Download rotational kinetic energy

Physics
Session
Rotational Mechanics - 4
Session Objectives
Session Objective
1. Angular Momentum of a Particle
2. Conservation of angular momentum
3. Angular Impulse
Angular Momentum
Angular momentum of a particle
Particle P rotates around Point O
Angular Momentum
L  of P
v
L  r  p (p  mv)

 mr  (  r)
P(m)
L is an axial vector units : Kg m2 / s
L  mvr sin 
L  I ( I  mr2 )
r
O
Angular Momentum for a System
of Particles
For a system of particles (i = 1 to n)
L
n
r
i 1
i
 pi  I
( is a cons tan t. So I   m i ri2 )
Angular momentum depends on
the position of the axis
Angular Momentum for a Rigid
Body
For a rigid body
L  I
( I : moment of inertia around
the axis of rotation)
Z

Y
r
O
m
X
Conservation of Angular Momentum
L  l
dL
d
l
 l
dt
dt
ext 
dL
dt
If, ext

d 




dt 
ext  l 
dL
 0 then,
0
dt
 L  cons tan t
If ext  0 lii  lf f
Conservation of Angular Momentum

If no external torque  acts on a body
 dL

 0  so the total angular momentum


 dt

of the body (or system of bodies) remains
a constant , and the vector sum of all torques
add up to zero.
Equilibrium of a Rigid Body
Two conditions are necessary.
(i) Total force acting on the body
must add up to zero (equilibrium of
linear motion)
(ii) Total torque acting on the body
must add up to zero (equilibrium of
rotational motion)
Comparison
LINEAR
Mass
Momentum
m
P=Mv
ANGULAR
I
L  I
Newton’s
Second Law
F=ma
Work
W   Fdx
W   d
Kinetic
Energy
1
K  mv 2
2
1 2
K  I
2
Power(const
ant force)
P=Fv
P  
  I
Class Test
Class Exercise - 1
A planet revolves round the sun as
shown. The KE is greatest at
(a) A
(b) B
(c) C
(d) D
C
B
A
D
Solution
Kinetic energy at any point:
2
2
L
L
K

2I 2mr 2
L is a constant for the system (as motion of sun
is negligible). L is constant for planet. So K is

1
maximum for smallest r  K 
which is A.


r2 
Hence, answer is (a).
Class Exercise - 2
A horizontal platform with a mass of
100 kg rotates at 10 rpm about a
vertical axis passing through its centre.
A man weighing 60 kg is standing on
the edge. With what velocity will the
platform begin to rotate if the man
moves from the edge to the centre?
(a) 22 rpm
(b) 11 rpm
(c) 44 rpm
(d) 66 rpm
Solution
Moment of inertia of platform:
1
2
mPr  I1
2
Moment of inertia of man at edge:
mmr 2  I2
Moment of inertia of man at centre: O (
r = 0)
Solution contd..
L is conserved:

I1  I2  1  I12
1
1

   100  60   10   100  2
2
2


1100  502

2  22 rpm
Hence, answer is (a).
Class Exercise - 3
A mass moves about the Y-axis
with acceleration ay = (by2 – c); b
and c are constants. The value of y
for which the angular momentum
is zero, is
3c
(a)
b
(c)
3c
b
(b)
c
b
(d)
3b
c
Solution
dv y
Acceleration  a y 
 by 2  c
dt
dv y
 vy 
 by 2  c
dy
vy


0

y
v y dv y 


2
by  c dy
0
3


by
2 
vy 
 cy  2
 3



by3

 cy  y 
3
3c
b
L (= mvyr) is zero when vy = 0
Hence, the answer is (c).
Class Exercise - 4
A thin circular ring of mass M and radius
R is rotating about its axis with a constant
angular velocity . Two objects each of
mass m are attached gently to the ring.
The ring now rotates with an angular
velocity.
M
(a)
Mm
(b)
M
(c)
M  2m
(d)
 M  2m
M  2m
 M  2m
M
Solution
I (ring) = MR2 , L = I = MR2 
when two masses are attached,
M.I. becomes
I'  MR2  mR2  mR2  (M  2m)R2
L  I  I' '
I
M
 ' 

I' M  2m
Hence, answer is (c).
Class Exercise - 5
A mass M is moving with a constant
velocity parallel to the X-axis. Its
angular momentum with respect to
the origin
(a) is zero
(b) remains constant
(c) goes on increasing
(d) goes on decreasing
Solution
Let the motion of P be in xy plane.
y
z=0
P
Vx
b
x
O
Then y = b is constant.
 Velocity Vx is along x-axis (constant)




 L  k L Z  k xVy  yVx  m

  k mbVx (Constant)
(Vy, Vz = 0)
Hence, answer is (b).
Class Exercise - 6
If the rotational kinetic energy and
translation kinetic energy of a
rolling body are same, the body is
(a) disc
(b) sphere
(c) cylinder
(d) ring
Solution
1
KE (translatory) = mv2
2
2
1 2 1 v
KE(rotatary)  I  I
2
2 R2
I always has the form kmR2, where k is a fraction or unity.
2
1
1
2v
2
 KE (rotatory)  kmR
 kmv
2
2
2
R
KE (rotatory) = KE (translatory) if k = 1 or I = mR2.
This is true for a ring.
Hence, answer is (d).
Class Exercise - 7
Two gear wheels, A and B, whose radii
are in the ratio RA : RB = 1 : 2, are
attached to each other by an endless
belt. They are mounted with their axes
parallel to each other. The system of
two wheels is set into rotation. It is
seen that both have the same angular
momentum. What is the ratio of their
moments of inertia? (Belts do not slip)
(a) 1 : 2
(b) 1 : 1
(c) 1 : 4
(d) 1: 2
Solution
Linear speed v of both wheels is the same.
v  ARA  BRB

A
B

RB
RA
2
 A  2B 
L  IAA  IBB
IA
B
1



IB A 2
Hence, answer is (a).
Class Exercise - 8
Consider the previous problem. If the
mass ratio mA : mB = 1 : 4, what is
the ratio of their rotational kinetic
energies?
(a) 4 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1 : 4
Solution
IA : IB = mARA2 : mBRB2
= mARA2 : 4mA(2RA)2
= 1 : 16
A = 2B
2
 A 
1
1
2 1
2
 IA A , IBB   16IA  

2
2
2
2


 (KE)A  4 KEB 
Hence, answer is (d).

1 16
2
 IA A
2 4
Class Exercise - 9
Two masses mA and mB are attached to
each other by a rigid, mass less rod of
length 2r.They are set to rotation about
the centre of the rod with an angular speed
. Then their angular momentum is
(a)
1
mA  mB  r 2  along the x  axis

2
z
(b) mA  mB  r  along the y  axis
2
1
(c) mA  mB  r 2  along the z  axis
2
2
(d) mA  mB  r  in the x  y plane
MA
2r
r
r

MB
x
Solution
1
1
2
IA  mAr , IB  mBr 2
2
2
1
2
 I  mA  mB  r
2
1
L  mA  mB  r 2
2
As the rotation is in the xy plane, L is along the z-axis.
Hence, answer is (c).
Thank you