Physics 106: Mechanics
Lecture 06
Wenda Cao
NJIT Physics Department
Conservation of Angular
Momentum



Cross Product
Comparison: definitions
of single particle torque
and angular momentum
Angular Momentum




of a system of particles
of a rigid object
Conservation of angular
momentum
Examples
http://www.youtube.com/watch?v=AQLtcEAG9v0
February 24, 2011
Cross Product
 
C  A B




B

B sin 
The cross product of two vectors says
something about how perpendicular they are.

Magnitude:
 

A

A sin 

C  A  B  AB sin 




 is smaller angle between the vectors
Cross product of any parallel vectors = zero
Cross product is maximum for perpendicular
vectors
Cross products of Cartesian unit vectors:
iˆ  ˆj  kˆ; iˆ  kˆ   ˆj; ˆj  kˆ  iˆ
iˆ  iˆ  0; ˆj  ˆj  0; kˆ  kˆ  0
y
j
i
x
k
z
i
j
k
February 24, 2011
Cross Product

Direction: C perpendicular to
both A and B (right-hand rule)






Place A and B tail to tail
Right hand, not left hand
Four fingers are pointed along
the first vector A
“sweep” from first vector A
into second vector B through
the smaller angle between
them
Your outstretched thumb
points the direction of C
First practice
   
A B  B  A ?
 
 
A B  - B  A
   
A B  B  A ?
February 24, 2011
More about Cross Product






The quantity ABsin is the area of the
parallelogram formed by A and B
The direction of C is perpendicular to
the plane formed by A and B
Cross product is not commutative
 
 
A B  - B  A
The distributive law
  
   
A (B  C)  A B  A C
The derivative of cross product
obeys the chain rule
Calculate cross product


d   dA   dB
A B 
 B  A
dt
dt
dt


 
A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ
February 24, 2011
Torque and Angular
Momentum for a Single Particle
Torque
  
  r F
  rF  rF sin()
  r F  rF sin()
Angular Momentum
 
L  r p
L  rp  rp sin() p  mv
L  rp  rp sin() p  mv
February 24, 2011
Angular momentum of
a system of particles

Angular momentum of a system of particles

 

Lnet  L 1  L 2  ...  L n 


Li 
all i




ri  p i
all i
angular momenta add as vectors
be careful of sign of each angular momentum
for this case:

     
Lnet  L1  L2  r1  p1  r2  p2

| Lnet |   r1 p1 - r2 p2
February 24, 2011
Angular Momentum of a Rigid Body

Angular momentum of a rotating rigid object


L  I




L


L has the same direction as 
L is positive when object rotates in CCW
L is negative when object rotates in CW

Angular momentum SI unit: kgm2/s

Calculate L of a 10 kg disc when  = 320 rad/s, R = 9 cm = 0.09 m
L = I and I = MR2/2 for disc
L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s


February 24, 2011
Finding angular momentum

A solid sphere and a hollow sphere have the same mass
and radius. They are rotating with the same angular
speed. Which one has the higher angular momentum?
A) the solid sphere
B) the hollow sphere
C) both have the same angular momentum
D) impossible to determine
L  I
February 24, 2011
Linear Momentum and Force



Linear motion: apply force to a mass
The force causes the linear momentum to change
The net force acting on a body is the time rate of
change of its linear momentum





dv dp
Fnet  F  ma  m

dt dt


p  mv
 
 

IL
 Fnet t  p
t
February 24, 2011
Angular Momentum and Torque



Rotational motion: apply torque to a rigid body
The torque causes the angular momentum to change
The net torque acting on a body is the time rate of
change of its angular momentum



 dp

 dL
Fnet  F 
 net   
dt
dt

  and L


to be measured about the same origin
The origin should not be accelerating, should be an
inertial frame
February 24, 2011
Demonstration





 dp

 dL
Fnet  F 
 net   
dt
dt

Start from dL  d (r  p )  m d (r  v )
dt dt
dt
Expand using derivative chain rule



   
dL
d  
 dr   dv 
 m (r  v )  m  v  r    mv  v  r  a 
dt
dt
dt 
 dt

   
  
  

dL
 mv  v  r  a   mr  a  r  (ma )  r  Fnet   net
dt
February 24, 2011
What about SYSTEMS of Rigid
Bodies?

 dL i
Rotational 2 law for a single body : i 
dt
 • individual
Total angular momentum 
angular momenta Li
Lsys   L i • all about same
of a system of bodies:
origin


dLsys
• i = net torque on particle “i”

dLi


  i
• internal torque pairs are
dt
dt
included in sum
i
nd
BUT… internal torques in the sum cancel in Newton 3rd law
pairs. Only External Torques contribute to Lsys

dLsys
dt


  i ,ext   net
net external torque on the system
i
Nonisolated System: If a system interacts with its environment in the
sense that there is an external torque on the system, the net external
torque acting on a system is equal to the time rate of change of its
angular momentum.
February 24, 2011
Example: A Non-isolated System

