sin 53.1
=
T2 sin 36.9

0.800 4
= =
0.600 3

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T1 cos 36.9 × d = T2 cos 53.1(2 − d)

T1 × 0.800d = T2 0.600(2 − d)

4
× T2 × 0.800d = T2 [0.600 × 2 − 0.600d]
3

1.067d + 0.6d = 1.2

1.2
∴ d =
1.67

= 0.72m

Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m frorn its left end.

Page : 178 , Block Name : Exercise

Q7.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of
gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on
each front wheel and each back wheel.

Answer.
Answer: Let F1 and F2 be the forces exerted by the level ground on front wheels

and back wheels respectively. Considering rotational equilibrium about the front

wheels, F2 × 1.8 = mg × 1.05 or F2 = 1.05/1.8 × 1800 × 9.8N = 10299N Force on

each back wheel is = 10290/2Nor5145N .

F1 × 1.8 = mg(1.8 − 1.05) = 0.75 × 1800 × 9.8

or F1 = 0.75 × 1800 × 9.8/1.8 = 7350N

Force on each front wheel is 7350/2 N or 3675N .

Page : 178 , Block Name : Exercise

Q7.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the
moment of inertia of the sphere about any of its diameters to be 2MR /5, where M is the
2

mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to
1 be 1/4MR , nd the moment of inertia about an axis normal to the disc passing through a
2

point on its edge.

Answer. (a) 7

5
MR
2

The moment of inertia (M.I.) of a sphere about its diameter = 2

5
MR
2

According to the theorem of parallel axes, the moment of inertia of a body about any axis is
equal to the sum of the moment of inertia of the body about a parallel axis
passing through its centre of mass and the product of its mass and the square of the distance
between the two parallel axes.
The M.I. about a tangent of the sphere = M R + M R = M R
2 2 2 7 2

5 5

(b) 3

2
MR
2

The moment of inertia of a disc about its diameter = 1

4
MR
2

According to the theorem of perpendicular axis, the moment of inertia of a planar body
(lamina) about an axis perpendicular to its plane is equal to the sum of its moments of Inertia

about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the
body.
1 2 1 2 1 2
The M.I. of the disc about its centre = MR + MR = MR
4 4 2

The situation is shown in the given figure.

Applying the theorem of parallel axes:
The moment of inertia of out axis normal to the disc md passing through a point on
1 2 2 3 2
= MR + MR = MR
2 2

Page : 178 , Block Name : Exercise

Q7.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both
having the same mass and radius. The cylinder is free to rotate about its standard axis of
symmetry, and the sphere Is free to rotate about an axis passing through its centre. Which Of
the two will acquire a greater angular speed after a given time?

Answer.
Let M be the mass and R the radius of the hollow cylinder, and also of

2
the solid sphere. Their moments of inertia about the respective axes are 11 = MR

2
and 12 = 2/5MR

Let T be the magnitude of the torque applied to the cylinder and the sphere,
producing angular accelerations α and α respectively. Then T = | a = I a The angular
1 2 1 1 2 2

acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular
speed after a given time

Page : 179 , Block Name : Exercise

Q7.12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s . The −1

radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the
cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Answer.

M = 20kg

−1
Angular speed, w = 100rads ; R = 0.25m

Moment of inertia of the cylinder about its axis

2 2 2 2
= 1/2MR = 1/2 × 20(0.25) kgm = 0.625kgm

Rotational kinetic energy.

2
Er = 1/2 lw 2 = 1/2 × 0.625 × (100) J = 3125J

Angular momentum,

L = Iw = 0.625 × 100Js = 62.5Js

Page : 179 , Block Name : Exercise

Q7.13 (a) A child stands at the centre of a turntable with his arms outstretched. The turntable
is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child
if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial
value? Assume that the turntable rotates without friction, (b) Show that the child's new
kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you
account of this increase in kinetic energy?

Answer. (a) Suppose, initial moment of inertia of the child is II Then nal moment of inertia,
2
I2 = Il
5
−1

v1 = 40rev min

I1 ω1 = I2 ω2 or I1 (2πv1 ) = I2 (2πv2 )

I1 v1 I1 × 40 −1

v2 = = = 100rev min
2
I2 × I1
5

(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, E = I ω
1 2
F 2 2
2

Initial kinetic rotation, E 1 =
1

2
I1 ω
2

1

1 2
EF I2 (t)
2 2
=
E1 1 2
I1 ω
2 1

2
2 I1 (100)
=
2
5 I1 (40)
2 100×100
= ×
5 40×40

5
= = 2.5
2

∴ EF = 2.5E1

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

Page : 179 , Block Name : Exercise

Q7.14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40

cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?
What is the linear acceleration of the rope? Assume that there is no slipping.

