Chapter 10 : Rotational Mechanics - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

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JEE : Chapter 10 : Rotational Mechanics - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


10.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 10
1. ?
0
= 0 ; ? = 100 rev/s ; ? = 2 ? ; ? = 200 ? rad/s
? ? = ?
0
= ?t
? ? = ?t
? ? = (200 ?)/4 = 50 ? rad /s
2
or 25 rev/s
2
? ? = ?
0
t + 1/2 ?t
2
= 8 × 50 ? = 400 ? rad
? ? = 50 ? rad/s
2
or 25 rev/s
s
? = 400 ? rad. ?
2. ? = 100 ?; t = 5 sec
? = 1/2 ?t
2
? 100 ? = 1/2 ? 25
? ? = 8 ? × 5 = 40 ? rad/s = 20 rev/s
? ? = 8 ? rad/s
2
= 4 rev/s
2
? = 40 ? rad/s
2
= 20 rev/s
2
. ?
3. Area under the curve will decide the total angle rotated 
? maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
? Total angle rotated = 800 rad.
4. ? = 1 rad/s
2
, ?
0
= 5 rad/s; ? = 15 rad/s
? w = w
0
+ ?t
? t = ( ? – ?
0
)/ ? = (15 – 5)/1 = 10 sec
Also, ? = ?
0
t + 1/2 ?t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad. ?
5. ? = 5 rev, ? = 2 rev/s
2
, ?
0
= 0 ; ? = ?
?
2
= (2 ? ?)
? ? = 5 2 2 ? ? = 5 2 rev/s.
or ? = 10 ? rad, ? = 4 ? rad/s
2
, ?
0
= 0, ? = ?
? = ?? 2 = 2 × 4 ? × 10 ?
= 5 4 ? rad/s = 5 2 rev/s.
?
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
? Linear velocity on the rim = ?r = 20 × 0.1 = 2 m/s
? Linear velocity at the middle of radius = ?r/2 = 20 × (0.1)/2 = 1 m/s. ?
7. t = 1 sec, r = 1 cm = 0.01 m
? = 4 rd/s
2
Therefore ? = ?t = 4 rad/s
Therefore radial acceleration, 
A
n
= ?
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, a
r
= ?r = 0.04 m/s
2
= 4 cm/s
2
. ?
8. The Block is moving the rim of the pulley 
The pulley is moving at a ? = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r ?
= 20 × 20 = 200 cm/s = 2 m/s. ?
T 
10 ? ?
10 20
Page 2


10.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 10
1. ?
0
= 0 ; ? = 100 rev/s ; ? = 2 ? ; ? = 200 ? rad/s
? ? = ?
0
= ?t
? ? = ?t
? ? = (200 ?)/4 = 50 ? rad /s
2
or 25 rev/s
2
? ? = ?
0
t + 1/2 ?t
2
= 8 × 50 ? = 400 ? rad
? ? = 50 ? rad/s
2
or 25 rev/s
s
? = 400 ? rad. ?
2. ? = 100 ?; t = 5 sec
? = 1/2 ?t
2
? 100 ? = 1/2 ? 25
? ? = 8 ? × 5 = 40 ? rad/s = 20 rev/s
? ? = 8 ? rad/s
2
= 4 rev/s
2
? = 40 ? rad/s
2
= 20 rev/s
2
. ?
3. Area under the curve will decide the total angle rotated 
? maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
? Total angle rotated = 800 rad.
4. ? = 1 rad/s
2
, ?
0
= 5 rad/s; ? = 15 rad/s
? w = w
0
+ ?t
? t = ( ? – ?
0
)/ ? = (15 – 5)/1 = 10 sec
Also, ? = ?
0
t + 1/2 ?t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad. ?
5. ? = 5 rev, ? = 2 rev/s
2
, ?
0
= 0 ; ? = ?
?
2
= (2 ? ?)
? ? = 5 2 2 ? ? = 5 2 rev/s.
or ? = 10 ? rad, ? = 4 ? rad/s
2
, ?
0
= 0, ? = ?
? = ?? 2 = 2 × 4 ? × 10 ?
= 5 4 ? rad/s = 5 2 rev/s.
?
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
? Linear velocity on the rim = ?r = 20 × 0.1 = 2 m/s
? Linear velocity at the middle of radius = ?r/2 = 20 × (0.1)/2 = 1 m/s. ?
7. t = 1 sec, r = 1 cm = 0.01 m
? = 4 rd/s
2
Therefore ? = ?t = 4 rad/s
Therefore radial acceleration, 
A
n
= ?
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, a
r
= ?r = 0.04 m/s
2
= 4 cm/s
2
. ?
8. The Block is moving the rim of the pulley 
The pulley is moving at a ? = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r ?
= 20 × 20 = 200 cm/s = 2 m/s. ?
T 
10 ? ?
10 20
Chapter-10
10.2
9. Therefore, the ? distance from the axis (AD) = 3 5 10 2 / 3 ? ? cm.
Therefore moment of inertia about the axis BC will be 
I = mr
2
= 200 K 
2
) 3 5 ( = 200 × 25 × 3
= 15000 gm – cm
2
= 1.5 × 10
–3
kg – m
2
.
b) The axis of rotation let pass through A and ? to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr
2
+ mr
2
= 2 × 200 ×10
2
= 40000 gm-cm
2
= 4 ×10
–3
kg-m
2
. 
10. Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis 
respectively. A perpendicular axis is passed at the 50
th
particle.
Therefore on the L.H.S. side of the axis there will be 49 particles and on 
the R.H.S. side there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment inertial due to these two particle will be = 
49 × 1
2
+ 51 + 1
2
= 100 gm-cm
2
Similarly if we consider 48
th
and 52
nd
term we will get 100 ×2
2
gm-cm
2
Therefore we will get 49 such set and one lone particle at 100 cm.
