Physics
Pre-Medical
©
2
COPYRIGHT DISCLAIMER
© ALLEN CAREER INSTITUTE 2019. All Rights Reserved.
This study material is prepared by and for the exclusive use of ALLEN CAREER INSTITUE. All
rights including copyrights, right of translation, etc. are reserved and vested exclusively with
ALLEN CAREER INSTITUTE.
No part of this publication may be copied, reproduced, adapted, translated, abridged or stored in any
retrieval system, computer system, photographic or otherwise transmitted in any form or by any
means including, but not limited to; electronic, mechanical, digital, photocopying, recording or
stored in any drive/storage system of any nature without the prior written permission of the copyright
owner. Any breach shall entail legal action and prosecution without further notice.
No unauthorised sale, re-sale, distribution, lending, display, advertisement, or circulation of this
material in any form is permitted, without the prior written consent of the owner.
The contents of this study material, and all ALLEN logos, taglines etc. are exclusive intellectual
property right of ALLEN CAREER INSTITUTE protected under the Indian Copyright and
Trademark Laws. Any violation or infringement of these rights shall be punishable under the
Trademark Act, 1999 and the Copyright Act, 1957.
, ®
ALLEN Pre-Medical : Physics 33
Solutions Rotational Motion
BEGINNER'S BOX-1 v 36
4. (a) Angular speed w = = = 120 rad/s
r 0.30
1. The given equation is w = at + b , therefore motion
is rotation with uniform angular acceleration. (b) (i) Q w2 = w 20 + 2aq
Initial angular velocity = w0 = b = 1.0 rad/s
90
\ 0 = (120)2 + 2 a (40 × 2p) Þ a = – p rad/s2
Angular acceleration a = a
1 2 (ii) distance covered during braking
q= at + w0t + c
2
= circumference of wheel × number of revolution
Since at t = 0, w = 1.0 rad/s, we obtain the constant
c. = 2pr × 40 = 2p × 0.30 × 40 = 24p m
Initial angular position = q0 = c = 2.0 rad w t
5. ò dw = ò adt Þ w –2 = (t4 – t3 + t2)0t
w - w0 5 -1 2 0
a= = = 2 rad/s2
t 2
Þ w – 2 = t4 – t3 + t2
1 2
q = w0t + at Þ w = 2 + t2 – t3 + t4
2
q t t
æ t3 t 4 t5 ö
q = 1× 4 +
1
× 2(4)2
ò1 d q = ò0 w dt Þ q - 1 = ç
è
2t + - + ÷
3 4 5 ø0
2
t3 t4 t5
q = 20 rad. Þ q = 1 + 2t + - +
3 4 5
Final angular position = 20 + 2 = 22 rad. 7. w = w0 + at Þ36 = 0 + a × 6 Þ a = 6
2. Q Initial angular speed w0 = 2pn0 q = 1/2 × 6 × 36 = 108 rad.
F 60 ´ 1 IJ
= 2p G
8. w2 = w 20 + 2aq ; q = 1/2 × a × 4 = 2a
H p 60 K = 2 rad/s
æ 600 ´ 2p ö
2
÷ = 2× a × 2a Þ a = 10p rad/s
and ç 2
\ Angular speed w = w0 + at è 60 ø
= 2 + 10 × 2 = 22 rad/s BEGINNER'S BOX-2
Z:\NODE02\B0B0-BA\TARGET\PHY\SOLUTIONS\MODULE_2\04-ROTATION MOTION.P65
1 2
Angular displacement q = w0t + at
2
1 1.
= 2 ×2 + × 10 (2)2 = 24 radian y
2
(0, 3) 3 kg
3. w = A – Bt
1 kg 2 kg
x
(0,0) (2,0)
dw 10 kg
a= = –B
dt (–2, –2)
at t = 0 ; w0 = A
I = 1 × (0)2 + 2 × (2)2 + 3 × (0)2 + 10(2)2
2
w - w02 0-A 2
A 2
q= = = = 48 kg-m2
2a -2B 2B
, ®
34 Pre-Medical : Physics ALLEN
2. Iring = Idisc MR 2
11. Moment of inertia I = MR2 +
MR12 MR22 R 1 2
= Þ 1 =
2 4 R2 2
2MR 2 + MR 2 3
2 2 = = MR2
I1 M1R1 4 M1 æ 4 ö M1 1 2 2
3. = Þ = ´ç ÷ Þ =
I2 M2 R 2
2
1 M2 è 1 ø M2 4
2
12. It will be again = MR 2 , because the axis in
2 2 5
æLö æLö
4. I = m1 ç ÷ + m2 ç ÷
è2ø è2ø question is also diameter of the sphere.