A sphere mass m1 and a
block of mass m2 are
connected by a light cord
that passes over a pulley.
The radius of the pulley is R,
and the mass of the thin rim
is M. The spokes of the
pulley have negligible mass.
The block slides on a
frictionless, horizontal
surface. Find an expression
for the linear acceleration of
the two objects.
a

a
 ext  m1 gR
February 24, 2011

a
Masses are connected by a light cord
Find the linear acceleration a.
• Use angular momentum approach
• No friction between m2 and table
• Treat block, pulley and sphere as a nonisolated system rotating about pulley axis. As
sphere falls, pulley rotates, block slides
• Constraints: Equal v' s and a' s for block and sphere
I

a
v  ωR for pulley α  d / dt
a  αR  dv/dt
• Ignore internal forces, consider external forces only
• Net external torque on system:
  m gR
net
• Angular momentum of system:
(not constant)
1
about center of wheel
Lsys  m1vR  m2vR  Iω  m1vR  m2vR  MR 2ω
dLsys
 m1aR  m2 aR  MR 2α  (m1R  m2 R  MR)a  τ net  m1 gR
dt
m1 g
same result followed from earlier
a 
method using 3 FBD’s & 2nd law
M  m1  m2
February 24, 2011
Isolated System

Isolated system: net external torque acting on
a system is ZERO


no external forces
net external force acting on a system is ZERO


dLtot
 ext 
0
dt

Ltot  constant
or
 
Li  L f
February 24, 2011
Angular Momentum Conservation

Ltot  constant
or
 
Li  L f
where i denotes initial state, f is final state
 L is conserved separately for x, y, z direction
 For an isolated system consisting of particles,


  
Ltot   Ln  L1  L2  L3    constant


For an isolated system is deformable
I ii  I f  f  constant
February 24, 2011
First Example



A puck of mass m = 0.5 kg is
attached to a taut cord passing
through a small hole in a
frictionless, horizontal surface. The
puck is initially orbiting with speed
vi = 2 m/s in a circle of radius ri =
0.2 m. The cord is then slowly
pulled from below, decreasing the
radius of the circle to r = 0.1 m.
What is the puck’s speed at the
smaller radius?
Find the tension in the cord at the
smaller radius.
February 24, 2011
Angular Momentum Conservation
m = 0.5 kg, vi = 2 m/s, ri = 0.2 m,
rf = 0.1 m, vf = ?
 Isolated system?
 Tension force on m exert zero
torque about hole, why?

 
Li  L f
   

L  r  p  r  (mv )
Li  mri vi sin 90  mri vi L f  mrf v f sin 90  mrf v f
ri
0.2
v f  vi 
2  4m/s
rf
0.1
v 2f
42
T  mac  m  0.5
 80 N
rf
0.1
February 24, 2011
Isolated
System

τ net  0

L
about z - axis

 L  constant
I ω  I
i
initial
i
final
f
ωf
Moment of inertia
changes
February 24, 2011
How fast should the student spin?
The student on a platform is rotating (no friction) with angular speed 1.2
rad/s.
•
•
His arms are outstretched and he holds a brick in each hand.
The rotational inertia of the system consisting of the professor, the
bricks, and the platform about the central axis is 6.0 kg·m2.
By moving the bricks the student decreases the rotational inertia of the
system to 2.0 kg·m2.
(a) what is the resulting angular speed of the platform?
(b) what is the ratio of the system’s new kinetic energy to the
original kinetic energy?
L is constant… while moment of inertia changes
February 24, 2011
Ii = 6 kg-m2
If = 2 kg-m2
i = 1.2 rad/s
f = ? rad/s
L is constant… while moment of inertia changes,
Zero external torque
... about a fixed axis
Solution (a):
Solution (b):
f 
Kf
Ki

 L final  L initial  L

L  I ii  I f  f
Ii
6
i  1.2  3.6 rad/s
If
2
1
2
1
2
I f  2f
I f  f 2 I f Ii 2 Ii
 ( ) 
( ) 
 3
2
Ii  f
Ii I f
If
I ii
KE has increased!!
February 24, 2011
Controlling spin () by changing I (moment of inertia)
In the air, net = 0
L is constant
L  I i i  I f  f
Change I by curling up or stretching out
- spin rate  must adjust
Moment of inertia changes
February 24, 2011
Example: A merry-go-round problem
A 40-kg child running at 4.0
m/s jumps tangentially onto a
stationary circular merry-goround platform whose radius is
2.0 m and whose moment of
inertia is 20 kg-m2. There is
no friction.
Find the angular velocity of
the platform after the child
has jumped on.
February 24, 2011
The Merry-Go-Round
The moment of inertia of the
system = the moment of
inertia of the platform plus the
moment of inertia of the
person.
 Assume the person can be
treated as a particle
 As the person moves toward
the center of the rotating
platform the moment of inertia
decreases.
 The angular speed must
increase since the angular
momentum is constant.

February 24, 2011
Solution: A merry-go-round problem

Ltot   I i ωi 
I
f
f
Li  I ii  I0  mc vT r  mc vT r
L f  I f ω f  ( I  mc r 2 )ω f
I = 20 kg.m2
VT = 4.0 m/s
mc = 40 kg
r = 2.0 m
0 = 0
( I  mc r 2 )ω f  mc vT r
ωf 
mc vT r
40  4  2

 1.78 rad/s
2
2
I  mc r
10  40  2
February 24, 2011