Rotational Motion– 1 Practice Questions - DPP for NEET

 Page 1


1. (a) Initial angular momentum of ring. L = Iw =Mr
2
w
Final angular momentum of ring and four particles
system 
( )
2 2 2'
M
Mr Mr 4mr
M 4m
w
w= + w=
+
2. (b) The angular momontum of a system of particles is con
served when no external torque acts on the system.
3. (c) Rotational kinetic energy 
2
L
E L 2EI
2l
\=
A AA
B BB
L EI 1
1005
L EI4
Þ = ´ = ´=
4. (c) Angular momentum L = Iw constant
\ I increases and w decreases
5. (c) Conservation of angular momentum
I
1
w
1
+I
2
w
2
 = (I
1
w
1
+I
2
)
w
Angular velocity of system 
1 1 22
12
II
II
w+w
w=
+
\  Rotational kinetic energy = ( )
2
12
1
II
2
+w
( )
( )
( )
2 2
1 1 22 1 1 22
12
12 12
II II 1
II
2 II 2II
w+w æö w+w
=+=
ç÷
+ + èø
6. (d) Kinetic energy 
11
E L L 2n
22
= w= ´p
2 21
1 12
L En
E Ln
L En
\¥´Þ =´
21 12
2
1 11
LE/2nLL
L
L E 2n 44
é ùéù
= ´ Þ ==
ê úêú
ë ûëû
7. (b)
2
L
E
2I
= . If boy stretches hgis arm then moment of
inertia increases and accordingly kinetic energy of the system
decreases because L = constant and 
1
E
I
¥
8. (c)According  to conservation of angular momentum
( )
1
11 22 1 1 2 22
12
I
I I I II
II
w
\w = w Þ w= + w Þw=
+
9. (a)
7 32
L 2EI 2 10 8 10 4 10 kg m / s
--
= = ´ ´ ´ =´
10. (d) Angular momentum, of earth about its axis of rotation,
2
2
2 2 4 MR
L l MR
5 T 5T
pp
=w= ´=
11. (d) Angular  momentum, L = mvr = 
22
2
mrmr
T
p
w = ´´
= 
( )
2
24 11
402
7
2 3.14 6 10 1.5 10
2.7 10 kg m / s
3.14 10
´ ´´ ´´
=´-
´
12. (c)
2 1800
2 60
60
´
=== n
p
wpp
rad/s
=´ P tw
Þ
3
100 10
531
60
P
Nm
´
== =- t
wp
13. (a)
0 00 21
43
44
A AA LL dL
dtt
- -
t== ==
D
ur
r
14. (c)
21
60
20
2() 60
60
æö
-
ç÷
èø -
==
nn
t
p
p
a
2
60 30
--
==
pp
 rad / sec
2
\t=a I
= 
2
30
´p
 = 
15
Nm
p
-
15. (a)
$ $ $ $
(7 3 ) ( 3 5) r f i j k i jk
r r ur
$$
t= ´ = + + ´- ++
$ $
7 31
3 15
=
-
$
r
i jk
t
$ $
(15 1) (35 3) (7 9) = - - + ++
$
i jk
$$
14 38 16 i jk = -+
$
16. (a)
2
1000
5 ad / sec
200
r
I
t
a===
From 
0
0 5 3 15 = + = + ´= t w wa rad/s
17. (a)
2
30
15 rad/s
2
= ==
I
t
a
Q
2
0
1
2
q=w +a tt
Page 2