Therefore total moment of inertia = 
100 {1
2
+ 2
2
+ 3
2
+ … + 49
2
} + 100(50)
2
.
= 100 × (50 × 51 × 101)/6 = 4292500 gm-cm
2
= 0.429 kg-m
2
= 0.43 kg-m
2
.
11. The two bodies of mass m and radius r are moving along the common tangent.
Therefore moment of inertia of the first body about XY tangent.
= mr
2
+ 2/5 mr
2
– Moment of inertia of the second body XY tangent = mr
2
+ 2/5 mr
2
= 7/5 mr
2
Therefore, net moment of inertia = 7/5 mr
2
+ 7/5 mr
2
= 14/5 mr
2
units.
12. Length of the rod = 1 m, mass of the rod = 0.5 kg
Let at a distance d from the center the rod is moving
Applying parallel axis theorem :
The moment of inertial about that point 
? (mL
2
/ 12) + md
2
= 0.10 
? (0.5 × 1
2
)/12 + 0.5 × d
2
= 0.10
? d
2
= 0.2 – 0.082 = 0.118
? d = 0.342 m from the centre.
13. Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis ? to the plane of the ring, the 
moment of inertia
= mR
2
+ mR
2
= 2mR
2
(parallel axis theorem)
? mK
2
= 2mR
2
(K = radius of the gyration)
? K = R 2 R 2
2
? .
14. The moment of inertia about the center and ? to the plane of the disc of 
radius r and mass m is = mr
2
.
According to the question the radius of gyration of the disc about a point = 
radius of the disc.
Therefore mk
2
= ½ mr
2
+ md
2
(K = radius of gyration about acceleration point, d = distance of that point 
from the centre)
? K
2
= r
2
/2 + d
2
? r
2
= r
2
/2 + d
2
( ? K = r)
? r
2
/2 = d
2
? d = 2 / r .
C B
A
I 1
D
1 10 20 30 40 50 60 70 80 90 100
48 49 51 52
1
x
2
y
r r
1m
ml
2
/12
d
(ml
2
/12)+md
2
R 
1/2 mr
2
r
k
d
1/2 mr
2
+md
2
Page 3


10.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 10
1. ?
0
= 0 ; ? = 100 rev/s ; ? = 2 ? ; ? = 200 ? rad/s
? ? = ?
0
= ?t
? ? = ?t
? ? = (200 ?)/4 = 50 ? rad /s
2
or 25 rev/s
2
? ? = ?
0
t + 1/2 ?t
2
= 8 × 50 ? = 400 ? rad
? ? = 50 ? rad/s
2
or 25 rev/s
s
? = 400 ? rad. ?
2. ? = 100 ?; t = 5 sec
? = 1/2 ?t
2
? 100 ? = 1/2 ? 25
? ? = 8 ? × 5 = 40 ? rad/s = 20 rev/s
? ? = 8 ? rad/s
2
= 4 rev/s
2
? = 40 ? rad/s
2
= 20 rev/s
2
. ?
3. Area under the curve will decide the total angle rotated 
? maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
? Total angle rotated = 800 rad.
4. ? = 1 rad/s
2
, ?
0
= 5 rad/s; ? = 15 rad/s
? w = w
0
+ ?t
? t = ( ? – ?
0
)/ ? = (15 – 5)/1 = 10 sec
Also, ? = ?
0
t + 1/2 ?t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad. ?
5. ? = 5 rev, ? = 2 rev/s
2
, ?
0
= 0 ; ? = ?
?
2
= (2 ? ?)
? ? = 5 2 2 ? ? = 5 2 rev/s.
or ? = 10 ? rad, ? = 4 ? rad/s
2
, ?
0
= 0, ? = ?
? = ?? 2 = 2 × 4 ? × 10 ?
= 5 4 ? rad/s = 5 2 rev/s.
?
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
? Linear velocity on the rim = ?r = 20 × 0.1 = 2 m/s
? Linear velocity at the middle of radius = ?r/2 = 20 × (0.1)/2 = 1 m/s. ?
7. t = 1 sec, r = 1 cm = 0.01 m
? = 4 rd/s
2
Therefore ? = ?t = 4 rad/s
Therefore radial acceleration, 
A
n
= ?
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, a
r
= ?r = 0.04 m/s
2
= 4 cm/s
2
. ?
8. The Block is moving the rim of the pulley 
The pulley is moving at a ? = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r ?
= 20 × 20 = 200 cm/s = 2 m/s. ?
T 
10 ? ?
10 20
Chapter-10
10.2
9. Therefore, the ? distance from the axis (AD) = 3 5 10 2 / 3 ? ? cm.
Therefore moment of inertia about the axis BC will be 
I = mr
2
= 200 K 
2
) 3 5 ( = 200 × 25 × 3
= 15000 gm – cm
2
= 1.5 × 10
–3
kg – m
2
.
b) The axis of rotation let pass through A and ? to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr
2
+ mr
2
= 2 × 200 ×10
2
= 40000 gm-cm
2
= 4 ×10
–3
kg-m
2
. 
10. Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis 
respectively. A perpendicular axis is passed at the 50
th
particle.
Therefore on the L.H.S. side of the axis there will be 49 particles and on 
the R.H.S. side there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment inertial due to these two particle will be = 
49 × 1
2
+ 51 + 1
2
= 100 gm-cm
2
Similarly if we consider 48
th
and 52
nd
term we will get 100 ×2
2
gm-cm
2
Therefore we will get 49 such set and one lone particle at 100 cm.
Therefore total moment of inertia = 
100 {1
2
+ 2
2
+ 3
2
+ … + 49
2
} + 100(50)
2
.
