Transcript Monday, Nov. 11, 2002

PHYS 1443 – Section 003
Lecture #16
Monday, Nov. 11, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Angular Momentum
Angular Momentum and Torque
Angular Momentum of a System of Particles
Angular Momentum of a Rotating Rigid Body
Angular Momentum Conservation
Today’s homework is homework #16 due 12:00pm, Monday, Nov. 18!!
Monday, Nov. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Similarity Between Linear and Rotational Motions
All physical quantities in linear and rotational motions show striking similarity.
Similar Quantity
Mass
Length of motion
Speed
Acceleration
Force
Work
Power
Momentum
Kinetic Energy
Monday, Nov. 11, 2002
Linear
Mass
Rotational
Moment of Inertia
M
I   r 2 dm
Displacement
r
Angle  (Radian)
v
dr
dt

d
dt
a
dv
dt
 
d
dt
Force F  ma
Work W   Fdx
xf
xi
Kinetic
Torque
Work
  I
f
W   d 
i
P  F v
P  
p  mv
L  I
K
1
mv 2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Rotational
KR 
1
I 2
2
2
Angular Momentum of a Particle
If you grab onto a pole while running, your body will rotate about the pole, gaining
angular momentum. We’ve used linear momentum to solve physical problems
with linear motions, angular momentum will do the same for rotational motions.
z
Let’s consider a point-like object ( particle) with mass m located
at the vector location r and moving with linear velocity v
L=rxp
The instantaneous angular momentum
L  r p
L of this particle relative to origin O is
O
y
m
r
What is the unit and dimension of angular momentum? kg m2 / s 2
 p
Note that L depends on origin O. Why? Because r changes
x
What else do you learn? The direction of L is +z
Since p is mv, the magnitude of L becomes L  mvr sin 
What do you learn from this?
The point O has
to be inertial.
Monday, Nov. 11, 2002
If the direction of linear velocity points to the origin of
rotation, the particle does not have any angular momentum.
If the linear velocity is perpendicular to position vector, the
particle
moves exactly the same way as a point on a 3rim.
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
Angular Momentum and Torque
Can you remember how net force exerting on a particle
and the change of its linear momentum are related?
F 
dp
dt
Total external forces exerting on a particle is the same as the change of its linear momentum.
The same analogy works in rotational motion between torque and angular momentum.


Net torque acting on a particle is
F  r d p
  r
dt
z
d r p
dr
dp
dp
dL


 pr
 0r
L=rxp
dt
dt
dt
dt
dt

O
x
r
y
m


Why does this work?
p
Thus the torque-angular
momentum relationship
Because v is parallel to
the linear momentum
 
dL
dt
The net torque acting on a particle is the same as the time rate change of its angular momentum
Monday, Nov. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
4
Angular Momentum of a System of Particles
The total angular momentum of a system of particles about some point
is the vector sum of the angular momenta of the individual particles
L  L1  L2  ......  Ln   L
Since the individual angular momentum can change, the total
angular momentum of the system can change.
Both internal and external forces can provide torque to individual particles. However,
the internal forces do not generate net torque due to Newton’s third law.
Let’s consider a two particle
system where the two exert
forces on each other.
Since these forces are action and reaction forces with
directions lie on the line connecting the two particles, the
vector sum of the torque from these two becomes 0.
Thus the time rate change of the angular momentum of a
system of particles is equal to the net external torque
acting on the system
Monday, Nov. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu

ext
dL

dt
5
Example 11.4
A particle of mass m is moving in the xy plane in a circular path of radius r and linear
velocity v about the origin O. Find the magnitude and direction of angular momentum with
respect to O.
Using the definition of angular momentum
y
v
L  r  p  r  mv  mr  v
r
O
x
Since both the vectors, r and v, are on x-y plane and
using right-hand rule, the direction of the angular
momentum vector is +z (coming out of the screen)
mrrvv  mrv sin   mrv sin 90  mrv
The magnitude of the angular momentum is LL  m
So the angular momentum vector can be expressed as
L  mrvk
Find the angular momentum in terms of angular velocity .
Using the relationship between linear and angular speed
2
2
L  mrvk  mr  k  mr   I 
Monday, Nov. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
6
Angular Momentum of a Rotating Rigid Body
z
Let’s consider a rigid body rotating about a fixed axis
Each particle of the object rotates in the xy plane about the zaxis at the same angular speed, 
L=rxp
O
y
m
r

x
Magnitude of the angular momentum of a particle of mass mi
about origin O is miviri
2
Li  mi ri vi  mi ri 
p
Summing over all particle’s angular momentum about z axis
Lz   Li   mi ri 2 
i
i
Since I is constant for a rigid body
Thus the torque-angular momentum
relationship becomes
What do
you see?
Lz 
ext
2
i i
i
dL z
d
 I
dt
dt

 m r   I
 I
 is angular
acceleration
dLz

 I
dt
Thus the net external torque acting on a rigid body rotating about a fixed axis is equal to the moment
of inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.
Monday, Nov. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
7
Example 11.6
A rigid rod of mass M and length l pivoted without friction at its center. Two particles of mass
m1 and m2 are connected to its ends. The combination rotates in a vertical plane with an
angular speed of . Find an expression for the magnitude of the angular momentum.
y
m2
l

m2 g
O
m1
m1 g
x
The moment of inertia of this system is
1
1
1
2
2

I

I

I

Ml

m
l

m2l 2
I
rod
m1
m2
1
12
4
4
2
l2  1


l
1

  M  m1  m2  L  I 
 M  m1  m2 
4 3
4 3


Find an expression for the magnitude of the angular acceleration of the
system when the rod makes an angle  with the horizon.
If m1 = m2, no angular
momentum because net
torque is 0.
If /-p/2, at equilibrium
so no angular momentum.
Monday, Nov. 11, 2002
First compute net
external torque
   m1 g
l
cos
2
 ext     2

 2  -m2 g
gl cos  m1 - m2 
2
1
m1 - m1 gl cos 
2



Thus 
ext


2
l
1


I
 M  m1  m2 
becomesPHYS 1443-003, Fall
2002 4  3

Dr. Jaehoon Yu
l
cos
2
2m1 - m1  cos 
g /l
1

 M  m1  m2 
3
8 
Conservation of Angular Momentum
Remember under what condition the linear momentum is conserved?
Linear momentum is conserved when the net external force is 0.  F  0 
dp
dt
p  const
By the same token, the angular momentum of a system
is constant in both magnitude and direction, if the
resultant external torque acting on the system is 0.
What does this mean?
 ext 
dL
0
dt
L  const
Angular momentum of the system before and
after a certain change is the same.
Li  L f  constant
Three important conservation laws
for isolated system that does not get
affected by external forces
Monday, Nov. 11, 2002
 K i  U i  K f  U f Mechanical Energy


Linear Momentum
 pi  p f


Angular Momentum
Li  L f
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
9
Example 11.8
A start rotates with a period of 30days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses
into a neutron start of radius 3.0km. Determine the period of rotation of the neutron star.
What is your guess about the answer?
Let’s make some assumptions:
Using angular momentum
conservation
The period will be significantly shorter,
because its radius got smaller.
1. There is no torque acting on it
2. The shape remains spherical
3. Its mass remains constant
Li  L f
I i  I f  f
The angular speed of the star with the period T is
Thus

Tf 
I i
mri 2 2p


f
If
mrf2 Ti
2p
f
Monday, Nov. 11, 2002
 r f2
 2
r
 i
2p

T
2

3
.
0


-6
Ti  

2
.
7

10
days  0.23s

30
days

4

1
.
0

10



PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
10