1. (a) Initial angular momentum of ring. L = Iw =Mr
2
w
Final angular momentum of ring and four particles
system 
( )
2 2 2'
M
Mr Mr 4mr
M 4m
w
w= + w=
+
2. (b) The angular momontum of a system of particles is con
served when no external torque acts on the system.
3. (c) Rotational kinetic energy 
2
L
E L 2EI
2l
\=
A AA
B BB
L EI 1
1005
L EI4
Þ = ´ = ´=
4. (c) Angular momentum L = Iw constant
\ I increases and w decreases
5. (c) Conservation of angular momentum
I
1
w
1
+I
2
w
2
 = (I
1
w
1
+I
2
)
w
Angular velocity of system 
1 1 22
12
II
II
w+w
w=
+
\  Rotational kinetic energy = ( )
2
12
1
II
2
+w
( )
( )
( )
2 2
1 1 22 1 1 22
12
12 12
II II 1
II
2 II 2II
w+w æö w+w
=+=
ç÷
+ + èø
6. (d) Kinetic energy 
11
E L L 2n
22
= w= ´p
2 21
1 12
L En
E Ln
L En
\¥´Þ =´
21 12
2
1 11
LE/2nLL
L
L E 2n 44
é ùéù
= ´ Þ ==
ê úêú
ë ûëû
7. (b)
2
L
E
2I
= . If boy stretches hgis arm then moment of
inertia increases and accordingly kinetic energy of the system
decreases because L = constant and 
1
E
I
¥
8. (c)According  to conservation of angular momentum
( )
1
11 22 1 1 2 22
12
I
I I I II
II
w
\w = w Þ w= + w Þw=
+
9. (a)
7 32
L 2EI 2 10 8 10 4 10 kg m / s
--
= = ´ ´ ´ =´
10. (d) Angular momentum, of earth about its axis of rotation,
2
2
2 2 4 MR
L l MR
5 T 5T
pp
=w= ´=
11. (d) Angular  momentum, L = mvr = 
22
2
mrmr
T
p
w = ´´
= 
( )
2
24 11
402
7
2 3.14 6 10 1.5 10
2.7 10 kg m / s
3.14 10
´ ´´ ´´
=´-
´
12. (c)
2 1800
2 60
60
´
=== n
p
wpp
rad/s
=´ P tw
Þ
3
100 10
531
60
P
Nm
´
== =- t
wp
13. (a)
0 00 21
43
44
A AA LL dL
dtt
- -
t== ==
D
ur
r
14. (c)
21
60
20
2() 60
60
æö
-
ç÷
èø -
==
nn
t
p
p
a
2
60 30
--
==
pp
 rad / sec
2
\t=a I
= 
2
30
´p
 = 
15
Nm
p
-
15. (a)
$ $ $ $
(7 3 ) ( 3 5) r f i j k i jk
r r ur
$$
t= ´ = + + ´- ++
$ $
7 31
3 15
=
-
$
r
i jk
t
$ $
(15 1) (35 3) (7 9) = - - + ++
$
i jk
$$
14 38 16 i jk = -+
$
16. (a)
2
1000
5 ad / sec
200
r
I
t
a===
From 
0
0 5 3 15 = + = + ´= t w wa rad/s
17. (a)
2
30
15 rad/s
2
= ==
I
t
a
Q
2
0
1
2
q=w +a tt
DPP/ P 15
45
2
1
0 (15) (10)
2
=+ ´´
= 750 rad
18. (d) As the block remains stationary therefore
For translatory equilibrium
0
x
F FN = \=
å
and 0
y
F f mg = \=
å
mg
f
O
N
F
For rotational equilibrium 0 t=
å
By taking the torque of different forces about point 0
0
F f N mg
uu u r uu r uu u r uuu u r
t +t +t +t =
As F and mg passing through point O
0
fN
\t +t=
uur u uu r
As 
00
fN
t ¹ \t¹
and torque by friction and normal reaction will be in
opposite direction.
19. (c) The velocity of the top point of the wheel is twice that
of centre of mass and the speed of centre of mass is
same for both the wheels (Angular speeds are different).
20. (d)
2 21
4500 1200
2
2() 60
rad/s
10
- æö
ç÷
èø -
==
nn
t
p
p
a
2
3300
2
3 60 de gre e
60
102
s
p
=´
p
2
1980 degree/s a=
21. (b)
2
0
1
2
=+tt qwa
Þ = q 100 rad
\
 Number of revolution 
100
16
2
==
p
(approx.)
22. (a) As mechanical contact is not made, total angular
momentum remains constant.
\ Iw
0
=  constant
Differentiating both sides,
D (Iw
0
) = 0
00
I I0 Þ Dw +w D=
I
0
I
DwD
Þ +=
w
 