= 100 × (50 × 51 × 101)/6 = 4292500 gm-cm
2
= 0.429 kg-m
2
= 0.43 kg-m
2
.
11. The two bodies of mass m and radius r are moving along the common tangent.
Therefore moment of inertia of the first body about XY tangent.
= mr
2
+ 2/5 mr
2
– Moment of inertia of the second body XY tangent = mr
2
+ 2/5 mr
2
= 7/5 mr
2
Therefore, net moment of inertia = 7/5 mr
2
+ 7/5 mr
2
= 14/5 mr
2
units.
12. Length of the rod = 1 m, mass of the rod = 0.5 kg
Let at a distance d from the center the rod is moving
Applying parallel axis theorem :
The moment of inertial about that point 
? (mL
2
/ 12) + md
2
= 0.10 
? (0.5 × 1
2
)/12 + 0.5 × d
2
= 0.10
? d
2
= 0.2 – 0.082 = 0.118
? d = 0.342 m from the centre.
13. Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis ? to the plane of the ring, the 
moment of inertia
= mR
2
+ mR
2
= 2mR
2
(parallel axis theorem)
? mK
2
= 2mR
2
(K = radius of the gyration)
? K = R 2 R 2
2
? .
14. The moment of inertia about the center and ? to the plane of the disc of 
radius r and mass m is = mr
2
.
According to the question the radius of gyration of the disc about a point = 
radius of the disc.
Therefore mk
2
= ½ mr
2
+ md
2
(K = radius of gyration about acceleration point, d = distance of that point 
from the centre)
? K
2
= r
2
/2 + d
2
? r
2
= r
2
/2 + d
2
( ? K = r)
? r
2
/2 = d
2
? d = 2 / r .
C B
A
I 1
D
1 10 20 30 40 50 60 70 80 90 100
48 49 51 52
1
x
2
y
r r
1m
ml
2
/12
d
(ml
2
/12)+md
2
R 
1/2 mr
2
r
k
d
1/2 mr
2
+md
2
Chapter-10
10.3
15. Let a small cross sectional area is at a distance x from xx axis.
Therefore mass of that small section = m/a
2
× ax dx
Therefore moment of inertia about xx axis
= I
xx
= 
2 / a
0
3
2 / a
0
2 2
)] 3 / x )( a / m ( 2 ( x ) adx ( ) a / m ( 2 ? ? ? ?
?
= ma
2
/ 12 
Therefore I
xx
= I
xx
+ I
yy
= 2 × *ma
2
/12)= ma
2
/6
Since the two diagonals are ? to each other 
Therefore I
zz
= I
x’x’
+ I
y’y’
? ma
2
/6 = 2 × I
x’x’
( because I
x’x’ 
= I
y’y’
) ? I
x’x’
= ma
2
/12
16. The surface density of a circular disc of radius a depends upon the distance from the centre as 
P(r) = A + Br
Therefore the mass of the ring of radius r will be 
? = (A + Br) × 2 ?r dr × r
2
Therefore moment of inertia about the centre will be
= 
? ? ?
? ? ? ? ? ? ?
a
0
4
a
0
3
a
0
dr Br 2 dr Ar 2 dr r 2 ) Br A (
= 2 ?A (r
4
/4) + 2 ? B(r
5
/5)
a
0
] = 2 ?a
4
[(A/4) + (Ba/5)]. ?
17. At the highest point total force acting on the particle id its weight acting downward.
Range of the particle = u
2
sin 2 ? / g
Therefore force is at a ? distance, ? (total range)/2 = (v
2
sin 2 ?)/2g
(From the initial point)
Therefore ? = F × r ( ? = angle of projection)
= mg × v
2
sin 2 ?/2g (v = initial velocity)
= mv
2
sin 2 ? / 2 = mv
2
sin ? cos ?. ?
18. A simple of pendulum of length l is suspended from a rigid support. A bob of weight W is hanging on the 
other point.
When the bob is at an angle ? with the vertical, then total torque acting on the point of 
suspension = i = F × r
? W r sin ? = W l sin ?
At the lowest point of suspension the torque will be zero as the force acting on the body 
passes through the point of suspension. ?
19. A force of 6 N acting at an angle of 30° is just able to loosen the wrench at a distance 8 cm from it.
Therefore total torque acting at A about the point 0
= 6 sin 30° × (8/100)
Therefore total torque required at B about the point 0
= F × 16/100 ? F × 16/100 = 6 sin 30° × 8/100
? F = (8 × 3) / 16 = 1.5 N.
20. Torque about a point = Total force × perpendicular distance from the point to that force.
Let anticlockwise torque = + ve
And clockwise acting torque = –ve
Force acting at the point B is 15 N
Therefore torque at O due to this force 
= 15 × 6 × 10
–2
× sin 37°
= 15 × 6 × 10
–2
× 3/5 = 0.54 N-m (anticlock wise)
Force acting at the point C is 10 N
Therefore, torque at O due to this force 
= 10 × 4 × 10
–2
= 0.4 N-m (clockwise)
Force acting at the point A is 20 N
Therefore, Torque at O due to this force = 20 × 4 × 10
–2
× sin30°
= 20 × 4 × 10
–2
× 1/2 = 0.4 N-m (anticlockwise)
Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.
y
O
y
x
D C
B A
x
x ?
x ?
y ?
y ?
dx
r
? ?
(v
2
sin2 ?) /2 ? ?
I
B
? ?
A
A  
10N ?
4cm
E B
C
18/5
30°
6cm
3cm
4cm
15N
2cm
5N
20N
30°
Page 4


10.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 10
1. ?
0
= 0 ; ? = 100 rev/s ; ? = 2 ? ; ? = 200 ? rad/s
? ? = ?
