HINTS AND SOLUTIONS
CONCEPTUAL QUESTION
1.
Torque =
r F
aiˆ bjˆ
Aiˆ Bjˆ
= Ba
Ab kˆ
Arm
of the force
r F
Ba Ab kˆ
A2
F
B2
m1m 2
r2 ; I
m1 + m 2 ‚
2.
I=
3.
L
7.1 10 23 kg m 2 / s
K
2.6 10 29 joule , t
4.
2.79 10
47
kg-m 2
2.35 109 year
When the rod begins to slip, its position changes from time to time. At any instant, two
successive positions (one a little before that instant and the other a little later) can be
represented by the rotation of the rod about the certain point called the instantaneous centre.
At any instant, the end A slides along the vertical direction, while the end B slides along the
horizontal. This motion may be imagined to be rotation about some point O ' as follows:
OA
cos
;
v1
d
OA
dt
sin
d
dt
OB
sin
;
v2
d
OB
dt
cos
d
dt
v1 A
O’
10m
60°
O
B
v2
Clearly, v 1
. sin
.O ' A
cos
.O ' B
and, v 2
if, O ' A O B and O ' B O A .
At any position of the rod, the instantaneous centre is the point of intersection of the vertical
line through the lower end and the horizontal line through the upper end. Therefore, the
coordinates of the instantaneous centre of rotation are 10 cos 60 0 and 10 sin 600 or 5 m and
5 3 m.
5m and 5 3 m from vertical & horizontal walls respectively.
5.
We know that acceleration a of a body rolling down an inclined plane is given by
a
g sin
1
K2
R2
2 2
R
5
For a sphere, K 2
where R = radius of the sphere.
5 g sin
7
a
Substituting the given values, we have
1
2
5 9.8
a
5 4.9
=
7
7
The angular acceleration
1
mv 2
2
E
6.
1
I
2
2
3.5 m/s2.
a / R, while the K.E is :
7
mv 2 .
10
The disc is shown in figure.
C
axis
b
a
O1
Let m be the mass of the cut out portion of the disc, then m
Mb 2
a2
We know that moment of inertia of a disc about an axis passing through its centre and
perpendicular to its plane is given by
1
Ma 2
2
Let I 1 be the moment of inertia of the uncut disc about an axis passing through O 1 and
perpendicular to the plane of disc. Then applying theorem of parallel axes, we have
I1
1
Ma 2
2
Mc 2
If I 2 be the moment of inertia of the cut-off portion about an axis passing through O 1 and
perpendicular to its plane, then
I2
1
mb 2
2
1 Mb 2 2
b
2 a2
Mb 4
2a 2
Moment of inertia of the holed disc about an axis passing through O1 and perpendicular to its
plane
I1 I 2
1
Ma 2
2
Mc 2
1 Mb 4
2 a2
=
1
M a2
2
(i) 35 rad/s2 , (ii) 3.5 m/s2, (iii)
1
M a2
2
7.
2c
2
To prevent slipping
4
b
a2
+
10 J
2c
2
b4
.
a2
F
a
a
o
30
h
mg
a
N
F
N
a
O
a/2
mg
the body will topple about point O, if
a
2
here h 2a cos30
a
mga 3 mg
2
1
2 3
F .h mg
a 3
0.21
8.
r F , so
9.
B
10.
I AB
11.
12.
0
.F 0
1.6 Ma 2
2
M 4a
2
0 rotational KE can’t be transferred from A to B as surfaces are frictionless.
Conserving Angular momentum before and after collision
mv
Now, I
a
I
2
m ( 2 a )2
12
ma 2
1
6
I
I1
I2
I3
a
2
2
1
2
2 2
ma .
3
mva
2
13.
m
3
2ma 2
3v
.
4a
I4
(this equation violates
r
axes theorem)
14.
Speed of the bottommost point is zero but acceleration is not zero. Friction force may be there
if it is an accelerated motion but work done by fritional is always zero.
15.
Velocity of particle is v
r and centripetal acceleration is ac
r
correct option are (a) and (c)
v
16.
In case of pure rolling
KR
KT
1 for a ring and 2/5 for a solid sphere.
Here K R rotational kinetic energy and KT transnational kinetic energy.