0
0
I
I
Dw D
Þ =-
w
Also,  
0
0
I
I
Dw D
=-
w
2R I 2R
R IR
D DD æö
=-=
ç÷
èø
Q
2T =- aD
23. (a)
2
2
L
EK
I
== (given)
1
K
I
\µ (If L = constant)
When child stretches his arms the moment of inertia of
system get doubled so kinetic energy will becomes half i.e.
K/2.
24.    (c). Angular impulse = change in angular momentum : Frt =
L Þ L
1
 < L
2
K = 
2
L
2I
Þ K
1
 = K
2
25. (b);  26. (a);  27. (c)
Drawing the F.B. D of the plank and the cylinder.
N
1
f
1
mg
F cos
F sin
q
q
 
N
1
N
2
f
2
f
1
Mg
Equations of motion are
F cos q – f
1
 = ma ....(1)
F sinq + N
1
 = mg ....(2)
f
1
+ f
2
 = MA .....(3)
f
1
R – f
2
R = Ia .....(4)
A = Ra .....(5)
( )( ) [ ]
2
1
4 55
4 cos
2
10 m/s
38 3 1 81
F
a
Mm
´´
q
= ==
+ ´ +´
Page 3


1. (a) Initial angular momentum of ring. L = Iw =Mr
2
w
Final angular momentum of ring and four particles
system 
( )
2 2 2'
M
Mr Mr 4mr
M 4m
w
w= + w=
+
2. (b) The angular momontum of a system of particles is con
served when no external torque acts on the system.
3. (c) Rotational kinetic energy 
2
L
E L 2EI
2l
\=
A AA
B BB
L EI 1
1005
L EI4
Þ = ´ = ´=
4. (c) Angular momentum L = Iw constant
\ I increases and w decreases
5. (c) Conservation of angular momentum
I
1
w
1
+I
2
w
2
 = (I
1
w
1
+I
2
)
w
Angular velocity of system 
1 1 22
12
II
II
w+w
w=
+
\  Rotational kinetic energy = ( )
2
12
1
II
2
+w
( )
( )
( )
2 2
1 1 22 1 1 22
12
12 12
II II 1
II
2 II 2II
w+w æö w+w
=+=
ç÷
+ + èø
6. (d) Kinetic energy 
11
E L L 2n
22
= w= ´p
2 21
1 12
L En
E Ln
L En
\¥´Þ =´
21 12
2
1 11
LE/2nLL
L
L E 2n 44
é ùéù
= ´ Þ ==
ê úêú
ë ûëû
7. (b)
2
L
E
2I
= . If boy stretches hgis arm then moment of
inertia increases and accordingly kinetic energy of the system
decreases because L = constant and 
1
E
I
¥
8. (c)According  to conservation of angular momentum
( )
1
11 22 1 1 2 22
12
I
I I I II
II
w
\w = w Þ w= + w Þw=
+
9. (a)
7 32
L 2EI 2 10 8 10 4 10 kg m / s
--
= = ´ ´ ´ =´
10. (d) Angular momentum, of earth about its axis of rotation,
2
2
2 2 4 MR
L l MR
5 T 5T
pp
=w= ´=
11. (d) Angular  momentum, L = mvr = 
22
2
mrmr
T
p
w = ´´
= 
( )
2
24 11
402
7
2 3.14 6 10 1.5 10
2.7 10 kg m / s
3.14 10
´ ´´ ´´
=´-
´
12. (c)
2 1800
2 60
60
´
=== n
p
wpp
rad/s
=´ P tw
Þ
3
100 10
531
60
P
Nm
´
== =- t
wp
13. (a)
0 00 21
43
44
A AA LL dL
dtt
- -
t== ==
D
ur
r
14. (c)
21
60
20
2() 60
60
æö
-
ç÷
èø -
==
nn
t
p
p
a
2
60 30
--
==
pp
 rad / sec
2
\t=a I
= 
2
30
´p
 = 
15
Nm
p
-
15. (a)
$ $ $ $
(7 3 ) ( 3 5) r f i j k i jk
r r ur
$$
t= ´ = + + ´- ++
$ $
7 31
3 15
=
-
$
r
i jk
t
$ $
(15 1) (35 3) (7 9) = - - + ++
$
i jk
$$
14 38 16 i jk = -+
$
16. (a)
2
1000
5 ad / sec
200
r
I
t
a===
From 
0
0 5 3 15 = + = + ´= t w wa rad/s
17. (a)
2
30
15 rad/s
2
= ==
I
t
a
Q
2
0
1
2
q=w +a tt
DPP/ P 15
45
2
1
0 (15) (10)
2
=+ ´´
= 750 rad
18. (d) As the block remains stationary therefore
For translatory equilibrium
0
x
F FN = \=
å
and 0
y
F f mg = \=
å
mg
f
O
N
F
For rotational equilibrium 0 t=
å
By taking the torque of different forces about point 0
0
F f N mg
uu u r uu r uu u r uuu u r
t +t +t +t =
As F and mg passing through point O
0
fN
\t +t=
uur u uu r
As 
00
fN
t ¹ \t¹
and torque by friction and normal reaction will be in
opposite direction.
19. (c) The velocity of the top point of the wheel is twice that
of centre of mass and the speed of centre of mass is
same for both the wheels (Angular speeds are different).
20. (d)
2 21
4500 1200
2
2() 60
rad/s
10
- æö
ç÷
èø -
==
nn
t
p
p
a
2
3300
2
3 60 de gre e
60
102
s
p
=´
p
2
1980 degree/s a=
21. (b)
2
0
1
2
=+tt qwa
Þ = q 100 rad
\
 Number of revolution 
100
16
2
==
p
(approx.)
22. (a) As mechanical contact is not made, total angular
momentum remains constant.
\ Iw
0
=  constant
Differentiating both sides,
D (Iw
0
) = 0
00
I I0 Þ Dw +w D=
I
0
I
DwD
Þ +=
w
 
