1. Chapt. 10: Angular Momentum
Angular momentum conservation
And applications
4/27/2017
Phys 201, 2011

2. Angular Momentum Conservation
where and
In the absence of external torques
Ii ωi = If ωf
Total angular momentum is conserved.
(demos)
4/27/2017
Phys 201, 2011

3. Because the torque exerted by the ice is
Spinning skater:
Because the torque exerted by the ice is
small, the angular momentum of the skater
is approximately constant.
When she reduces her moment of inertia
by drawing in her arms, her angular speed
increases.
4/27/2017
Phys 201, 2011

4. A student sitting on a stool that rests on
a turntable with frictionless bearings
is holding a rapidly spinning bicycle
Wheel (a). The rotation axis of the wheel
is initially horizontal, and the magnitude
of the spin-angular-momentum vector
of the spinning wheel is
What will happen if the student
suddenly tips the axle of the wheel (b)
so that after the rotation the spin axis
of the wheel is vertical and the wheel
is spinning counterclockwise
(when viewed from above)?
Answer: The turntable, stool, and student will be rotating clockwise with an
angular momentum about the vertical axis of the turntable of magnitude
4/27/2017
Phys 201, 2011

5. Example: Two Disks
A disk of mass M and radius R rotates around the z axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity
What is the relation ship between and ?
analogous to an inelastic collision: p1 + p2(=0)  p3 z z
4/27/2017
Phys 201, 2011

6. Example: Two Disks
First realize that there are no external torques acting on the two-disk system of combined mass M.
Angular momentum will be conserved!
Initially, the total angular momentum is due only to the disk on the bottom: z z 2 1
4/27/2017
Phys 201, 2011

7. Example: Two Disks Since Li = Lf An inelastic collision,
since E is not
conserved (friction)! z z Li Lf
4/27/2017
Phys 201, 2011

8. Example: bullet hitting a stick
A uniform stick of mass m and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. Knowing that the initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed ωf of the stick immediately after the collision? (Ignore gravity)
4/27/2017
Phys 201, 2011

9. Example: bullet hitting a stick
Initial angular momentum:
Final angular momentum:
where
Conservation of angular momentum around pivot axis:
4/27/2017
Phys 201, 2011

10. Example: throw ball from stool
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Phys 201, 2011

11. Angular momentum
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Phys 201, 2011

12. Angular momentum
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Phys 201, 2011

13. A 25-kg child in a playground runs with an initial
speed of 2.5 m/s along a path tangent to the rim
of a merry-go-round, whose radius is 2.0 m.
The merry-go-round, which is initially at rest,
has a moment of inertia of 500 kg · m2
The child then jumps on to it.
Find the final angular velocity of the child and
the merry-go-round together.
4/27/2017
Phys 201, 2011

14. Summary of Dynamics Dynamics (cause – effect)
Force – Linear acceleration
Change in linear momentum, Fnet = dp/dt
Torque – angular acceleration
Change in angular momentum, τ = dL/dt
Conservation of momentum
When net external force = zero
Conservation of angular momentum
When net external torque = zero
Conservation of Energy
Mechanical energy = kinetic + potential
Non-conservative forces e.g. friction
Energy lost to the environment, e.g., heat
4/27/2017
Phys 201, 2011

15. Dynamical Reprise: Linear
The change of motion of an object is described by Newton as F=ma where F is the force, m is the mass, and a is the acceleration.
For a set of discrete point particles, all forces act on the center of mass
The center of mass is a calculable property of the object.
If F=0, then there is no change of motion -- if at rest, the object will remain at rest.
4/27/2017
Phys 201, 2011

16. Dynamical Reprise: Rotation
About a fixed rotation axis, you can always write where is the torque, I is the moment of inertia, and is the angular acceleration.
For a set of discrete point particles,
The parallel axis theorem lets you calculate the moment of inertia about an axis parallel to an axis through the CM if you know ICM :
IPARALLEL = ICM + MD2 L D M x CM
ICM
IPARALLEL
4/27/2017
Phys 201, 2011