momen mômen angular momentum

Lecture 7 Angular Momentum

Tran Thi Ngoc Dung

HCMUT


Outline
• Angular Momentum
• Conservation of Angular Momentum


O
d


r


F


M



Torque of a force about a fixed point O


F /O  r  F



F / O  plane(OM, F)

The direction of τ is outward
F / O  F r sin   F.d

O
d


r


p


M



Angular momentum about a fixed point O

  

L /O  r  p  r  mv


L / O  plane(OM , p)

The direction of L is outward

L  mvr sin   mv.d


Angular momentum and Torque

Angular momentumabout a fixed point O

  

L /O  r  p  r  mv
Differentiate by time :



  
dL /O d r   dp 

p r 
 ( v  mv)  
r F






dt
dt
dt


0


dL /O 
dL / 
 F / O
 F / 
dt
dt
The net torque acting on a particle equals the rate of change
of the angular momentum of the particle.


For a system of particles

Angular momentum of i th particel about a fixed point O

  

Li/O  ri  pi  ri  mvi
Angular momentum of the system of particles




L / O   Li/O   ri  mvi
i

i

Differenti ate by time :




 


dL /O
 d ri  
  dpi 
    pi     ri 
vi  mvi )   ri  (Fi,inter  Fi,ext )
   (
dt
dt  i  i
 i 
i  dt
0







 ri  Fi,inter  0,  ri  Fi,ext   i,ext  net.ext
i

i

i



dL /O 
dL / 
 net ,ext / 0
 net ,ext / 
dt
dt




dL / 
d
I
 L /  I  
dt
dt


net ,ext / 



d
 I  I
dt

ANGULAR MOMENTUM OF A
RIGID OBJECT


d 
I  I
 net .ext / 

 dt
L   I A 
dL  
 net .ext . / 
dt
The net external torque acting on a system equals the rate of change
of the angular momentum of the system.


dL / 
 net ,ext / 
dt



L /  I  


Example 10-2
An Atwood' s machine has two blocks of mass

m1 and m 2 (m1 > m 2 ), connected by a string of
negligible mass that passes over a pulley wit h
frictionle ss bearings. The pulley is a uniform disk
of mass M and radius R. The string does not
slip on the pulley.

dL 
Apply Equation
 net ,ext
dt
to the system consisting of both blocks plus pulley
to find the angular acceleration of the pulley
and the acceleration of the blocks.


Angular momentum of the system
{Block m1 , block m 2 , pulley M) about
the axis z through t he center of pulley
L  m1vR  m 2 vR  I
because that the rope is not sliding on the pulley.
v  R  a  R
dL
I 

 m1aR  m 2aR  I   m1  m 2  2 aR
dt
R 

Net torque of external forces :
net  m1gR  m 2gR

m1  m 2  m1 moves down,
m 2 moves up with the speed v
 pulley rotates counter - clockwise

with  outward.

dL
 net,ext
dt
I 

 m1  m 2  2 aR  m1gR  m 2gR
R 

m1  m 2
a
g
I
m1  m 2  2
R
Applying


Conservation of Angular Momentum


dL 
 net,ext
dt



if net,ext  0 L  const
If the net external torque acting on a system is zero, the total angular
momentum of the system is constant.


Example
A disk is rotating with an initial angular speed i about a frictionle ss shaft
through its symmetry ax is as shown. Its moment of inertia
about this axis is I1 . It drops onto another disk of moment of inertia I 2 that
is initially at rest on the same shaft. Because of surface friction, the two
disks eventually attain a common angular speed f . Find f
External force : weight m1g parallel to the rotation axis


 torque net  0  L  const
initial Angular momentum  final angular momentum
Li  L f



I1i  (I1  I 2 )f  f 
1
L2
2
Initial KE : Ti  I1i 
2
2I1

I1 

i
I1  I 2

1
L2
2
Final KE : Tf  (I1  I 2 )f 
2
2(I1  I 2 )

The change in KE :
L2  1
1
L2
I2
T  Tf  Ti  
   
0
2  I1  I 2 I1 
2 (I1  I 2 )I1


Example
A merry - go - round of radius 2 m and moment of inertia 500 kg.m 2 is
rotating about a frictionless pivot, making one revolution every 5 s. A child
of mass 25 kg originally standing at the center walks out to the rim.
Find the new angular speed of the merry - go - round.

External force is the gravitatio nal force mg, parallel to the rotation axis. The torque is zero.
Applying the law of Conservation of Angular momentumLi  Lf

(I merry  Ichild,i )i  (I merry  Ichild,f )f
f 

I merry  Ichild,i
I merry  Ichild,f

Ichild,i  0
f 

i

Ichild,f  mR 2

I merry
I merry  mR

2

i

500
2
f 
 1.05rad / s
2
500  25  2 5


Problem

In the figure the incline is frictionless and the string passes through the
center of mass of each block. The pulley has a moment of inertia I and a
radius r. (a) Find the net torque acting on the system (the two masses,
string, and pulley) about the center of the pulley. (b) Write an expression for
the total angular momentum of the system about the center of the pulley
when the masses are moving with a speed v. (c) Find the acceleration of the
masses from your results for parts (a) and (b) by setting the net torque
equal to the rate of change of the angular momentum of the system.

a )   m 2g sin R  m1gR
b) L  m1vR  m 2 vR  I
c)

dL

dt

I
(m1  m 2  2 )aR  m 2g sin R  m1gR
R
m sin   m1
a 2
g
I
m1  m 2  2
R