Therefore, friction of its total energy associated with rotation is :
1
1
for ring and
1 1 2
2
2
for solid sphere.
2 5 7
17.
On smooth horizontal surface it can roll with out slipping if v R
acting on it.
18.
L1
I
MK 2
L2
I
MRv
MK 2
M
K2
MR
R2
…(i)
R
as v
R
…(ii)
From equation (i) and (ii) we can see that
L2
19.
20.
2 L1
when K
R
and L2
2 L1
when K
R
B, D
Angular momentum L mr v
if net =0, then L constant
LEVEL – 1
and no external force is
Model Question
= 6 rad , = 6 rad / s ,
m1m 2
I=
r 2 ; I 2.79 10
m1 + m 2 ‚
21.
22.
23.
47
kg-m 2
M=10kg
a
R=0.1 m
T2
T1
a
m2 =8 kg
m1 =12kg
For m1, 12g T1
For m 2 , T2
8g 8a
Torque,
…(ii)
I
T1R T2 R
and a
….(i)
12a
R
…(iii)
I
….(iv)
From (i), (ii), (iii) and (iv)
a 1.57ms 2
T1 99N and T2
24.
max
Fmax
25.
(a)
(b)
91N
20 5
rad / s
3
16mg
=
3
=
The forces acting on the ladder are shown in figure. They are:
its weight W,
normal force N 1 exerted by the vertical wall,
(c) normal force N 2 exerted by the floor and
(d) frictional force f exerted by the floor.
N1
A
o
53
N2
W
B
f
c
Taking horizontal and vertical components, we get,
N1 f
… (i)
and
N2 W
… (ii)
Taking torque about B,
N 1 AO W CB
or,
N1
CB
W
AO
O
BO W 4 W
BO
AO
AO 2 3 2
The normal force by the floor is,
N 2 W 10 kg 10 m / s 2 =100 N.
or,
tan 530
4
3
… (iii)
The frictional force is,
200
f = N1 =
N.
3
26.
2u sin
=2 sec.
g
Time of flight = T =
So after 1 sec, the projectile will be at its maximum height
L
mv cos .H (directed along –ve z-axis)
L 10
kˆ kgm 2 / s
y
mv cos
H
x
1.
27.
Given that kinetic energy of ring is equal to the kinetic energy of the cylinder. Let it be x joule.
Now,
K .E
K .E
ring
1
1
m1v12
I
2
2
1
2
0.3 v1
2
0.3v12
So, 0.3v12
1
m1v12
2
2
1
1
0.3 v12
2
rotational
1 2
mr1
2
v
2
1
r
x
x
0.3
or, v1
K .E
translational
1/ 2
… (1)
Similarly,
K .E
1
m2 v22
2
3
m2 v22
4
0.3v22
So, 0.3v22
or, v2
K .E
cylinder
translational
K .E
rotational
1
m2 v22
4
3
0.4 v22
4
x
x
0.3
1/ 2
… (2)
As the velocity of ring and cylinder is the same and hence they will reach the wall
simultaneously.
28.
1
acceleration
MR 2
2
R
acceleration R
I
acceleration
2
A sin t
or, acceleration
1 MR 2 2 A
2
R
max
29.
Mg T
1
MRA
2
1
I
2
1 2
mv
2
3 2
mv
2
2
2
1 MD 2
v2
. 2
2
4
D /4
1 2
mv
2
mv 2
A
Ma
1 2
mv
2
mgh
2
max
mgh
mgh
mgh
h
T
a
mg
v is independent of D
10
2
30.
I
MR 2
rad/s2 = 5 rad/s2
1
0.2
2
2
0.02 kg-m 2
5 0.02 0.10 Nm .
I
PRACTICE QUESTION
31.
32.
33.
a = 1.57m / s 2 , T1 = 99N,T2 = 91 N
-----IA
I AB
I AC
I BC
4
m
3
4
m
3
2
2
1
m
3
2
3
m
2
6m
2
If
is the angular velocity in the second position, then using conservation of mechanical
energy, we have
3
3mg
3
1
6m
2
2
2
2
3mg
3
3
g 3
or,
Now, velocity of C at this instant is 2
34.
or 2 g
3 and maximum.
