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Physics
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Session
Rotational Mechanics - 4
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Session Objectives
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Session Objective
1. Angular Momentum of a Particle
2. Conservation of angular momentum
3. Angular Impulse
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Angular Momentum
Angular momentum of a particle
Angular Momentum of P L
L r p (p mv)
mr ( r)
2
L is an axial vector units : Kg m /s
2
L mvr sin
L I ( I mr )
Particle P rotates around Point O
Or
P(m)
v
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Angular Momentum for a Systemof Particles
For a system of particles (i = 1 to n)
n
i i
i 1
L r p I
2
i i( is a constant. So I m r )
Angular momentum depends on
the position of the axis
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Angular Momentum for a RigidBody
For a rigid body
L I
( I : moment of inertia around
the axis of rotation)
X
Y
O m
r
Z
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L l
dL dl l
dt dt
d
dt
ext extdL ldt
ext i i f f If 0 l l
ext
dLIf, 0 then, 0
dt
L cons tant
Conservation of Angular Momentum
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Conservation of Angular Momentum
If no external torque acts on a body
so the total angular momentum
of the body (or system of bodies) remainsa constant , and the vector sum of all torquesadd up to zero.
dL
0d t
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Equilibrium of a Rigid Body
Two conditions are necessary.
(i) Total force acting on the bodymust add up to zero (equilibrium oflinear motion)
(ii) Total torque acting on the bodymust add up to zero (equilibrium of
rotational motion)
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LINEAR ANGULAR
Mass
Momentum
Newtons
Second Law
Work
KineticEnergy
Power(constant force)
FdxW dW
2
mv2
1
K
2I
2
1K
Comparison
ILP=Mv
IF=ma
m I
PP=Fv
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Class Test
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Class Exercise - 1
A planet revolves round the sun as
shown. The KE is greatest at
A
B
C
D
(a)A (b) B
(c) C (d) D
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Solution
Kinetic energy at any point:
2 2
2
L LK
2I 2mr
L is a constant for the system (as motion of sunis negligible). L is constant for planet. So K is
maximum for smallest r which is A.
2
1K
r
Hence, answer is (a).
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Class Exercise - 2
A horizontal platform with a mass of
100 kg rotates at 10 rpm about avertical axis passing through its centre.A man weighing 60 kg is standing onthe edge. With what velocity will theplatform 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
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Solution
Moment of inertia of platform:
2P 11
m r I2
Moment of inertia of man at edge:
2m 2m r I
Moment of inertia of man at centre: O ( r = 0)
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Solution contd..
L is conserved:
1 2 1 1 2
2
2
2
I I I
1 1
100 60 10 1002 2
1100 50
22 rpm
Hence, answer is (a).
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Class Exercise - 3
A mass moves about the Y-axis
with acceleration ay = (by2 c); band c are constants. The value of yfor which the angular momentumis zero, is
3c c(a) (b)b b
3c 3b(c) (d)
b c
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Solution
y 2
y
dvAcceleration a by c
dt
y 2y
vy y
2y y
0 0
32
y
dvv by c
dy
v dv by c dy
by
v cy 23 L (= mvyr) is zero when vy = 0
3
by 3ccy y
3 b
Hence, the answer is (c).
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Class Exercise - 4
A thin circular ring of mass M and radius
R is rotating about its axis with a constantangular velocity . Two objects each ofmass m are attached gently to the ring.The ring now rotates with an angularvelocity.
M 2mM(a) (b)
M m M 2m
M 2mM(c) (d)
M 2m M
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Solution
2 2I (ring) = MR , L = I = MR
when two masses are attached,M.I. becomes
2 2 2 2I' MR mR mR (M 2m)R
L I I' '
I M'
I' M 2m
Hence, answer is (c).
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Class Exercise - 5
A mass M is moving with a constant
velocity parallel to the X-axis. Itsangular momentum with respect tothe origin
(a) is zero (b) remains constant
(c) goes on increasing (d) goes on decreasing
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Solution
Let the motion of P be in xy plane.
z = 0
x
y
P
O
b
Vx
Then y = b is constant.
Velocity Vx is along x-axis (constant) (Vy, Vz = 0)
Z y x
x
L kL k xV yV m
k mbV (Constant)
Hence, answer is (b).
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Class Exercise - 6
If the rotational kinetic energy and
translation kinetic energy of arolling body are same, the body is
(a)disc (b) sphere
(c) cylinder (d) ring
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Solution
KE (translatory) = mv21
2
2
2
2
1 1 vKE(rotatary) I I
2 2 R
I always has the form kmR2
, where k is a fraction or unity.
2
2 2
2
1 v 1KE (rotatory) kmR kmv
2 2R
KE (rotatory) = KE (translatory) if k = 1 or I = mR2.This is true for a ring.
Hence, answer is (d).
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Class Exercise - 7
Two gear wheels, A and B, whose radii
are in the ratio RA : RB = 1 : 2, areattached to each other by an endlessbelt. They are mounted with their axesparallel to each other. The system oftwo wheels is set into rotation. It is
seen that both have the same angularmomentum. What is the ratio of theirmoments of inertia? (Belts do not slip)
(a)1 : 2 (b) 1 : 1
(c) 1 : 4 (d)1: 2
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Solution
Linear speed v of both wheels is the same.
A A B Bv R R
A BA B
B A
R2 2
R
A A B BL I I
A B
B A
I 1
I 2
Hence, answer is (a).
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Class Exercise - 8
Consider the previous problem. If the
mass ratio mA : mB = 1 : 4, what isthe ratio of their rotational kineticenergies?
(a)4 : 1 (b) 2 : 1
(c) 1 : 2 (d) 1 : 4
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Solution
IA : IB = mARA2 : mBRB
2
= mARA2 : 4mA(2RA)
2
= 1 : 16
A B= 2
22 2 A
A A B B A1 1 1
I , I 16I2 2 2 2
2A A1 16
I2 4
A B(KE) 4 KE
Hence, answer is (d).
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Class Exercise - 9
Two masses mA and mB are attached to
each other by a rigid, mass less rod oflength 2r.They are set to rotation aboutthe centre of the rod with an angular speed. Then their angular momentum is
2 r
r r
z
xM
A MB
2
A B
1
(a) m m r along the x axis2
2A B(b) m m r along the y axis
2A B1
(c) m m r along the z axis
2 2A B(d) m m r in the x y plane
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Solution
2 2
A A B B
1 1
I m r , I m r2 2
2A B1
I m m r2
2A B1
L m m r2
As the rotation is in the xy plane, is along the z-axis.L
Hence, answer is (c).
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Thank you
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