Answer.
Here, M = 3kg, R = 40cm = 0.4m

Moment of inertia of the hollow cylinder about its axis.

2 2 2
1 = MR = 3(0.4) = 0.48kgm

Force applied F = 30N

∴ Torque, τ = F × R = 30 × 0.4 = 12N − m.

If α is angular acceleration produced, then from τ = I α

−2
Linear acceleration, a = Rα = 0.4 × 25 = 10ms

Page : 179 , Block Name : Exercise

Q7.15 To maintain a rotor at a uniform angular speed of 200 rad , an engine needs to transmit
a torque of 180 Nm. What is the power required by the engine?
(Note: Uniform angular velocity in the absence of friction implies zero torque. In practice,
applied torque is needed to counter frictional torque). Assume that
the engine is 100 ef cient.

Answer.
−1
a = 200 rad s ; Torque, = 180N − m
1

since, Power, P = Torque (T) × angular speed (w)

= 180 × 200 = 36000watt = 36KW

Page : 179 , Block Name : Exercise

Q7.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the
hole is at R/2 from the centre of the original disc.Locate the centre of gravity of the resulting
at body.

Answer. R/6; from the original centre of the body and opposite to the centre of the cut portion.

Mass per unit area of the original disc = o
Radius of the original disc = R
Mass of the original disc. M = nR σ 2

The disc with the cut portim is 10wn in the following gure

R
Radius of the smaller disc =
2

2
R 1 2 M
π( ) σ = πR σ =
2 4 4

Let O and O' be the respective cultures of the original disc and the disc cut off from the
original. As p« the de nition of the centre of mass, the center of mass of the original disc is
supposed to be concentrated at O, while that of the smaller disc is supposed to be
concentrated at O'
It is given that
′ R
OO =
2

After the smaller disc has been cut from the original, the remaining portion is considered to be
a system of two masses. The two masses are: M (concentrated at O), and
′ M
−M (= )
4 ′
concentrated at 0

(The negative indicates that this portion has removed from the original disc. )
Let x be the distance through which the centre of mass of the remaining portion shifts
from point O.
The relation between the centres of masses of two masses is given as:
m1 r1 +m2 r2
x =
m1 +m2

For the given system, we can write:

R
M × 0 − M × ( )
2

−M
x = ′
M +(−M )

−M −M R 4 −R
= M
= × =
M− 8 3M 6
4

(The negative sign indicates that the centre of mass gets 4-,shifted toward the left of point o.)

Page : 179 , Block Name : Exercise

Q7.17 A metre stick is balanced on a knife edge at its centre. V.hen two coins, each of mass 5g
ue put cone on top of the other at the 12.0 cm rnn, the stick is found to be balanced at 45.0 cm.
What is the mass of the metre stick?

Answer. Let W and W be the respective weights of the metre stick and the coin.
′

For equilibrium about C, the 45cm mark,

10g(45 − 12) = mg(50 − 45)

10g × 33 = mg × 5

⇒ m = 10 × 33/5

or m = 66 grams.

Page : 179 , Block Name : Exercise

Q7.18 A solid sphere rolls down two different inclined planes of the same heights but different
angles of inclination,
(a) Will it reach the bottom with the same speed in each case?
(b) Will it take longer to roll down one plane than the other?
(c) If so, which one and why?

Answer.
(a) Using law of conservation of energy,

1 2 1 2
mv + Iω = mgh
2 2

2
1 2 1 2 2 v
or mv + ( mR ) = mgh
2 2 5 2
R

7 10gh
2
or v = gh or v = √
10 7

since h is same for both the inclined planes therefore v is the same.

(b) 1 g sin θ 2 g sin θ 2 5g sin θ 2
l = ( )t = t = t
2 2
K 2 14
1+ 2(1+ )
5
R2

14l
or t = √
5g sin θ

h h
Now, sin θ = or l =
l sin θ
h h
Now, sin θ = or l =
l sin θ

1 14h
∴ t = √
sin θ 5g

Lesser the value of θ, more will be t

(c) Clearly, the solid sphere will take longer to roll down the plane with smaller inclination.