The total K.E. of the loop is given by
K .E
1 2
mv
2
loop
1
I
2
2
1 2
mv
2
1
mr 2
2
1 2
mv
2
1 2
mv
2
v r
1 2
mv
2
1 mr 2
2 2
1 2
mv
4
3 2
mv … (2)
4
K .E
disc
1 2
mv
2
2
… (1)
mv 2
2
Substituting the value from equation (1) in equation (2), we get
K .E
35.
C
36.
LA
LB
L
3
8 6J .
4
disc
2 2
mr
5
2 2
mr
5
LA
LB
ˆj (parallel to y-axis)
kˆ
2 3
mr
5
ˆj kˆ
There is no angular momentum about x-axis
Lx
37.
0
When the rod rotates through an angle
BG ' C
, the centre of gravity falls through a distance h. From
A
1
A
G
h
2
cos
2
or,
h
1 cos
2
Decrease in
P.E
mg
2
… (i)
1 cos
The decrease in P.E is equal to the kinetic energy of rotatation
K .E
1 m 2
2 3
rotation
2
… (ii)
From equations (i) and (ii), we get
1 m 2
2 3
2
mg
6g
38.
sin
2
2
1 cos
.
Rita of the disc and small mass about the axis of rotation
MR 2
2
3MR 2
2
MR 2
When the small body comes to bottom, its kinetic energy is given by
K .E
1
I
2
2
,
angular velocity
Loss of potential energy
1
I
2
or,
2
2
M gR
M gR
2M g R
I
Substituting the value of I, we get
4g
3R
LEVEL – 2
Model Question
1
I
2
2
39.
Here, the linear velocity increases due to friction and the angular velocity decreases due to the
torque of frictional force.
mg
ma
or,
a
g
0
gt
or,
v
gt
v
mg r
mr 2
I
g
r
g
t
r
0
v
r
0
When there is no slipping v
v
r
1
2
v
2v
r
v
r
0
0
r
,
0
r
0
r
v
mg(kinetic friction)
40.
The situation of the rod at an angle
is shown in figure.
x
Here, r iˆ cos jˆ sin
ˆ
and, F jF
(Force is always perpendicular to the initial position of the rod, which is
axis)
r F
ˆ
( i cos
ˆ )
jˆ sin ) ( jF
F cos kˆ
or, | | F cos
We know that
I
1
m
3
Where, I
F cos
1
m
3
2
d
d
2
Integrating within proper limits, we have
F
0
cos d
F sin
=
1
m
3
6F sin
m
1
m
3
2
0
2
2
2
‚
.
d
0 i.e., along the x-
41.
From conservation of mechanical energy,
1
I
2
2
mgh
1 1
mr 2
2 4
42.
I1
1
I1
I2
I2
g
5r
4
=
mr 2
2
mg 2 r
.
2
1
ma 2
2
1
1
ma 2
2
a
2m
2
=
2
2
.
a
t
t a 1
2
2
1
1
And the moment of inertia, I
ma 2
mr
2
2
The radial portion is : r
or,
43.
1
1
2
2
.
;
t /2
For m : mg T macm , where acm is the acceleration of the centre of mass of the wheel =
acceleration of m .
For wheels T f fr M 4 m ' acm
f fr r
and
f fr
1
m 'r 2
2
2m ' acm
=
1.4
IAB
2m ' r 2
4
acm
r
mg
6m ' m
Hence ac m
44.
ma 2
I
Further,
a
M
0.5 g
6 0.4 + 0.5
=
2
1.15 m/s .
1.6 Ma 2
2
[theorem of perpendicular axis]
I AB
45.
IAB
M 2a
2
[theorem of parallel axis] =4.8 Ma2
Applying the conservation of angular momentum about initial point of contact, we have
I cm
0
I cm
2
mR 2
5
2
mR 2
5
mvR
0
mvR
… (1)
When pure rolling starts and friction force vanishes
v R … (2)
Solving equations (a) and (b), we have
2
0
7
Now
mv 2
2
E
2
2
7
2
m
10
7
2
E0 ,
7
R
0
1 2
mR 2
2 5
where E0
46.
47.
2
I cm
2
0
.