0
= ?t
? ? = ?t
? ? = (200 ?)/4 = 50 ? rad /s
2
or 25 rev/s
2
? ? = ?
0
t + 1/2 ?t
2
= 8 × 50 ? = 400 ? rad
? ? = 50 ? rad/s
2
or 25 rev/s
s
? = 400 ? rad. ?
2. ? = 100 ?; t = 5 sec
? = 1/2 ?t
2
? 100 ? = 1/2 ? 25
? ? = 8 ? × 5 = 40 ? rad/s = 20 rev/s
? ? = 8 ? rad/s
2
= 4 rev/s
2
? = 40 ? rad/s
2
= 20 rev/s
2
. ?
3. Area under the curve will decide the total angle rotated 
? maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
? Total angle rotated = 800 rad.
4. ? = 1 rad/s
2
, ?
0
= 5 rad/s; ? = 15 rad/s
? w = w
0
+ ?t
? t = ( ? – ?
0
)/ ? = (15 – 5)/1 = 10 sec
Also, ? = ?
0
t + 1/2 ?t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad. ?
5. ? = 5 rev, ? = 2 rev/s
2
, ?
0
= 0 ; ? = ?
?
2
= (2 ? ?)
? ? = 5 2 2 ? ? = 5 2 rev/s.
or ? = 10 ? rad, ? = 4 ? rad/s
2
, ?
0
= 0, ? = ?
? = ?? 2 = 2 × 4 ? × 10 ?
= 5 4 ? rad/s = 5 2 rev/s.
?
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
? Linear velocity on the rim = ?r = 20 × 0.1 = 2 m/s
? Linear velocity at the middle of radius = ?r/2 = 20 × (0.1)/2 = 1 m/s. ?
7. t = 1 sec, r = 1 cm = 0.01 m
? = 4 rd/s
2
Therefore ? = ?t = 4 rad/s
Therefore radial acceleration, 
A
n
= ?
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, a
r
= ?r = 0.04 m/s
2
= 4 cm/s
2
. ?
8. The Block is moving the rim of the pulley 
The pulley is moving at a ? = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r ?
= 20 × 20 = 200 cm/s = 2 m/s. ?
T 
10 ? ?
10 20
Chapter-10
10.2
9. Therefore, the ? distance from the axis (AD) = 3 5 10 2 / 3 ? ? cm.
Therefore moment of inertia about the axis BC will be 
I = mr
2
= 200 K 
2
) 3 5 ( = 200 × 25 × 3
= 15000 gm – cm
2
= 1.5 × 10
–3
kg – m
2
.
b) The axis of rotation let pass through A and ? to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr
2
+ mr
2
= 2 × 200 ×10
2
= 40000 gm-cm
2
= 4 ×10
–3
kg-m
2
. 
10. Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis 
respectively. A perpendicular axis is passed at the 50
th
particle.
Therefore on the L.H.S. side of the axis there will be 49 particles and on 
the R.H.S. side there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment inertial due to these two particle will be = 
49 × 1
2
+ 51 + 1
2
= 100 gm-cm
2
Similarly if we consider 48
th
and 52
nd
term we will get 100 ×2
2
gm-cm
2
Therefore we will get 49 such set and one lone particle at 100 cm.
Therefore total moment of inertia = 
100 {1
2
+ 2
2
+ 3
2
+ … + 49
2
} + 100(50)
2
.
= 100 × (50 × 51 × 101)/6 = 4292500 gm-cm
2
= 0.429 kg-m
2
= 0.43 kg-m
2
.
11. The two bodies of mass m and radius r are moving along the common tangent.
Therefore moment of inertia of the first body about XY tangent.
= mr
2
+ 2/5 mr
2
– Moment of inertia of the second body XY tangent = mr
2
+ 2/5 mr
2
= 7/5 mr
2
Therefore, net moment of inertia = 7/5 mr
2
+ 7/5 mr
2
= 14/5 mr
2
units.
12. Length of the rod = 1 m, mass of the rod = 0.5 kg
Let at a distance d from the center the rod is moving
Applying parallel axis theorem :
The moment of inertial about that point 
? (mL
2
/ 12) + md
2
= 0.10 
? (0.5 × 1
2
)/12 + 0.5 × d
2
= 0.10
? d
2
= 0.2 – 0.082 = 0.118
? d = 0.342 m from the centre.
13. Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis ? to the plane of the ring, the 
moment of inertia
= mR
2
+ mR
2
= 2mR
2
(parallel axis theorem)
? mK
2
= 2mR
2
(K = radius of the gyration)
? K = R 2 R 2
2
? .
14. The moment of inertia about the center and ? to the plane of the disc of 
radius r and mass m is = mr
2
.
According to the question the radius of gyration of the disc about a point = 
radius of the disc.
Therefore mk
2
= ½ mr
2
+ md
2
(K = radius of gyration about acceleration point, d = distance of that point 
from the centre)
? K
2
= r
2
/2 + d
2
? r
2
= r
2
/2 + d
2
( ? K = r)
? r
2
/2 = d
2
? d = 2 / r .
C B
A
I 1
D
1 10 20 30 40 50 60 70 80 90 100
48 49 51 52
1
x
2
y
r r
1m
ml
2
/12
d
(ml
2
/12)+md
2
R 
1/2 mr
2
r
k
d
1/2 mr
2
+md
2
Chapter-10
10.3
15. Let a small cross sectional area is at a distance x from xx axis.
Therefore mass of that small section = m/a
2
× ax dx
Therefore moment of inertia about xx axis
= I
xx
= 
2 / a
0
3
2 / a
0
2 2
)] 3 / x )( a / m ( 2 ( x ) adx ( ) a / m ( 2 ? ? ? ?
?