Page : 179 , Block Name : Exercise

Q7.19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal oor so that its centre of
mass has a speed of 20 cm/s. How much work has to be done to stop it?

Answer.
Here, R = 2m, M = 100kg

v = 20cm/s = 0.2m/s

2 2
Total energy of the hoop = 1/2Mv + 1/2lw

2 2 2
= 1/2Mv + 1/2 (MR ) w

2 2 2
= 1/2Mv + 1/2Mv = Mv
2 2
Work required to stop the hoop = total energy of the hoop W = M v = 100(0.2) =

4 Joule.

Page : 179 , Block Name : Exercise

Q7.20 The oxygen molecule has a mass of 5.30 × 10 and a moment of inertia of
−26

kgm about an axis through its centre perpendicular to the lines joining the two
−45 2
1.94 × 10

atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic
energy of rotation is two thirds of its kinetic energy of translation. Find the average angular
velocity of the molecule.

Answer.
−26
Mass of an oxygen molecule, m = 5.30 × 10 kg

−46 2
Moment of inertia, I = 1.94 × 10 kgm

Velocity of the oxygen molecule, v = 500m/s

The separation between the two atoms of the oxygen molecule = 2r
m
Mass of each oxygen atom =
2

Hence, moment of inertia I , is calculated as:

m 2 m 2 2
( )r + ( )r = mr
2 2

i
r = √
m

−46
1.94×10 −10
√ = 0.60 × 10 m
−66
5.36×10

−66
K ⋅ 36 × 10
1 2 2 1 2
Iω = × × mv
2 3 2

2 2 2 2
mr ω = mv
3

2 v
ω = √
3 r

2 500
= √ × −10
3 0.6×10

12
= 6.80 × 10 rad/s

Page : 179 , Block Name : Exercise

Q7.21 A solid cylinder rolls up an inclined plane of angle of inclination 300. At the bottom of
the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?

Answer. A solid cylinder rolling up an inclination is shown in the follow ng gure.

Initial velocity of the solid cylinder, v = 5 m/s
∘
Angle of inclination, θ = 30

Helght reached by the cylinder = h

(a) Energy of the cylinder at point A:

KEmt = KErmas
1 2 1 2
Iω = mv = mgh
2 2

1 2
Moment of inertia of the solid cylinder, I = mr
2

1 1 2 2 1 2
∴ ( mr ) ω + mv = mgh
2 2 2

1 2 2 1 2
mr ω + mv = mgh
4 2
1 2 1 2
∴ v + v = gh
4 2

3 2
v = g/r
4

2
3 v
∴ h =
4 g

3 5×5
= × = 1.91m
4 9.8

BC
sin θ =
AB

∘ h
sin 30 =
AB

1.91
AB = = 3.82m
0.5

(b) For radius of gyration K, the velocity of the cylinder at the instance when it rolls
back to the bottom is given by the relation:
1

2

2gh
v = ( )
2
K
1+
R2

1

2

2gΔB sin θ
∴ v = ( )
2
K
1+
R2

2
2 R
For the solid cylinder, K =
2

1

2
2gAB sin θ
∴ v = ( )
1
1 +
2
1

4 2

= ( gAB sin θ)
3

AB
t =
v
AB
=
1 1

2 2
4 3AB
( gAB sin θ) = ( )
3 4g sin θ

1

11.46 2

= ( ) = 0.764s
19.6

Therefore, the total time taken by the cylinder to return to the bottom is (2 x 0.754)
1.53 s.

Page : 179 , Block Name : Exercise

Q7.22 As shown in Fig. the two sides of a step ladder BA and CA are 1.6 m long and hinged at
A. A rope DE, 0.5 m is tied halfway up. A weight 40 kg is suspended from a point F, 1.2 m from
B along the ladder BA Assuming the oor to be frictionless and neglecting the weight of the
ladder, nd the tension in the rope and forces exerted by the oor on the ladder. (Take g = 9.8
m )(Hint: Consider the equilibrium of each side of the ladder separately.)
2

Answer.

NB = Force exerted on the ladder by the floor point B

Nc = Force exerted on the ladder by the floor point C

BA = CA = 1.6m

DE = 0.5m

BF = 1.2m

Mass of the welght, m = 40kg

△ABI and ΔAIC are similar

∴ BI = IC

Hence, I is the midpoint of BC.