D
Net external force on the system is zero. Therefore, linear momentum will remain conserved.
Initially both the particle have equal momentum in opposite directions. Hence initial
momentum is zero i.e. final linear momentum is also zero or velocity of center of mass vc 0 .
Let
be the angular velocity about centre of mass. Net external torque on system is also zero. Hence
angular momentum about center of mass will also be conserved i.e.
Li
or
Lf
2mva 4mva
6a
8m
or
2
2m a
12
24ma 2
6mva
I
2
m 2a
2ma 2
4ma 2
1
30ma 2
2
v
5a
2
30ma 2
v
5a
or
1
I
2
Finally E
2
2
3 2
mv
5
48.
L
mv0 r sin
y
v0 dt
b
r1
r2
x
L
r1
mv0b costant
r2
Now dA
r
1
v0 dt r sin
2
or
49.
dA
dt
mv0 r sin
2m
L
2m
Particle P will have a velocity in vertical direction if
R
R cos
or
Q
v
cos
1
v
R
Required angle
and
P v
cos
1
cos
1
is :
v
R
v
R
Point P corresponds to
cos
while Q corresponds to
v
R
1
cos
1
v
R
and it has velocity in vertically upward direction,
and it has velocity in vertically downward direction.
Paragraph Question 50 to 52
50.
51.
52.
We can again apply conservation of angular momentum principle
final
initial
R
R
V0= 0R
Li
Li
I
mV0 R
mV0 R
;
Li
2
V
0
Lf ;
V
V0
2
Friction will stop acting
PRACTICE QUESTION
53.
54.
55.
56.
57.
58.
59.
2R/3 above the centre
R /3
3v / 5
0.05Kgm2/s, 50cm/s
2.66 10-7
19.7 rad/s
2.4rad/s
60.
61.
mvr
( M m) R 2
I
10 gh / 7
62.
3v 2
4g
63.
64.
10 / 7 gl sin
17mg / 7
65.
27
g (R r)
7
PARAGRAPH QUESTION 66 TO 68
66.
A
67.
A
68.
C
From angular momentum conservation about axis of cone.
r0
C
D
h
E
r
m v0 r0 = m v r cos
v0 r0 = v r cos
…(i)
from energy conservation
E1 = E2
1 2
mv 0
2
1
mv 2
2
mgh
v 20
v
r0 – r = h tan
0
;
2gh
r = (r0 – h tan )
Nsin
N
Ncos
mg
From equation (i)
Eliminating v and substituting r = r0 – h tan
v0 r0
v 20
v
N cos
2
0
2gh (r0
= mg
mv
r0
2
0
Solving (ii) and (iii)
tan
h tan ) cos
v 0 r0
cos
N sin
2gh (r0
gr0
.
v 02
h tan )
…(ii)
…(iii)
LEVEL – 3
MODEL QUESTION
69.
Applying the law of conservation of angular momentum before and after impact, we have
mv c R cos
I
mk 2
Given that I
Further, v c
mk 2
vc
R
R and v C
mv c' R
vc
This gives, v c'
v c'
R
mv c R cos
v c'
R
mk 2
' mv c' R
I
(
vc 1
1
k2
R
R cos
R
k2
R
cos
)
2
.
‚
VC
1+
k
R2
VC
70.
Let m0= initial mass of the wheel at t = 0
m = mass of the wheel at time t
R = initial radius of the wheel
r = radius of the wheel at time t
Now
R2
r2
m0
m
R2
r2
r
where I
F
or ,
d
dt
R
m
m0
F
r
… (1)
d
dt
I
1
mr 2
2
d
1
r
mr 2
2
dt
2F
mr
Further,
2F m 0
d
dt
m 3/2
R
d dm
.
dm dt
d
dm
2F m 0
d
dm
R
m
3/2
.
Integrating this with proper limits for
, i.e.,
=
0
and
f
and m
=
m 0 and m
=
m0
2
,
we get the desired result.
71.
The hemisphere is moving with uniform velocity. Hence, it is in equilibrium (both rotational
as well as translational)
N
mg
.... (i)
O
G
N
mg
N
F
N
.... (ii)
Taking the torque due to all the forces about the point P, we get,
Net clockwise torque of applied force F = anticlockwise torque of mg about bottom most point
r sin
F r
mg
3r
sin
8
... (iii)
Solving these three equations, we get sin
72.