= ma
2
/ 12 
Therefore I
xx
= I
xx
+ I
yy
= 2 × *ma
2
/12)= ma
2
/6
Since the two diagonals are ? to each other 
Therefore I
zz
= I
x’x’
+ I
y’y’
? ma
2
/6 = 2 × I
x’x’
( because I
x’x’ 
= I
y’y’
) ? I
x’x’
= ma
2
/12
16. The surface density of a circular disc of radius a depends upon the distance from the centre as 
P(r) = A + Br
Therefore the mass of the ring of radius r will be 
? = (A + Br) × 2 ?r dr × r
2
Therefore moment of inertia about the centre will be
= 
? ? ?
? ? ? ? ? ? ?
a
0
4
a
0
3
a
0
dr Br 2 dr Ar 2 dr r 2 ) Br A (
= 2 ?A (r
4
/4) + 2 ? B(r
5
/5)
a
0
] = 2 ?a
4
[(A/4) + (Ba/5)]. ?
17. At the highest point total force acting on the particle id its weight acting downward.
Range of the particle = u
2
sin 2 ? / g
Therefore force is at a ? distance, ? (total range)/2 = (v
2
sin 2 ?)/2g
(From the initial point)
Therefore ? = F × r ( ? = angle of projection)
= mg × v
2
sin 2 ?/2g (v = initial velocity)
= mv
2
sin 2 ? / 2 = mv
2
sin ? cos ?. ?
18. A simple of pendulum of length l is suspended from a rigid support. A bob of weight W is hanging on the 
other point.
When the bob is at an angle ? with the vertical, then total torque acting on the point of 
suspension = i = F × r
? W r sin ? = W l sin ?
At the lowest point of suspension the torque will be zero as the force acting on the body 
passes through the point of suspension. ?
19. A force of 6 N acting at an angle of 30° is just able to loosen the wrench at a distance 8 cm from it.
Therefore total torque acting at A about the point 0
= 6 sin 30° × (8/100)
Therefore total torque required at B about the point 0
= F × 16/100 ? F × 16/100 = 6 sin 30° × 8/100
? F = (8 × 3) / 16 = 1.5 N.
20. Torque about a point = Total force × perpendicular distance from the point to that force.
Let anticlockwise torque = + ve
And clockwise acting torque = –ve
Force acting at the point B is 15 N
Therefore torque at O due to this force 
= 15 × 6 × 10
–2
× sin 37°
= 15 × 6 × 10
–2
× 3/5 = 0.54 N-m (anticlock wise)
Force acting at the point C is 10 N
Therefore, torque at O due to this force 
= 10 × 4 × 10
–2
= 0.4 N-m (clockwise)
Force acting at the point A is 20 N
Therefore, Torque at O due to this force = 20 × 4 × 10
–2
× sin30°
= 20 × 4 × 10
–2
× 1/2 = 0.4 N-m (anticlockwise)
Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.
y
O
y
x
D C
B A
x
x ?
x ?
y ?
y ?
dx
r
? ?
(v
2
sin2 ?) /2 ? ?
I
B
? ?
A
A  
10N ?
4cm
E B
C
18/5
30°
6cm
3cm
4cm
15N
2cm
5N
20N
30°
Chapter-10
10.4
21. The force mg acting on the body has two components mg sin ? and mg cos ?
and the body will exert a normal reaction. Let R = 
Since R and mg cos ? pass through the centre of the cube, there will be no torque 
due to R and mg cos ?. The only torque will be produced by mg sin ?.
? i = F × r (r = a/2) (a = ages of the cube)
? i = mg sin ? × a/2
= 1/2 mg a sin ?. ?
22. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis passing through its 
centre.
A force F is acting perpendicular to the rod at a distance L/4 from the centre.
Therefore torque about the centre due to this force 
i
i
= F × r = FL/4.
This torque will produce a angular acceleration ?.
Therefore ?
c
= I
c
× ?
? i
c
= (mL
2
/ 12) × ? (I
c
of a rod = mL
2
/ 12)
? F i/4 = (mL
2
/ 12) × ? ? ? = 3F/ml
Therefore ? = 1/2 ?t
2
(initially at rest)
? ? = 1/2 × (3F / ml)t
2
= (3F/2ml)t
2
. ?
23. A square plate of mass 120 gm and edge 5 cm rotates about one of the edge.
Let take a small area of the square of width dx and length a which is at a distance x from the axis of 
rotation.
Therefore mass of that small area
m/a
2
× a dx (m = mass of the square ; a = side of the plate)
I = 
a
0
3
a
0
2 2
)] 3 / x )( a / m ( dx ax ) a / m ( ? ?
?
= ma
2
/3
Therefore torque produced = I × ? = (ma
2
/3) × ?
= {(120 × 10
–3
× 5
2
× 10
–4
)/3} 0.2
= 0.2 × 10
–4
= 2 × 10
–5
N-m. ?
24. Moment of inertial of a square plate about its diagonal is ma
2
/12 (m = mass of the square plate)
a = edges of the square
Therefore torque produced = (ma
2
/12) × ?
= {(120 × 10
–3
× 5
2
× 10
–4
)/12 × 0.2
= 0.5 × 10
–5
N-m. ?
25. A flywheel of moment of inertia 5 kg m is rotated at a speed of 60 rad/s. The flywheel comes to rest due 
to the friction at the axle after 5 minutes.
Therefore, the angular deceleration produced due to frictional force = ? = ?
0
+ ?t
? ?
0
= – ?t ( ? = 0+
? ? = –(60/5 × 60) = –1/5 rad/s
2
.
a) Therefore total workdone in stopping the wheel by frictional force 
W = 1/2 i ?
