1. AP Physics C: Mechanics Chapter 11
Angular Momentum

2. The Vector Product
There are instances where the product of two vectors is another vector.
Earlier we saw where the product of two vectors was a scalar.
This was called the dot product.
The vector product of two vectors is called the cross product.
Section 11.1

3. The Vector Product and Torque
The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector.
The torque is the vector (or cross) product of the position vector and the force vector.
Section 11.1

4. The Vector Product Defined
Given two vectors, and
The vector (cross) product of and is defined as a third vector,
C is read as “A cross B”.
The magnitude of vector C is AB sin q
q is the angle between and.
Section 11.1

5. More About the Vector Product
The quantity AB sin q is equal to the area of the parallelogram formed by .
The direction of is perpendicular to the plane formed by
The best way to determine this direction is to use the right-hand rule.
Section 11.1

6. Properties of the Vector Product
The vector product is not commutative. The order in which the vectors are multiplied is important.
To account for order, remember
If is parallel to (q = 0o or 180o), then
Therefore
If is perpendicular to , then
The vector product obeys the distributive law.
Section 11.1

7. Final Properties of the Vector Product
The derivative of the cross product with respect to some variable such as t is
where it is important to preserve the multiplicative order of the vectors.
Section 11.1

8. Vector Products of Unit Vectors
Section 11.1

9. Signs in Cross Products
Signs are interchangeable in cross products
and
Section 11.1

10. Using Determinants The cross product can be expressed as
Expanding the determinants gives
Section 11.1

11. Vector Product Example
Given
Find
Result
Section 11.1

12. Torque Vector Example Given the force and location
Find the torque produced
Section 11.1

13. Angular Momentum
Consider a particle of mass m located at the vector position and moving with linear momentum .
Find the net torque.
This looks very similar to the equation for the net force in terms of the linear momentum since the torque plays the same role in rotational motion that force plays in translational motion.
Section 11.2

14. Angular Momentum
Same basic techniques that were used in linear motion can be applied to rotational motion.
F becomes 
m becomes I
a becomes 
v becomes ω
x becomes θ
Linear momentum defined as
What if mass of center of object is not moving, but it is rotating?
Angular momentum
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15. Angular Momentum I Angular momentum of a rotating rigid object
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: kg m2/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) = kgm2/s
Both the magnitude and direction of the angular momentum depend on the choice of origin.
The magnitude is L = mvr sin theta
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16. Angular Momentum II Angular momentum of a particle
r is the particle’s instantaneous position vector
p is its instantaneous linear momentum
Only tangential momentum component contribute
r and p tail to tail form a plane, L is perpendicular to this plane
The instantaneous angular momentum of a particle relative to the origin O is defined as the cross product of the particle’s instantaneous position vector and its instantaneous linear momentum.
f is the angle between r and p .
The direction of L is perpendicular to the plane formed by r and p.
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17. Angular Momentum of a Particle in Uniform Circular Motion
Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v.
The angular momentum vector points out of the diagram
The magnitude is
L = rp sin = mvr sin (90o) = mvr
A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path O
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18. Angular momentum III Angular momentum of a system of particles
angular momenta add as vectors
be careful of sign of each angular momentum
for this case:
The total angular momentum of a system of particles is defined as the vector sum of the angular momenta of the individual particles.
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19. Example: calculating angular momentum for particles
Two objects are moving as shown in the figure . What is their total angular momentum about point O? m2 m1
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20. 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
How about this equation in rotational motion?
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21. 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
and to be measured about the same origin
The origin should not be accelerating, should be an inertial frame
The torque is related to the angular momentum.
Similar to the way force is related to linear momentum.
The torque acting on a particle is equal to the time rate of change of the particle’s angular momentum.
This is the rotational analog of Newton’s Second Law .
must be measured about the same origin.
This is valid for any origin fixed in an inertial frame.
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22. Demonstration Start from Expand using derivative chain rule
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23. What about SYSTEMS of Rigid Bodies?
Total angular momentum
of a system of bodies:
individual angular momenta Li
all about same origin
ti = net torque on particle “i”
internal torque pairs are
included in sum
Any torques associated with the internal forces acting in a system of particles are zero.
Therefore,
The net external torque acting on a system about some axis passing through an origin in an inertial frame equals the time rate of change of the total angular momentum of the system about that origin.
This is the mathematical representation of the angular momentum version of the non-isolated system model.
Rearranging the equation gives
This is the angular impulse-angular momentum theorem.
The resultant torque acting on a system about an axis through the center of mass equals the time rate of change of angular momentum of the system regardless of the motion of the center of mass.
This applies even if the center of mass is accelerating, provided torque and L are evaluated relative to the center of mass.
BUT… internal torques in the sum cancel in Newton 3rd law pairs. Only External Torques contribute to Lsys
net external torque on the system
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.
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24. 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.
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25. I 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 non- isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides
Constraints: I
Add page 344 a
Ignore internal forces, consider external forces only
Net external torque on system:
Angular momentum of system:
(not constant)
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26. 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
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27. Angular Momentum Conservation
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,
For an isolated system is deformable
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28. 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.
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29. 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?
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30. Moment of inertia changes
Isolated System
Moment of inertia changes
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31. Change I by curling up or stretching out - spin rate w must adjust
Controlling spin (w) by changing I (moment of inertia)
In the air, tnet = 0
L is constant
Change I by curling up or stretching out
- spin rate w must adjust
Moment of inertia changes
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32. Example: A merry-go-round problem
A 40-kg child running at 4.0 m/s jumps tangentially onto a stationary circular merry-go-round 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.
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33. 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.
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34. Solution: A merry-go-round problem
I = 20 kg.m2
VT = 4.0 m/s
mc = 40 kg
r = 2.0 m
w0 = 0
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