=
8
3
8
.
The free body diagrams of plank and sphere are shown in figure.
N
mg
Vcm
mg
mg
VP
Let a p and v p t be the acceleration and velocity of the plank respectively. Then
ap
g
and v p t
For sphere, acm
gt
g and v cm
gt
The torque acting on the sphere about centre of mass is
mgr
or,
Now,
t
mgr
I
5 g
I
2r
5 gt
0
2r
For rolling, v p v cm
t
I
2
mr 2
5
0
5 gt
r
2r
2 0r
9 g
The distance travelled by the plank ( s p ) is given by
sp
1
2
2 02 r 2
81 g
2
2 0r
9 g
g
.
Initially v 0 r 0 . Therefore, slipping is in forward. Hence, friction will be downward direction
to the incline. Once, v r , force of friction becomes upward.
73.
a
mg cos
mg sin
8
g sin
7
m
1
tan
7
1
tan
7
rmg cos
2
mr 2
5
5 g sin
14 r
This will last till v
(v 0 at 1 )
r
(
t 1 )r
0
Substituting the values
and
t1
2 (v 0
0r )
3 g sin
v
16
21
5
v0.
21
r
0
v0
a
0
sin
mg
Afterwards: Minimum value of
tan
mr 2
1
I
tan
1 5/2
f
mg cos
require for pure rolling in
tan
3.5
Since this value is greater than
tan
7
, slipping will occur even after that. Force of friction is
upwards but maximum. Hence, linear retardation
a
g sin
6
g sin
7
cos
Further time of rise
t2
v
a
16
21
0
r
5
v0
21
6
g sin
7
t1
and
74.
75.
76.
77.
t2
+
4 0 r + 7v 0
.
18 g sin
=
D
B
A
A
PRACTICE QUESTION
78.
The situation is as shown in the figure.
Here
we have,
and,
F
f
f
m 2 ac
Further, f
and,
m1 a p
r
ap
… (1)
… (2)
2
m 2r 2
5
I
ac
… (3)
r
… (4)
m2
ac
f
m1
F
ap
Substituting the value of f from equation (2) in equation (1), we get
F
m 2 ac
m 1a p
… (5)
Now we substitute the value of a p from equation (4) in equation (5), hence
F
m 2 ac
m 1 ac
r
… (6)
From equations (2) and (3)
r
or ,
5
ac
2
Substituting this value in equation (6), we have
or,
F
m 2 ac
or,
F
m2
or,
ac
Now
79.
7
m 1ac
2
7
m 1 ac
2
F
7
m2
m
2 1
ap
=
ac
+
r
=
7
ac
2
=
F
.
2
m1 + m 2
7
The free body diagram is shown in the figure.
The initial energy
1
I
2
2
0
where I is the moment of inertia of the cylinder and is given by
1
Mr 2
2
I
(M = Mass of the cylinder and r = Radius)
Initial Kinetic energy of cylinder
1
Mr 2
4
2
0
… (1)
f1
R
N
f2
Mg
Here, there is no motion of the centre of gravity of the cylinder, hence,
R
N
N
Mg
… (2)
… (3)
R
Solving for R and N,
R
Mg
1 2
… (4)
Mg
N
… (5)
2
1
The total initial energy is dissipated against frictional forces.
1
Mr 2
4
2
0
(
N
R ) .2 n ;
where n is the number of turns made by the cylinder before it stops.
Putting the values of N and R, and solving for n gives the final result.
80.
At angle
1
I
2
or,
2
2
3g
mg
2
(1 cos )
(1 cos )
…(i)
Differentiating w.r.t. ,
sin
2
m 2
3
3 g sin
… (ii)
2
3g
2
an
(1 cos )
2
2
3
at
g sin
2
4
f ma x m (at cos a n sin )
mg
and
m
3
3 g sin
g sin cos
(1 cos )
4
2
3
3
mg sin
cos 1
2
2
x
C
an
at
y
Further, mg N ma y
or,
N m ( g at )
m[g
(at sin
an cos )]
3
g sin 2
4
m g
3 g cos
(1 cos )
2
mg
[4 3 sin 2
6 cos
6 cos 2 ]
4
mg
(1 3 cos ) 2 .