2
= 1/2 × 5 × (60 × 60) = 9000 Joule = 9 KJ.
b) Therefore torque produced by the frictional force (R) is
I
R
= I × ? = 5 × (–1/5) = IN – m opposite to the rotation of wheel.
c) Angular velocity after 4 minutes 
? ? = ?
0
+ ?t = 60 – 240/5 = 12 rad/s
Therefore angular momentum about the centre = 1 × ? = 5 × 12 = 60 kg-m
2
/s. ?
mg sin ? ?
? ?
mg cos ? ?
mg sin ? ?
R ? ?
mg cos ? ?
? ?
A ?
B ?
B
t =sec
A
F
1/4
A 
x 
dx 
x 
x ?
R  
1/2iW
2
Page 5


10.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 10
1. ?
0
= 0 ; ? = 100 rev/s ; ? = 2 ? ; ? = 200 ? rad/s
? ? = ?
0
= ?t
? ? = ?t
? ? = (200 ?)/4 = 50 ? rad /s
2
or 25 rev/s
2
? ? = ?
0
t + 1/2 ?t
2
= 8 × 50 ? = 400 ? rad
? ? = 50 ? rad/s
2
or 25 rev/s
s
? = 400 ? rad. ?
2. ? = 100 ?; t = 5 sec
? = 1/2 ?t
2
? 100 ? = 1/2 ? 25
? ? = 8 ? × 5 = 40 ? rad/s = 20 rev/s
? ? = 8 ? rad/s
2
= 4 rev/s
2
? = 40 ? rad/s
2
= 20 rev/s
2
. ?
3. Area under the curve will decide the total angle rotated 
? maximum angular velocity = 4 × 10 = 40 rad/s
Therefore, area under the curve = 1/2 × 10 × 40 + 40 × 10 + 1/2 × 40 × 10
= 800 rad
? Total angle rotated = 800 rad.
4. ? = 1 rad/s
2
, ?
0
= 5 rad/s; ? = 15 rad/s
? w = w
0
+ ?t
? t = ( ? – ?
0
)/ ? = (15 – 5)/1 = 10 sec
Also, ? = ?
0
t + 1/2 ?t
2
= 5 ×10 + 1/2 × 1 × 100 = 100 rad. ?
5. ? = 5 rev, ? = 2 rev/s
2
, ?
0
= 0 ; ? = ?
?
2
= (2 ? ?)
? ? = 5 2 2 ? ? = 5 2 rev/s.
or ? = 10 ? rad, ? = 4 ? rad/s
2
, ?
0
= 0, ? = ?
? = ?? 2 = 2 × 4 ? × 10 ?
= 5 4 ? rad/s = 5 2 rev/s.
?
6. A disc of radius = 10 cm = 0.1 m
Angular velocity = 20 rad/s
? Linear velocity on the rim = ?r = 20 × 0.1 = 2 m/s
? Linear velocity at the middle of radius = ?r/2 = 20 × (0.1)/2 = 1 m/s. ?
7. t = 1 sec, r = 1 cm = 0.01 m
? = 4 rd/s
2
Therefore ? = ?t = 4 rad/s
Therefore radial acceleration, 
A
n
= ?
2
r = 0.16 m/s
2
= 16 cm/s
2
Therefore tangential acceleration, a
r
= ?r = 0.04 m/s
2
= 4 cm/s
2
. ?
8. The Block is moving the rim of the pulley 
The pulley is moving at a ? = 10 rad/s
Therefore the radius of the pulley = 20 cm
Therefore linear velocity on the rim = tangential velocity = r ?
= 20 × 20 = 200 cm/s = 2 m/s. ?
T 
10 ? ?
10 20
Chapter-10
10.2
9. Therefore, the ? distance from the axis (AD) = 3 5 10 2 / 3 ? ? cm.
Therefore moment of inertia about the axis BC will be 
I = mr
2
= 200 K 
2
) 3 5 ( = 200 × 25 × 3
= 15000 gm – cm
2
= 1.5 × 10
–3
kg – m
2
.
b) The axis of rotation let pass through A and ? to the plane of triangle
Therefore the torque will be produced by mass B and C
Therefore net moment of inertia = I = mr
2
+ mr
2
= 2 × 200 ×10
2
= 40000 gm-cm
2
= 4 ×10
–3
kg-m
2
. 
10. Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis 
respectively. A perpendicular axis is passed at the 50
th
particle.
Therefore on the L.H.S. side of the axis there will be 49 particles and on 
the R.H.S. side there are 50 particles.
Consider the two particles at the position 49 cm and 51 cm.
Moment inertial due to these two particle will be = 
49 × 1
2
+ 51 + 1
2
= 100 gm-cm
2
Similarly if we consider 48
th
and 52
nd
term we will get 100 ×2
2
gm-cm
2
Therefore we will get 49 such set and one lone particle at 100 cm.
Therefore total moment of inertia = 
100 {1
2
+ 2
2
+ 3
2
+ … + 49
2
} + 100(50)
2
.
= 100 × (50 × 51 × 101)/6 = 4292500 gm-cm
2
= 0.429 kg-m
2
= 0.43 kg-m
2
.
11. The two bodies of mass m and radius r are moving along the common tangent.
Therefore moment of inertia of the first body about XY tangent.
= mr
2
+ 2/5 mr
2
– Moment of inertia of the second body XY tangent = mr
2
+ 2/5 mr
2
= 7/5 mr
2
Therefore, net moment of inertia = 7/5 mr
2
+ 7/5 mr
2
= 14/5 mr
2
units.
12. Length of the rod = 1 m, mass of the rod = 0.5 kg
Let at a distance d from the center the rod is moving
Applying parallel axis theorem :
The moment of inertial about that point 
? (mL
2
/ 12) + md
2
= 0.10 
? (0.5 × 1
2
)/12 + 0.5 × d
2
= 0.10
? d
2
= 0.2 – 0.082 = 0.118
? d = 0.342 m from the centre.
13. Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis ? to the plane of the ring, the 
moment of inertia
= mR
2
+ mR
2
= 2mR
2
(parallel axis theorem)
? mK
2
= 2mR
2
(K = radius of the gyration)
? K = R 2 R 2
2
? .
14. The moment of inertia about the center and ? to the plane of the disc of 
radius r and mass m is = mr
2
.
According to the question the radius of gyration of the disc about a point = 
radius of the disc.
Therefore mk
2
= ½ mr
2
+ md
2
(K = radius of gyration about acceleration point, d = distance of that point 
from the centre)
? K
2
= r
2
/2 + d
2
? r
2
= r
2
/2 + d
2
( ? K = r)
? r
2
/2 = d
2
? d = 2 / r .
C B
A
I 1
D
1 10 20 30 40 50 60 70 80 90 100
48 49 51 52
1
x
2
y
r r
1m
ml
2
/12
d
(ml
2
/12)+md
2
R 
1/2 mr
2
r
k
d
1/2 mr
2
+md
2
Chapter-10
10.3
15. Let a small cross sectional area is at a distance x from xx axis.
Therefore mass of that small section = m/a
2
× ax dx
Therefore moment of inertia about xx axis
= I
xx
= 
2 / a
0
3
2 / a
0
2 2
)] 3 / x )( a / m ( 2 ( x ) adx ( ) a / m ( 2 ? ? ? ?
?
= ma
2
/ 12 
Therefore I
xx
= I
xx
+ I
yy
= 2 × *ma
2
/12)= ma
2
/6
Since the two diagonals are ? to each other 
Therefore I
zz
= I
x’x’
+ I
y’y’
? ma
2
/6 = 2 × I
x’x’
( because I
x’x’ 
= I
y’y’
) ? I
x’x’
= ma
2
/12
16. The surface density of a circular disc of radius a depends upon the distance from the centre as 
P(r) = A + Br
Therefore the mass of the ring of radius r will be 
? = (A + Br) × 2 ?r dr × r
2
Therefore moment of inertia about the centre will be
= 
? ? ?
? ? ? ? ? ? ?
a
0
4
a
0
3
a
0
dr Br 2 dr Ar 2 dr r 2 ) Br A (
= 2 ?A (r
4
/4) + 2 ? B(r
5
/5)
a
0
] = 2 ?a
4
[(A/4) + (Ba/5)]. ?
17. At the highest point total force acting on the particle id its weight acting downward.
Range of the particle = u
2
sin 2 ? / g
Therefore force is at a ? distance, ? (total range)/2 = (v
2
sin 2 ?)/2g
(From the initial point)
Therefore ? = F × r ( ? = angle of projection)
= mg × v
2
sin 2 ?/2g (v = initial velocity)
= mv
2
sin 2 ? / 2 = mv
2
sin ? cos ?. ?
18. A simple of pendulum of length l is suspended from a rigid support. A bob of weight W is hanging on the 
other point.
When the bob is at an angle ? with the vertical, then total torque acting on the point of 
suspension = i = F × r
? W r sin ? = W l sin ?
At the lowest point of suspension the torque will be zero as the force acting on the body 
passes through the point of suspension. ?
19. A force of 6 N acting at an angle of 30° is just able to loosen the wrench at a distance 8 cm from it.
Therefore total torque acting at A about the point 0
= 6 sin 30° × (8/100)
Therefore total torque required at B about the point 0
= F × 16/100 ? F × 16/100 = 6 sin 30° × 8/100
? F = (8 × 3) / 16 = 1.5 N.
20. Torque about a point = Total force × perpendicular distance from the point to that force.
Let anticlockwise torque = + ve
And clockwise acting torque = –ve
Force acting at the point B is 15 N
Therefore torque at O due to this force 
= 15 × 6 × 10
–2
× sin 37°
= 15 × 6 × 10
–2
× 3/5 = 0.54 N-m (anticlock wise)
Force acting at the point C is 10 N
Therefore, torque at O due to this force 
= 10 × 4 × 10
–2
= 0.4 N-m (clockwise)
Force acting at the point A is 20 N
Therefore, Torque at O due to this force = 20 × 4 × 10
–2
× sin30°
= 20 × 4 × 10
–2
× 1/2 = 0.4 N-m (anticlockwise)
Therefore resultant torque acting at ‘O’ = 0.54 – 0.4 + 0.4 = 0.54 N-m.
y
O
y
x
D C
B A
x
x ?
x ?
y ?
y ?
dx
r
? ?
(v
2
sin2 ?) /2 ? ?
I
B
? ?
A
A  
10N ?
4cm
E B
C
18/5
30°
6cm
3cm
4cm
15N
2cm
5N
20N
30°
Chapter-10
10.4
21. The force mg acting on the body has two components mg sin ? and mg cos ?
and the body will exert a normal reaction. Let R = 
Since R and mg cos ? pass through the centre of the cube, there will be no torque 
due to R and mg cos ?. The only torque will be produced by mg sin ?.
? i = F × r (r = a/2) (a = ages of the cube)
? i = mg sin ? × a/2
= 1/2 mg a sin ?. ?
22. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis passing through its 
centre.
A force F is acting perpendicular to the rod at a distance L/4 from the centre.
Therefore torque about the centre due to this force 
i
i
= F × r = FL/4.
This torque will produce a angular acceleration ?.
Therefore ?
c
= I
c
× ?
? i
c
= (mL
2
/ 12) × ? (I
c
of a rod = mL
2
/ 12)
? F i/4 = (mL
2
/ 12) × ? ? ? = 3F/ml
Therefore ? = 1/2 ?t
2
(initially at rest)
? ? = 1/2 × (3F / ml)t
2
= (3F/2ml)t
2
. ?