DE∥BC

BC = 2 × DE = 1m

AF = BA − BF = 0.4m … (i)

D is the midpoint of AB .

Hence, we can write:
1
AD = × BA = 0.8m
2

Using equations (i) and (ii), we get:

FE = 0.4m

Hence, F is the midpoint of AD.

FG∥DH and F is the midpoint of AD. Hence, G will also be the mid-point of AH .

△AFG and △ADH are similar
FG AF
∴ =
DH AD

FG 0.4 1
= =
DH 0.8 2

1
FG = DH
2

1
FG = DH
2
1
= × 0.25 = 0.125m
2
In △ADH :

2 2
AH = √AD − DH

2 2
= √(0.8) − (0.25) = 0.76m

For translational equilibrium of the ladder, the upward force should be equal to the
downward force.
Nc + NB = mg = 392 … (iii)

For rotational equilibrium of the ladder, the net moment about A is:
−NB × BI + mg × FG + Nc × Cl + T × AG − T × AG = 0

−NB × 0.5 + 40 × 9.8 × 0.125 + NC × (0.5) = 0

(NC − NB ) × 0.5 = 49

NC − NB = 98

Adding equations ( iii) and (iv), we get:

NC = 245N

NB = 147N

For rotational equilibrium of the side AB, consider the moment about A.

−NB × BI + mg × FG + T × AG = 0

−245 × 0.5 + 40 + 9.8 × 0.125 + T × 0.76 = 0

0.76T = 122.5 − 49

∴ T = 96.7N

Page : 179 , Block Name : Additional Exercise

Q7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg
weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man
then brings his arms close to his body with the distance of each weight from the axis changing
from 90 cm to 20cm. he moment of inertia of the man together with the platform may be talæ-
l to be constant equal to 7.6 kg rn².
(a) What is his new angular speed? (Neglect friction.)
(b) Is kinetic energy conserved in the process? If not, from where does the change come
about?

Answer.
(a) 58.88 rev/min (b) No

2
(a)Moment of inertia of the man-platform system = 7.6kgm

Moment of inertia when the man stretches his hands to a distance of 90cm :

2
2 × mr

2
= 2 × 5 × (0.9)

2
= 8.1kgm

Initial moment of inertia of the system, l 1 = 7.6 + 8.1 = 15.7kgm
2

Angular speed, ω = 300 rev/min

Angular momentum, L1 = I1 ω1 = 15.7 × 30

Moment of inertia when the man folds his hands to a distance of 20 cm:

2
2 × mr

2 2
= 2 × 5(0.2) = 0.4kgm

2
Final moment of inertia, Iℓ = 7.6 + 0.4 = 8kgm

Final angular speed = ωf

From the conservation of angular momentum, we have:

I 1 ω1 = I 1 ωf

15.7×30
∴ ωr = = 58.88rev/min
8

(b)Kinetic energy is not conserved in the given process, In fact, with the decrease in the
moment of inertia, kinetic energy increases. The additional kinetic energy comes from
the work done by the man to fold his hands toward himself.

Page : 180 , Block Name : Additional Exercise

Q7.24 A bullet of mass 10 g and speed 500 m/s is red into a door and gets embedded exactly
at the centre of the door. The door is 1.0 m w de and weighs 12 kg. It is hinged at one end and
rotates about a vertical axis practically without friction. Find the angular speed of the door
just after the bullet embeds into it.

Answer.
−3
Mass of the bullet, m = 10g = 10 × 10 kg

velocity of the bullet, v = 500m/s

Thickness of the door, L = 1m
1
Radius of the door, r = m
2

Mass of the door, M = 12kg

Angular momentum imparted by the bullet on the door:
a = mvr
−3 1 2 −1
= (10 × 10 ) × (500) × = 2.5kgm s
2

Moment of inertia of the door:

1 2
I = ML
3
1 2 2
= × 12 × (1) = 4kgm
3

But α = I ω
α
∴ ω =
I

2.5 −1
= = 0.625rads
4

Page : 180 , Block Name : Additional Exercise

Q7.25 Two discs of moments of inertia I and I about their respective axes (normal to the
1 2

disc and passing through the centre), and rotating with angular speeds ω and ω are brought
1 2

into contact face to face with their axes of rotation coincident. (a) What is the angular speed of
the two-disc system? (b) Show that the kinetic energy of the combined system is less than the

sum of the initial kinetic energies of the two discs. How do you account for this loss in energy?
Take ω ≠ ω .
1 2

Answer.
(a)

Moment of inertia of disc I = I1

Angular speed of disc I = ω1

Angular speed of disc II = I2

Angular momentum of disc II = ω1

Angular momentum of disc I, L1 = I1 ω1

Angular momentum of disc II, L2 = I2 ω2

Total initial angular momentum, L1 = I1 ω1 + I2 ω2

When the two discs are joined together, their moments of inertia get added up.