4
The rod does not slip until
0
N
i.e.,
81.
=
cos
1
1
.
‚
3
= 60º
3 g (1 cos )
82.
w
83.
84.
85.
86.
87.
745N, 420N, .553
0.25m/s2
ADDITIONAL QUESTION’S HINT’S
Matching type questions
88.
89.
A
B
C
D
A
B
C
D
S
Q
P
R
S
P
P
Q, R
Assertion-Reason
90.
91.
92.
93.
B
C
D
A
94.
C
95.
A
96.
C
Please mention the type of Question (97-99)
P2
P1
97.
Apply F
98.
I1 I 2
2
3
MR 2
Mr 2
5
2
2
r
R
15
99.
Since part bc is frictionless, torque on the ball on this part is zero and hence its angular velocity
will be constant on this part. Further from conservation of mechanical energy hc < ha.
t
and principle of rotational equilibrium.
Passage
100.
101.
1 2 1
kx1
I(2 ) 2
2
2
1 2 1
kx 2
2I 2
2
2
From COAM
x1
x2
4
3
4I = 3I
Impulse = t
102.
103.
2
2I
4
3
2I
3t
1
16 2 1
8I 2
KE
3I
{2I 2 I 4 2}
2
9
2
3
2
2
3V
1
1 V
mV 2 ; But h
I
mgh
4g
2
2 r
3V 2 1 IV 2
4g 2 r 2
mV 2 1 V 2
I
4
2 r2
mg
3I
2
2
I
3
mV 2
2
mr 2
I
(Disc)
2
SUBJECTIVE QUESTION’S HINT’S
104.
If
is the density of the material of the carpet, initial mass of the carpet (cylinder) M will be
R L while when its radius becomes half the mass of cylindrical part will be
MF
( R / 2) 2 L
M /4
So initial PE of the carpet is MgR while final
( M / 4) g ( R / 2)
MgR / 8
Sol loss in potential energy when due to unrolling radius changes from R to R / 2
MgR (1/ 8) MgR
… (i)
(7 / 8)MgR
R
R/2
This loss in potential energy is equal to increase in KE which is
K
KT
1
Mv 2
2
KR
1
I
2
2
If v is the velocity when half the carpet has unrolled then as
v
and
i.e.,
I
R
M
, M
2
4
2
1 M R
2 4
2
K
1 M 2
v
2 4
K
1 2
Mv
8
1 MR 2
2 32
2
2v
R
1
Mv 2
16
3
Mv 2
16
… (ii)
so from equation (i) and (ii)
(3/16) Mv 2
i.e.,
105.
v
(7 / 8) MgR
(14 gR / 3) .
As moment of inertia of a disc about a diameter is
1 1
mR 2 , the moment of inertia of the disc
2 2
about the chord PQ by ‘ theorem of parallel axes’ will be
2
5
mR 2
16
and as particle of mass m is at a distance R ( R / 4) (5 / 4) R from PQ , the moment of inertia
( I D ) PQ
1
mR 2
4
1
R
4
m
of the system about PQ
I
( I D ) PQ
5
mR 2
16
( I P ) PQ
m
5
R
4
2
15
mR 2
8
m
A
Finally
P
D
Q
R/4 C'
P
C R/4
D
Initially
Q
R
A' m
Now if
is the angular speed of the system when A reaches the lowest point A on rotation
about the axis PQ , by ‘conservation of mechanical energy’.
1 2
I
mg ( AA ) mgCC
2
mg 2 AD 2CD
1 15
mR 2
2 8
i.e.,
2 mg
106.
R
1
R
4
g
5R
1
R
R
4
i.e.,
so,
2
1
R ,
4
4
(v ) A
r
g
5R
5gR .
(i) When truck accelerates along x-axis equation of translatory motion of a disc will be
F
f
ma
i.e.,
[ as F maT ]
9m f ma … (i)
While equation of rotational motion of a disc will be
I
i.e.,
f
r
1 2
mr ( a / r )
2
ma
… (ii)
i.e,
2f
so from equation (i) and (ii)
i.e,
3 f 9m
f 3 2 6N
or,
f
6iN .