0
0
I
I
Dw D
Þ =-
w
Also,  
0
0
I
I
Dw D
=-
w
2R I 2R
R IR
D DD æö
=-=
ç÷
èø
Q
2T =- aD
23. (a)
2
2
L
EK
I
== (given)
1
K
I
\µ (If L = constant)
When child stretches his arms the moment of inertia of
system get doubled so kinetic energy will becomes half i.e.
K/2.
24.    (c). Angular impulse = change in angular momentum : Frt =
L Þ L
1
 < L
2
K = 
2
L
2I
Þ K
1
 = K
2
25. (b);  26. (a);  27. (c)
Drawing the F.B. D of the plank and the cylinder.
N
1
f
1
mg
F cos
F sin
q
q
 
N
1
N
2
f
2
f
1
Mg
Equations of motion are
F cos q – f
1
 = ma ....(1)
F sinq + N
1
 = mg ....(2)
f
1
+ f
2
 = MA .....(3)
f
1
R – f
2
R = Ia .....(4)
A = Ra .....(5)
( )( ) [ ]
2
1
4 55
4 cos
2
10 m/s
38 3 1 81
F
a
Mm
´´
q
= ==
+ ´ +´
46
DPP/ P 15
1
1
3 1 55
3 cos
2
7.5
3 8 31 81
MF
fN
Mm
´´´
q
= ==
+ ´ +´
and 
2
1
1 55
cos
2
2.5
3 8 31 81
MF
fN
Mm
´´
q
= ==
+ ´ +´
28. (b) t=
ur
r
dL
dt
 and =w LI
29. (b) sin. t=q rF If q = 90° then 
max
t= rF
Unit of torque is N-m.
30. (d) Torque = Force × perpendicular distance of the line of
action of force from the axis of rotation (d).
Hence for a given applied force, torque or true tendency
of rotation will be high for large value of d. If distance
d is smaller, then greater force is required to cause the
same torque, hence it is harder to open or shut down
the door by applying a force near the hinge.