Moment of inertia of the system of two discs, I = I1 + I2

Moment of inertia of the system of two discs, I = I1 + I2

Let ω be the angular speed of the system.

Total final angular momentum, Lf = (I1 + I2 ) ω

L1 = L1

I1 ω1 + I2 ω2 = (I1 + I2 ) ω

I1 ω1 +I2 ω2
∴ ω =
I1 +I2

(b) Kinetic energy of disc E1
=
1

2
I1 ω
2
1

Kinetic energy of Disc II E 2 =
1

2
I2 ω
2

2

When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system I = I + I 1 2

Angular speed Of the system = ω
Final kinetic energy Ef :

1 2
= (I1 + I2 ) ω
2

2 2
1 I1 ω1 +I2 ω2 1 (I1 ω1 +I2 ω2 )
= (I1 + I2 ) ( ) =
2 I1 +I2 2 I1 +I2

∴ E1 − Ef
2

1 (I1 ω1 +I2 ω2 )
2 2
= (I1 ω + I2 e ) −
2 1 2 2(I1 +I2 )

2 2 2 2
I ω I ω 2I1 I2 ω1 ω2
1 2 1 2 1 1 1 1 2 2 1
= I1 ω + I2 ω − − −
2 1 2 2 2 (I1 +I2 ) 2 (I1 +I2 ) 2 (I1 +I2 )

1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2
= [ I ω + I1 I2 ω + I1 I2 ω + I ω − I ω − I ω − I 1 I 2 ω1 ω2 ]
1 1 1 2 2 1 1 2 2
(I1 + I2 ) 2 2 2 2 2 2

I1 I2
2 2
= [ω + ω − 2ω1 ω2 ]
1 2
2 (I1 + I2 )

2
l1 I2 (ω1 − ω2 )
=
2 (I1 + I2 )

∴ E1 − E1 > 0

E1 > Ef

The loss of KE can be attributed to the frictional force that comes into play when the two
discs come in contact with each other.

Page : 180 , Block Name : Additional Exercise

Q7.26 (a) prove the theorem Of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the X-Y plane from an axis through the
origin perpendicular to the plane x + y )
2 2

(b) Prove the theorem of parallel axes.

Answer. (a)The theorem of perpendicular axes states that the moment of inertia of a planar
body (lamina) about an axis perpendicular to its plane is equal to the sum of its
moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying
in the plane of the body.
A physical body with centre O and a point mass m,in the x y plane at (x, V) is shown in the
following gure.

2
Moment of inertia about x -axis, Ix = mx

2
Moment of inertia about y -axis, Iy = my
2 2
Ix + Iy = mx + my

2 2
= m (x + y )

2
2 2
= m(√x + y )

Ix + Iy = Ix

Hence, the theorem is proved.

(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is
equal to the sum of the moment of inertia of the body about a parallel axis
passing through its centre of mass and the product of its mass and the square of the distance
between the two parallel axes.

Suppose a rigid body is made up of n particles, having masses m m m … , m at 1, 2t 3, n

perpendicular distances r , r , r … , r respectively from the centre of mass O of the rigid
1 2 3, n

body .
The moment Of inertia about axis RS passing through the point O:
n 2
IRs = ∑ mr r
i=1 i

The perpendicular distance of mass from the axis QP = a + r,
Hence, the moment of inertia about axis QP:
n
2
IQP = ∑ mi (a + ri )

i=1

n
2
2 2
= ∑ mi (a + r + 2σri )
i

i=1
n 2 n 2 n
= ∑ mi a + ∑ mi r + ∑ mi 2ari
i=1 i=1 i i=1

n 2 n 2
= IRS + ∑ mi a + 2∑ ml ar
i=1 i=1 i

Now, at the centre Of mass, the moment Of inertia Of all the particles about the axis passing
through the centre of mass is zero, that is,
n
2∑ mi ari = 0
i=1

∵ a ≠ 0

∴ ∑ mi r i = 0

Also
n
∑ mi = M ; M = Total mass of the rigid body
i=1

2
∴ IQP = IRS + M a

Hence, the theorem is proved.