= 1 × (1)2 + 2 × (1)2 = 3 kg-m2 BEGINNER'S BOX-3
5. I=
5 4
MR2 Þ MR2 = I 1. r r r
4 5 t=r ´F
3 3 4 6
I' = MR 2 = ´ I = I ˆi ˆj kˆ
2 2 5 5 r
t= 2 1 3
6. Moment of inertia of hollow cylinder will be larger
1 -2 5
as compare to disc because most of its mass is located
away from the axis of rotation as compare to that
of disc. r ˆ -4 - 1)
t = ˆi(5 - ( -6)) - ˆj(10 - 3) + k(
7. Spokes do not carry much mass. Most of the mass
r
is located at the rim. This gives cycle wheel more t = 11iˆ - 7jˆ - 5kˆ
moment of inertia for same mass.
2. About point B moment of inertia is less than A so
t it is easier to rotate about point B.
8. MA = pr2tr and MB = p(4r)2 r Þ MB = 4 MA
4
3. t = Ia
IA MA r 2
1
\ = = Þ IB = 64IA 2
IB MB ´ 16r 2 64 20 2 æ 20 ö
2× = ´ç ÷ ´a
9. Mass of plate = (Mass per unit area) (Area) = ml2 100 2 è 100 ø
Ml 2 ml2 ´ l2 ml 4 a = 10 rad/sec
Z:\NODE02\B0B0-BA\TARGET\PHY\SOLUTIONS\MODULE_2\04-ROTATION MOTION.P65
Moment of inertia I = = =
6 6 6
20
at = ra = ´ 10 = 2 m/s2
100
4. r1F1 = r2F2
10. 1.6 × 1 = 0.4 × F2
Þ F2 = 4 N
Moment of inertia 5. Net force on rod Fnet = 0
FG a IJ 2
FG a IJ 2
tnet = – F × 20 + F × 40 – F × 60 + F × 80
I = m1× 0 + m2 ×
H 2K + m3 ×
H 2K
= 40 F
a2
= (m2 + m3)
4 so rod experinces a torque
Pre-Medical
©
2
COPYRIGHT DISCLAIMER
© ALLEN CAREER INSTITUTE 2019. All Rights Reserved.
This study material is prepared by and for the exclusive use of ALLEN CAREER INSTITUE. All
rights including copyrights, right of translation, etc. are reserved and vested exclusively with
ALLEN CAREER INSTITUTE.
No part of this publication may be copied, reproduced, adapted, translated, abridged or stored in any
retrieval system, computer system, photographic or otherwise transmitted in any form or by any
means including, but not limited to; electronic, mechanical, digital, photocopying, recording or
stored in any drive/storage system of any nature without the prior written permission of the copyright
owner. Any breach shall entail legal action and prosecution without further notice.
No unauthorised sale, re-sale, distribution, lending, display, advertisement, or circulation of this
material in any form is permitted, without the prior written consent of the owner.
The contents of this study material, and all ALLEN logos, taglines etc. are exclusive intellectual
property right of ALLEN CAREER INSTITUTE protected under the Indian Copyright and
Trademark Laws. Any violation or infringement of these rights shall be punishable under the
Trademark Act, 1999 and the Copyright Act, 1957.