Disc
y
F'=maT
x
0.1 m
(ii)
aT
f
Now as for wheels with centre O1 and O2
OP1
and
so
r1
0.1 j 0.1k
OP 2
1
r2
r1
0.1 j 0.1k
f
( 0.1 j 0.1k ) 6i
0.6 j 0.6k
and
2
r2
f
(0.1 j 0.1k ) 6i
0.6 j 0.6k
From this it is clear that
107.
1
2
0.6 2N×m and makes an angle of 45º with the rod.
(a) As X stops moving up, its acc. must be downwards along the plane. So equation of motion
of X will be
T mg sin 30
ma ,
i.e.,
T
1 1
g a
2 2
… (i)
Now or rotational motion of cylinder
T r I
a/r
1
here I
Mr 2
2
1
T r
Mr 2 (a / r )
2
1
T
2a
2
T a … (ii)
But as
and
so,
i.e,
or,
Solving equation (i) and (ii) for a and T we get
a
g
6
g
6
9.8
1.63 N .
6
and
T
(b)
When angular velocity of Y is 10 rad/s. the linear velocity of a point on it and so of X
will be
v r
0.2 10 2 m/s
Now as acc. of X is ( g / 6) down the plane, the distance moved by it, till it stops from
3rd equation of motion will be
108.
0 22 2( g / 6) s
i.e.,
s (12 / g ) 1.22m .
(a) As Fext 0 linear momentum of the system is conserved,
i.e.,
2m v m 2v 0
(2 m m 8m) V
or,
V 0
i.e.,
(b)
i.e.,
velocity of center of mass is zero.
As ext 0 angular momentum of the system is conserved,
m1v1r1 m2 v2 r2
( I1 I 2 I B )
2 mva m(2v)(2a)
[2m(a) 2
i.e.,
or,
(c)
m(2a) 2 8m (6a) 2 /12]
6mva 30ma 2
(v / 5a)
As from part (i) and (ii) it is clear that, the system has no translatory motion but only
rotatory motion,
E
1
I
2
2
1
v
(30ma 2 )
2
5a
109.
2
3 2
mv .
5
As Fext 0 , linear momentum of the system is conserved,
i.e.,
m1v1 m2 v2 (m1 m2 M )V
or,
0.08 10 0.08 6
i.e.,
(0.08 0.08 0.16)V
V (1.28 / 0.32) 4 m/s
… (i)
A
10 m/s
C
B
6 m/s
Also as ext 0 , angular momentum of the system about centre of mass is conserved
i.e.,
m1v1r1 m2 v2 r2 0
or,
(m1r12 m2 r22 I R )
0.08 10 0.5 0.08 6 0.5
2 0.08 (0.5)2
or,
i.e.,
0.16 ( 3) 2 /12
0.40 0.24 0.08
(0.16 / 0.08)
2 rad/s … (ii)
(-ve sign here indicates that rotation is clockwise)
From equation (i) and (ii) it is clear that the bar will translate and also rotate, so that its final
KE will be
KF
i.e.,
KF
1
1 2
MV 2
I
2
2
1
(0.08 0.08 0.16) 42
2
2.56 0.16 2.72 J
1
(0.08) ( 2) 2
2
i.e.,
KF
However, the initial KE of the system was
KI
1
1
0.08 10 2
0.08 (6) 2
2
2
4 1.44 5.44 J
0
KI
i.e.,
So there is loss in KE
KI
110.
5.44 2.72 2.72 J .
KF
When the cylinder rolls by an angle about the edge its centre of mass will go down by
h r (1 cos ) . So by conservation of ME,
mgr[1 cos ]
But here I
and
So
1 2
mv
2
1
I
2
2
1 2
mr
2
v r
mgr[1 cos
1 2 1
mv
2
2
gr[1 cos ]
]
1 2 2 2
mr (v / r )
2
(3 / 4)v 2 … (i)
i.e.,
Now, equation of cylinder about the edge,
mg cos
R
(mv 2 / r )
The cylinder will leave contact when R
… (ii)
g cos
(v 2 / r )
0
O
r
mg
R
E
(a)
O'
Eliminating v 2 between equation (i) and (ii)
gr[1 cos ] (3 / 4) gr cos
(b)
i.e.,
cos 1 (4 / 7)
Now substituting this value of cos in equation (ii),
v
(c)
(4 / 7) gr
At the time the cylinder leaves contact with the edge.