Page : 180 , Block Name : Additional Exercise

Q7.27 Prove the result that the velocity v of translation of a rolling body (like a ring, disc,
cylinder or sphere) at the bottom of an inclined plane of a height h is given by v
2gh
2
=
2 2
(1+k /R )

Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the
radius Of gyration Of the body about its symmetry axis, and R is the radius Of the body. The
body starts from rest at the top of the plane.

Answer. A body rolling on an inclined plane of height h,is shown in the following gure:

m = Mass of the body

R = Radius of the body

K = Radius of gyration of the body

v = Translational velocity of the body

h = Height of the inclined plane

g = Acceleration due to gravity

Total energy at the top of the plane, E1 = mgh
1 2 1 2
= Iω + mv
2 2

2 v
I = mk and ω =
R

2
1 2 v 1 2
∴ Eb = (mk ) ( ) + mv
2 2 2
R

2
1 2 k 1 2
= mv + mv
2 R2 2

2
1 2 k
= mv (1 + )
2 2
R

From the law of conservation of energy, we have:
ET = Eb

2
1 2 k
mgh = mv (1 + )
2 2
R

2gh
∴ v = 2 2
(1+k /R )

Hence, the given result is proved.

Page : 180 , Block Name : Additional Exercise

Q7.28 A disc rotating about its axis with angular speed ω Dais placed lightly (without any
o

translational push) on a perfectly frictionless table. The radius of the disc is R. What are the
linear velocities of the points A, 3 and C on the disc shown in Figure Will the disc roll in the
direction indicated?

Answer. v A = Rωoi ; vB = Rωoi

Angular speed of the disc = ω0

Radius of the disc = R

Using the relation for linear velocity, v = ω0 R

For point A:
vA = Rw ; in the direction tangential to the right
0

For point B:
vB = Rw ; in the direction tangential to the left
0

For point C:
vc = (
R

2
) ωo , in the direction same as that of vA
The directions of motion of points A, B, and C on the disc are shown in the following gure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of
friction is essential for the rolling of a body.

Page : 180 , Block Name : Additional Exercise

Q7.29 Explain why friction is necessary to make the disc in Figure roll in the direction
indicated.
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect
rolling begins.
(b) What is the force of friction after perfect rolling begins?

Answer. A torque is required to roll the given disc. As per the de nition of torque, the rotating
force should be tangential to the disc. Since the frictional force at point B is along the
tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear
velocity at point B is tangentially leftward. Hence, frictional force will act tangentially
rightward. The sense Of frictional torque before the start Of perfect rolling is perpendicular to
the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling
Rill begin when the velocity at that point becomes equal to zero. This will make the frictional
force acting on the disc zero.

Page : 180 , Block Name : Additional Exercise

Q7.30 A solid disc and a ring, both of radius 10 cm are placed on a horizontal table
simultaneously, with initial angular speed equal to 10 n rad Which of the two will
start to roll earlier 7 The coef cient of kinetic friction is = 0.2.

Answer.
Radili of the ring and the disc, r = 10cm = 0.1m

−1
Initial angular speed, ω0 = 10n rad s

coefficient of kinetic friction, μk = 0.2

Initial velocity of both the objects, u = 0

Motion of the two objects is caused by frictional force. As per Newton's second law of
motion, we have frictional force, f = ma
μb mg = ma

Where,

a = Acceleration produced in the objects

m = Mass

∴ a = μk g … (i)

As per the rst equation of motion, the nal velocity of the objects can be obtained as:
v = u + at

= 0 + μk gt

= μk gt … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and
cause reduction in the initial angular speed.
Torque, T = −I a

a = Angular acceleration

μx mgr = −I a

−μk mgr
∴ α =
I

Using the rst equation of rotational motion to obtain the nal angular speed

ω = ω0 + αt

−μ1 mgr
= ω0 +
I
ω = ω0 + αt

−μs mgr
= ω0 + t
I
Rolling starts when linear velocity, v = rω

μk gmrt
∴ v = r (ω0 − )
I

Equating equations (ii) and ( v ), we get:

