Transcript 25. Rotational Mechanics

Physics
Session
Rotational Mechanics -7
Session Objectives
Session Objective
1. Problems
Class Test
Class Exercise - 1
A wheel of mass 2 kg having practically all
the mass concentrated along the
circumference of a circle of radius 20 cm
is rotating on its axis with an angular
velocity of 100 rad/s. The rotational
kinetic energy of the wheel is
(a) 4 J
(b) 70 J
(c) 400 J
(d) 800 J
Solution
1 2
Rotational kinetic energy = I
2
I  MR2
[Ring]
I  2  (0.2)2  8  10 2 kg m2
  100 rad/sec
1
K   8  10 2  10 4
2
K  400 J
Hence, answer is (c).
Class Exercise - 2
A rod of length L is hinged from one end.
It is brought to the horizontal position
and released. The angular velocity of
the rod when it is in vertical position is
(a)
2g
L
(b)
3g
L
(c)
g
2L
(d)
g
L
Solution
Loss in potential energy = Gain in
kinetic
energy
L
Loss in potential energy =Mg
2
L

 Center of mass of rod is at 
2

1
Gain in kinetic energy = I2
2
Solution contd..
ML2
I = Moment of inertia of rod about fixed end =
3
L 1 ML2
 Mg  
 2
2 2
3
Hence, answer is (b).

3g
L
Class Exercise - 3
The moment of inertia of a square
plate about one of its diagonal is
(Assume to be the side of the
square and mass M)
M 2
(a)
6
M 2
(b)
12
M 2
(c)
3
M 2
(d)
24
Solution
MI of a square plate about an axis
passing through its centre and
M 2
perpendicular to its plane is
 I0
6
I0  ID1  ID2
ID1  ID2  (Diagonals are same and symmetric)
I0 = 2ID
M 2
ID 
12
I0
ID 
2
Hence, answer is (b).
Class Exercise - 4
A uniform disc of radius R and mass M
is mounted on an axle as shown in the
figure. A light cord is wrapped around
the rim of the wheel and a downward
pull T is exerted on the cord. The
angular acceleration of the wheel is
2T
(a)
MR
4T
(c)
MR
T
(b)
MR
2T
(d)
M
O
T
Solution
Applying torque equation, torque
of tension about O = TR
  I
TR  I
MR2
I
2
MR2
TR 

2
2T

MR
Hence, answer is (a).
Class Exercise - 5
A particle of mass m is released from
rest at A, falling parallel to y-axis. The
torque acting on the particle with respect
to origin and angular momentum at any
time with respect to origin is
(a) mgb and mgt
(b) mgt and mg
(c) mgb and mgbt (d) mgt and mgbt
O
b
A
Solution
Torque  Force  Perpendicular distance
Force  Only force due to gravity  mg
  mgb
... (i)
Angular momentum = Linear momentum × Perpendicular
distance from the axis
Velocity after time t of falling = gt
 Momentum = mgt
L = mgt × b
L = mgbt
… (ii)
Hence, answer is (c).
Class Exercise - 6
The ratio of the radius of gyration of a
circular disc and a circular ring of the
same mass and radius about a tangential
axis parallel to the plane is
(a) 6 : 5
(b) 1 : 2
(c) 5 : 6
(d) None of these
Solution
M.I. of circular disc about tangential
axis by theorem of parallel axis is
2
MR2
5MR
 MR2 
4
4
Radius of gyration =
5
R
4
M.I. of ring about tangential axis by theorem of
2
MR2
3MR
parallel axis is
 MR2 
2
2
3
Radius of gyration = R
2
Hence, answer is (d).
Ratio 
5
2


4
3
5
6
Class Exercise - 7
The moment of inertia of a cylinder
of radius a mass M and height h
about an axis parallel to the axis of
the cylinder and distant b from the
centre is
3
Ma2
4
3
(c) Ma2
2
(a)
(b)
2
Ma2
3
(d) None of these
Solution
According to theorem of parallel axis,
I  I0  Md2
Ma2
I
 Mb2
2
 a2

2
I  M
b 
 2

Hence, answer is (d).
Thank you