, ®
ALLEN Pre-Medical : Physics 33
Solutions Rotational Motion
BEGINNER'S BOX-1 v 36
4. (a) Angular speed w = = = 120 rad/s
r 0.30
1. The given equation is w = at + b , therefore motion
is rotation with uniform angular acceleration. (b) (i) Q w2 = w 20 + 2aq
Initial angular velocity = w0 = b = 1.0 rad/s
90
\ 0 = (120)2 + 2 a (40 × 2p) Þ a = – p rad/s2
Angular acceleration a = a
1 2 (ii) distance covered during braking
q= at + w0t + c
2
= circumference of wheel × number of revolution
Since at t = 0, w = 1.0 rad/s, we obtain the constant
c. = 2pr × 40 = 2p × 0.30 × 40 = 24p m
Initial angular position = q0 = c = 2.0 rad w t
5. ò dw = ò adt Þ w –2 = (t4 – t3 + t2)0t
w - w0 5 -1 2 0
a= = = 2 rad/s2
t 2
Þ w – 2 = t4 – t3 + t2
1 2
q = w0t + at Þ w = 2 + t2 – t3 + t4
2
q t t
æ t3 t 4 t5 ö
q = 1× 4 +
1
× 2(4)2
ò1 d q = ò0 w dt Þ q - 1 = ç
è
2t + - + ÷
3 4 5 ø0
2
t3 t4 t5
q = 20 rad. Þ q = 1 + 2t + - +
3 4 5
Final angular position = 20 + 2 = 22 rad. 7. w = w0 + at Þ36 = 0 + a × 6 Þ a = 6
2. Q Initial angular speed w0 = 2pn0 q = 1/2 × 6 × 36 = 108 rad.
F 60 ´ 1 IJ
= 2p G
8. w2 = w 20 + 2aq ; q = 1/2 × a × 4 = 2a
H p 60 K = 2 rad/s
æ 600 ´ 2p ö
2
÷ = 2× a × 2a Þ a = 10p rad/s
and ç 2
\ Angular speed w = w0 + at è 60 ø
= 2 + 10 × 2 = 22 rad/s BEGINNER'S BOX-2
Z:\NODE02\B0B0-BA\TARGET\PHY\SOLUTIONS\MODULE_2\04-ROTATION MOTION.P65
1 2
Angular displacement q = w0t + at
2
1 1.
= 2 ×2 + × 10 (2)2 = 24 radian y
2
(0, 3) 3 kg
3. w = A – Bt
1 kg 2 kg
x
(0,0) (2,0)
dw 10 kg
a= = –B
dt (–2, –2)
at t = 0 ; w0 = A
I = 1 × (0)2 + 2 × (2)2 + 3 × (0)2 + 10(2)2
2
w - w02 0-A 2
A 2
q= = = = 48 kg-m2
2a -2B 2B
, ®
34 Pre-Medical : Physics ALLEN
2. Iring = Idisc MR 2
11. Moment of inertia I = MR2 +
MR12 MR22 R 1 2
= Þ 1 =
2 4 R2 2
2MR 2 + MR 2 3
2 2 = = MR2
I1 M1R1 4 M1 æ 4 ö M1 1 2 2
3. = Þ = ´ç ÷ Þ =
I2 M2 R 2
2
1 M2 è 1 ø M2 4
2
12. It will be again = MR 2 , because the axis in
2 2 5
æLö æLö
4. I = m1 ç ÷ + m2 ç ÷
è2ø è2ø question is also diameter of the sphere.
= 1 × (1)2 + 2 × (1)2 = 3 kg-m2 BEGINNER'S BOX-3
5. I=
5 4
MR2 Þ MR2 = I 1. r r r
4 5 t=r ´F
3 3 4 6
I' = MR 2 = ´ I = I ˆi ˆj kˆ
2 2 5 5 r
t= 2 1 3
6. Moment of inertia of hollow cylinder will be larger
1 -2 5
as compare to disc because most of its mass is located
away from the axis of rotation as compare to that
of disc. r ˆ -4 - 1)
t = ˆi(5 - ( -6)) - ˆj(10 - 3) + k(
7. Spokes do not carry much mass. Most of the mass
r
is located at the rim. This gives cycle wheel more t = 11iˆ - 7jˆ - 5kˆ
moment of inertia for same mass.
2. About point B moment of inertia is less than A so
t it is easier to rotate about point B.
8. MA = pr2tr and MB = p(4r)2 r Þ MB = 4 MA
4
3. t = Ia
IA MA r 2
1
\ = = Þ IB = 64IA 2
IB MB ´ 16r 2 64 20 2 æ 20 ö
2× = ´ç ÷ ´a
9. Mass of plate = (Mass per unit area) (Area) = ml2 100 2 è 100 ø
Ml 2 ml2 ´ l2 ml 4 a = 10 rad/sec
Z:\NODE02\B0B0-BA\TARGET\PHY\SOLUTIONS\MODULE_2\04-ROTATION MOTION.P65
Moment of inertia I = = =
6 6 6
20
at = ra = ´ 10 = 2 m/s2
100
4. r1F1 = r2F2
10. 1.6 × 1 = 0.4 × F2
Þ F2 = 4 N
Moment of inertia 5. Net force on rod Fnet = 0
FG a IJ 2
FG a IJ 2
tnet = – F × 20 + F × 40 – F × 60 + F × 80
I = m1× 0 + m2 ×
H 2K + m3 ×
H 2K
= 40 F
a2
= (m2 + m3)
4 so rod experinces a torque