μ1 gmrt
μ1 gt = r (ω0 − )
l

2
μk gmr t
= rω0 −
I
2
For the ring: I = mr
2
μk gmr t
∴ μL gt = rω0 − 2
mr

2μk gt = rω0
rω0
∴ tr =
2μk g

0.1×10×3.14
= = 0.80s
2×0.2×9.8
1 2
I = mr
2
2
μk gmr f
∴ μk gtd = rω0
1 2
mr
2

= rω0 − 2μk gt
rω0
∴ td =
3μk g

0.1 × 10 × 3.14
= = 0.53s
3 × 0.2 × 9.8

Since t d > tr , the disc will start rolling before the ring.

Page : 180 , Block Name : Additional Exercise

Q7.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination
30 . The coef cient of static friction us = 0.25.
∘

(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination O of the plane is increased, at what value of 8 does the cylinder begin to
skid, and not roll perfectly?

Answer.
Mass of the cylinder, m = 10kg

Radius of the cylinder, r = 15cm = 0.15m

Coefficient of kinetic friction, μk = 0.25
∘
Angle of inclination, θ = 30

Moment of inertia of a solid cylinder about its geometric axis, I =
1 2
mr
2

The acceleration of the cylinder is given as:
mg sin θ
a =
I
m + 2
r

mg sin θ 2
∘
= = g sin 30
2
1 mr 3
m + 2
2 r

2
2
= × 9.8 × 0.5 = 3.27m/s
3

(a) Using Newton's second law of motion, we can write net force as:
f net = ma
∘
mg sin 30 − f = ma

∘
f = mg sin 30 − ma

= 10 × 9.8 × 0.5 − 10 × 3.27

= 49 − 32.7 = 16.3N

(b) During rolling, the instantaneous point Of contact with the plane comes to rest.
Hence, the work done against frictional force is zero.
(c) For rolling without skid, we have the relation:
1
μ = tan θ
3

tan θ = 3μ = 3 × 0.25

−1 ∘
∴ θ = tan (0.75) = 36.87

Page : 180 , Block Name : Additional Exercise

Q7.32 Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of
the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not
rolling) motion.

Answer. (a) False
Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the
case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional
force acts in the forward direction.
(b) True
Rolling can be considered as the rotation Of a body about an axis passing through the point of
contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False
When a body is rolling, its instantaneous acceleration is not equal to zero. It value.
(d) True
as some When perfect rolling begins, the frictional force acting at the lowermost point
becomes zero. Hence, the work done against friction is also zero.
(e) True
The rolling of a body occurs when a frictional force acts between the body and the surface.
This frictional force provides the torque necessary for rolling. In the absence of a frictional
force, the body slips from the inclined plane under the effect of its own weight.

Page : 180 , Block Name : Additional Exercise

Q7.33 Separation of Motion of a system of particles into motion of the centre of mass and
motion about the centre of mass:
(a) Show p = p + m V i
′

i i

Where pi is the momentum of the i particle (of mass mi) and p = m v . Note v is the th ′

i i
′

i
′

i

velocity of the i particle relative to the centre of mass.
th

Also, prove using the de nition of the centre of mass ∑ p = 0 i
′

i

(b) Show K = K + 1/2MV ′ 2

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of
the system when the particle velocities are taken with respect to the centre of mass and
MV /2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of
2

mass motion of the system). The result has been used in Sec. 7.14.
(c) Show L = L + R × MV
′

Where L = ∑ r × p is the angular momentum of the system about the centre of mass
′

i
′

i
′

with velocities taken relative to the centre of mass. Remember r = r − R, rest of the ′

i i

notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be
angular momenta, respectively, about and of the centre of mass of the system of particles.
′

(d) Show dL
= ∑ r
i
′

i
×
d
(p )
′

i
dt dt
′

Further, show that dL
= τ
′
ext
dt

where τ ′
ext
is the sum of all external torques acting on the system about the centre of mass.
(Hint: Use the de nition of centre of mass and Newton’s Third Law. Assume the internal
forces between any two particles act along the line joining the particles.)