1
I
2
KR
and then as
1 2
mv
4
2
0,
1
mgr
7
0 and so
1
mgr
7
constt. i.e., after leaving the edge K R
constt.
so when centre of mass comes in horizontal line with the edge, by conservation of ME,
mgr
111.
i.e.,
KT
so,
KT
KR
KR
KT
1
6
mgr
mgr
mgr
7
7
(6 / 7) mgr
6.
(1/ 7) mgr
(a) Initially the insect has an angular momentum about the point O and so when it comes in
contact with the rod both will rotate about O . If is the initial angular velocity, by
conservation of angular momentum, i.e.,
MVr
I
we have MV
or,
MV
i.e.,
(b)
ML2
12
L
4
L
4
12
7
M
L
4
2
7
ML2
48
V
… (i)
L
Since the weight of the insect will exert a torque, the angular momentum of the system will not
be conserved. Let at any time t , the insect be at a distance x from O on the rod and by then
the rod has rotated through an angle . Then angular momentum of the system will be
J
so
and,
dJ
dt
ML2
12
Mx 2
dx
dt
Mgx cos
… (ii)
2 Mx
[as
Mgx cos t … (iii)
constt. given]
[as
t]
S
L/2
L/2
B
A
L/4
C
O
x
Mg
(dJ / dt ) , equation (ii) and (iii) yields
dx
Mgx cos t 2 Mx
dt
g
or,
dx
cos t dt
2
According to given conditions, i.e., for x L / 4 , t
Since
above equation becomes
0 and x
L/2 , t
/2
[ as
t
/ 2 ], as
L/2
/2
g
(cos t )dt
2
0
g
/2
sin t 0
2
2
dx
L/4
i.e.,
x
or,
L
4
L/ 2
L/ 4
g
2
2
,
2g
L
i.e.,
Substituting this value of
equation (i)
2g
L
112.
i.e.,
V
So,
V
in
12 V
,
7 L
7
2 gL
12
7
2 10 1.8
12
7
2
3.5 m/s .
(a) Let the linear and angular velocities of centre of mass of rod just after collision be VCM and
respectively. Applying conservation of linear and angular momentum,
mv0 MvCM
… (i)
ML2
12
mv0 L
2
… (ii)
As the collision is elastic, so applying conservation of energy,
1 2
mv0
2
1
2
MvCM
2
1 ML2
2 12
from equation (i),
vCM
2
… (iii)
mv0
M
6mv0
,
ML
and from equation (ii),
Substituting in equation (iii),and simplifying,
m
M
1
.
4
m v0
A
x
L
vCM
P
B
(b)
For the velocity of point P immediately after collision w.r. to centre of mass of be
zero,
x
x
i.e.,
x
L
2
vCM
L 6 mv0
2 ML
2L
.
3
0
mv0
M
0,
(c)
Angle turned by rod in time
t
6mv0
ML
6 m
3 M
L
,
3v0
L
3v0
6 1
3 4
2
linear velocity of P in the direction perpendicular to rod (as
L
6
So,
L 6mv0
6 ML
v0
4
113.
2
vCM
speed of P
2
v0
4
2
/ 2 ),
v0
.
4
v0
4
2
v0
.
2 2
(a) From, Angular impulse = change in angular momentum
or,
(6 N s)(0.5 m) I AB (
) ...(i)
similarly, from impulse-momentum theorem,
Impulse = change in momentum
6N s m (
) … (ii)
Dividing equation (i) by (ii), we get
I AB 1
m
2
IC m 2 1
or,
[ IC
m
2
0.4 m or 0.1m.
Hence
Putting
0.4m in (ii), we get
1.2 kg-m 2 ]
0.5 rad/s which is not possible because the sheet has to
return back.
Putting
0.1m is equation (ii), we get
1 rad/s , which is possible for elastic collision.
Since the sheet returns back with same angular the velocity, it will never come to rest.