23. A square plate of mass 120 gm and edge 5 cm rotates about one of the edge.
Let take a small area of the square of width dx and length a which is at a distance x from the axis of 
rotation.
Therefore mass of that small area
m/a
2
× a dx (m = mass of the square ; a = side of the plate)
I = 
a
0
3
a
0
2 2
)] 3 / x )( a / m ( dx ax ) a / m ( ? ?
?
= ma
2
/3
Therefore torque produced = I × ? = (ma
2
/3) × ?
= {(120 × 10
–3
× 5
2
× 10
–4
)/3} 0.2
= 0.2 × 10
–4
= 2 × 10
–5
N-m. ?
24. Moment of inertial of a square plate about its diagonal is ma
2
/12 (m = mass of the square plate)
a = edges of the square
Therefore torque produced = (ma
2
/12) × ?
= {(120 × 10
–3
× 5
2
× 10
–4
)/12 × 0.2
= 0.5 × 10
–5
N-m. ?
25. A flywheel of moment of inertia 5 kg m is rotated at a speed of 60 rad/s. The flywheel comes to rest due 
to the friction at the axle after 5 minutes.
Therefore, the angular deceleration produced due to frictional force = ? = ?
0
+ ?t
? ?
0
= – ?t ( ? = 0+
? ? = –(60/5 × 60) = –1/5 rad/s
2
.
a) Therefore total workdone in stopping the wheel by frictional force 
W = 1/2 i ?
2
= 1/2 × 5 × (60 × 60) = 9000 Joule = 9 KJ.
b) Therefore torque produced by the frictional force (R) is
I
R
= I × ? = 5 × (–1/5) = IN – m opposite to the rotation of wheel.
c) Angular velocity after 4 minutes 
? ? = ?
0
+ ?t = 60 – 240/5 = 12 rad/s
Therefore angular momentum about the centre = 1 × ? = 5 × 12 = 60 kg-m
2
/s. ?
mg sin ? ?
? ?
mg cos ? ?
mg sin ? ?
R ? ?
mg cos ? ?
? ?
A ?
B ?
B
t =sec
A
F
1/4
A 
x 
dx 
x 
x ?
R  
1/2iW
2
Chapter-10
10.5
26. The earth’s angular speed decreases by 0.0016 rad/day in 100 years.
Therefore the torque produced by the ocean water in decreasing earth’s angular velocity
? = I ?
= 2/5 mr
2
× ( ? – ?
0
)/t
= 2/6 × 6 × 10
24
× 64
2
× 10
10
× [0.0016 /(26400
2
× 100 × 365)] (1 year = 365 days= 365 × 56400 sec)
= 5.678 × 10
20
N-m. ?
27. A wheel rotating at a speed of 600 rpm.
?
0
= 600 rpm = 10 revolutions per second.
T = 10 sec. (In 10 sec. it comes to rest)
? = 0
Therefore ?
0
= – ?t
? ? = –10/10 = –1 rev/s
2
? ? = ?
0
+ ?t = 10 – 1 × 5 = 5 rev/s.
Therefore angular deacceleration = 1 rev/s
2
and angular velocity of after 5 sec is 5 rev/s. ?
28. ? = 100 rev/min = 5/8 rev/s = 10 ?/3 rad/s
? = 10 rev = 20 ? rad, r = 0.2 m
After 10 revolutions the wheel will come to rest by a tangential force
Therefore the angular deacceleration produced by the force = ? = ?
2
/2 ?
Therefore the torque by which the wheel will come to an rest = I
cm
× ?
? F × r = I
cm
× ? ? F × 0.2 = 1/2 mr
2
× [(10 ?/3)
2
/ (2 × 20 ?)]
? F = 1/2 × 10 × 0.2 × 100 ?
2
/ (9 × 2 × 20 ?)
= 5 ? / 18 = 15.7/18 = 0.87 N. ?
29. A cylinder is moving with an angular velocity 50 rev/s brought in contact with another identical cylinder 
in rest. The first and second cylinder has common acceleration and deacceleration as 1 rad/s
2
respectively.
Let after t sec their angular velocity will be same ‘ ?’.
For the first cylinder ? = 50 – ?t 
? t = ( ? – 50)/–1
And for the 2
nd
cylinder ? = ?
2
t
? t = ?/I
So, ? = ( ? – 50)/–1 
? 2 ? = 50 ? ? = 25 rev/s.
? t = 25/1 sec = 25 sec. ?
30. Initial angular velocity = 20 rad/s
Therefore ? = 2 rad/s
2
? t
1
= ?/ ?
1
= 20/2 = 10 sec
Therefore 10 sec it will come to rest.
Since the same torque is continues to act on the body it will produce same angular acceleration and 
since the initial kinetic energy = the kinetic energy at a instant.
So initial angular velocity = angular velocity at that instant
Therefore time require to come to that angular velocity,
t
2
= ?
2
/ ?
2
= 20/2 = 10 sec
therefore time required = t
1
+ t
2
= 20 sec. ?
31. I
net
= I
net
× ?
? F
1
r
1
– F
2
r
2
= ) r m r m (
2
2 2
2
1 1
? × ? – 2 × 10 × 0.5
? 5 × 10 × 0.5 = (5 × (1/2)
2
+ 2 × (1/2)
2
) × ?
? 15 = 7/4 ?
? ? = 60/7 = 8.57 rad/s
2
. ?
32. In this problem the rod has a mass 1 kg
a) ?
net
= I
net
× ?
? 5 × 10 × 10.5 – 2 × 10 × 0.5
= (5 × (1/2)
2
+ 2 × (1/2)
2
+ 1/12) × ??
50 rev/s
2 kg 5 kg
2 kg 5 kg
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