Answer. (a)Take a system of i moving particles.
Mass of the i particle = m th
i

Velocity of the i particle = v th
i

Hence, momentum of the i particle, pi = m v th
i i

Velocity of the centre of mass = V
The velocity of the i particle with respect to the centre of mass of the system is given as:
th

′
v = vi − V … (1)
i

Multiplying mi throughout equation (1), we get:

′
mi v = m i vi − m i V
i

′
p = p − mi v
i i

Where,
p
′

i
= mi v
′

i
= Momentum of the i th
particle with respect to the centre of mass of the system
′
∴ p = p + mi V
i i

We have the relation: p = m v ′

i i
′

i

Taking the summation of momentum of all the particles with respect to the centre of mass of
the system, we get:
′
dr
′ ′ i
∑ p = ∑ mi v = ∑ mi
i i i i i dt

Where,

′
r = Position vector of i th particle with respect to the centre of mass
′
dr
′ i
v =
i dt

As per the definition of the centre of mass, we have:

′
∑ mr = 0
T i
′
dr
i
∴ ∑ mj = 0
i dt

′
∑ p = 0
f i

(b) We have the relation for velocity of the i th
particle as:
′
vi = v + V
i

∑ m1 v i = ∑ mi v
i i
′
i
+ ∑ mi V
i
...(2)
Taking the dot product of equation (2) with itself, we get:
′ ′
∑ mi vi ∑ mi vi = ∑ mi (v + V) ⋅ ∑ mi (v + v)
i i i i i i

2 2 2 ′2 2 2 ′ 2 2
M ∑v = M ∑ v + M ∑ vi ⋅ v′ + M ∑ v , vi + M V
i i i i , i

Here, for the centre of mass of the system of particles, ∑ v v = − ∑ v v i i
′
i i
′
i i

2 2 2 ′2 2 2
M ∑ v = M ∑ v + M V
i i i i

1 2 1 ′2 1 2
M ∑ v = M ∑ v + MV
2 i i 2 i i 2

′ 1 2
K = K + MV
2
1

Where k = 2 Total kinetic energy of the system of particles
2
M ∑ v
2 i i
=

κ
′
=
1

2
M ∑ v
i
′2
i
Total kinetic energy of the system of particles with respect to the centre of
mass
Kinetic energy of the translation of the system as a whole
1 2
MV
2

(c) Position vector of the i particle with respect to origin = r th
i

Position vector of the i particle with respect to the centre of mass = r’i th

Position vector of the centre of mass with respect to the origin = R
It is given that:

′
r = ri − R
i

′
ri = r + R
i

We have from part (a)

′
pi = p + mi V
i

Taking the cross product of this relation by ri, we get:
′
∑ ri × p = ∑ ri × p + ∑ r i × mi V
i i i i i

′ ′ ′
L = ∑ (r + R) × p + ∑ (r + R) × mi V
i i i i i

′ ′ ′ ′
= ∑ r × p + ∑ R × p + ∑ r × mi V + ∑ R × ml V
i i i i i i i i
′ ′ ′
= L + ∑ R × p + ∑ r × mi V + ∑ R × mi V
i i i i i
′
R × ∑p = 0 and
i

′
(∑ r ) × M V = 0
i i

∑ mi = M
i

′
∴ L = L + R × MV

(d) We have the relation:
′ ′ ′
L = ∑ r × p
i

′
dL d ′ ′
= (∑ r × p )
dt dt i

d ′ ′ ′ d ′
= (∑ r ) × p + ∑ r × (p )
dt i i i i i dt i

d ′ ′ ′ d ′
= (∑ mi r ) × v + ∑ r × (p )
dt i i i i dt i

Where, r', is the position vector with respect to the centre of mass of the system of particles.
′
∴ ∑ mi r = 0
i i

′
dL ′ d ′
∴ = ∑ r × (p )
dt i i dt i

We have the relation:
′
dL d
′ ′
= ∑ r × (p )
i i
dt dt
i

d
′ ′
= ∑ r × mi (v )
i i
dt
i

Where, is the rate of change of velocity of the i particle with respect ot the centre of
d ′ th
(v )
dt i

mass of the system
Therefore, according to Neuton's third law of motion, we can write:
(v ) = Extrenal force acting on the i particle = ∑ (τ )
d ′ th ′
m f
dt i i est

i.e , ∑ r × m i
′

i i
d
(v ) = τ
′

i
′
est
= External torque acting on the system as a whole
dt
′
dL ′
∴ = τ
dt est

Page : 180 , Block Name : Additional Exercise