PG Brainstormer - 9C (MECHANICS) - Solution Copy635526720976292613
PG Brainstormer - 9C (MECHANICS) - Solution Copy635526720976292613
PG Brainstormer - 9C (MECHANICS) - Solution Copy635526720976292613
PG Brainstormer - 9 C
RIGID BODY DYNAMICS & CONSERVATION LAWS
Solutions
Time Allowed
Maximum Marks
45 Min
45
Section - A
Single Choice Correct Type Questions (Q. No. 1 to 15)
1.
Sol.
GA
LA
XY
In this section 15 questions are there and each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. Marking Scheme - +3 for RIGHT answer and -1 for WRONG answer.
(A)
As shown in figure for pure rolling we use a = r
F + fr = mac
(1)
R(F fr) = I (
(2)
fr
2.
(A)
Sol.
3.
Sol.
(D)
The instantaneous axis of rotation will fall on the outside of the sphere. Since in case
of pure rotation, instantaneous axis of rotation will be on the centre. In case of pure
rolling, instantaneous axis of rotation will be on P.
1
v
mL2sin2 and =
3
L sin
YS
IC
S
(B)
As shown in figure we use
y2 +
vc
C
P
(x/2)2 = l2
dy x dx
2y
=0
dt 2 dt
x dx
l 3
dy
=
=
=
4
y
dt
dt
4 (l / 2)
But if vc > R, so instantaneous axis of rotation will fall out of the sphere.
4.
Sol.
30
A
3 /2 m/s
(A)
Using Impulse - momentum equation for rotational motion of rod we have MV(L/2) = 2 M (L/2)2
V/L
6.
Sol.
(B)
For equilibrium of hemisphere at maximum angle angle of inclie should be angle of repose and q should be such
that the line of action of Mg passes through the point of contact of the hemisphere with the incline plane.
PH
5.
Sol.
a
a/2
=
sin( )
sin
N
a/2
a
sin = 2 sin
= sin1 (2 sin )
PHYSICS
Mg
Page 1
PG Brainstormer - 9C Solutions
7.
Sol.
(A)
As shown in figure if point P has zero acceleration then we use 2r = R sin
r = R cos
tan = 2/ =
= 30
r = 3m
r = 3 sin 30 i 3 cos 30 j
= 1.5 ( i
R cos
3 j )
R sin
(C)
Figure shows the force exterted by the hinges on the door by arrows. These forces act
on door due to hinges to keep the door in equilibrium (translational as well as rotational)
under the effect of its weight.
9.
(B)
Sol.
Given that
d
= - k
d
For decrease in angular speed to half we use 0 / 2
2 n
kd
0
0
4n
GA
LA
XY
8.
Sol.
2R
Mg
0 / 2
10.
Sol.
YS
IC
S
Again from half value of angular speed to rest if it takes m revolutions we use 2m
4n d
m=n
(A)
Using angular momentum conservation just before and after contact with step, we have
2
3R MR 0 3
2
(M0R)
+ 2 = MR
4
2
11.
Sol.
(A)
Minimum angular speed is such that after impact the angular speed of cylinder must be sufficient enough to take the
cylinder to the top of wedge so by work energy theorem we use mg
mg
R 1 3
MR22
4 2 2
R 3
25
MR2 ( 02 )
4 4
36
02 =
PHYSICS
6 0
PH
36 g
25 R 3
6 g
2 3g
5 3R 5 R
Page 2
12.
Sol.
PG Brainstormer - 9C Solutions
(A)
On the cylinder, just after contact with the step normal reaction is maximum, if it becomes zero due to rotation of
cylinder about the edge of step, cylinder can break off / jump from the step, thus we use mg cos = mR2
mg
3
= mR2
4
3g
4R
13.
Sol.
3 3g
5 R
(C)
Using cosine rule in the triangle OAB we have l2 = r2 + x2 2rx cos t
x = r cos t + l
[assume OB = x]
r2
2
1 2 sin t
l
GA
LA
XY
0 =
dx
r 2
r
r sin t
sin t cos t r sin t 1 cos t
dt
l
l
14.
(A)
As point A is on the flywheel, it is in uniform circular motion which is periodic.
15.
Sol.
(B)
In the triangle OAB we can use -
YS
IC
S
l sin r sin t
Differentiating we get -
l cos
d r cos t
d
r cos t
dt
l
dt
* *
PH
PHYSICS
Page 3