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Physics 111: Lecture 22 Today’s Agenda

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1 Physics 111: Lecture 22 Today’s Agenda
Angular Momentum: Definitions & Derivations What does it mean? Rotation about a fixed axis L = I Example: Two disks Student on rotating stool Angular momentum of a freely moving particle Bullet hitting stick Student throwing ball

2 Lecture 22, Act 1 Rotations A girl is riding on the outside edge of a merry-go-round turning with constant w. She holds a ball at rest in her hand and releases it. Viewed from above, which of the paths shown below will the ball follow after she lets it go? (a) (b) (c) (d) w

3 Lecture 22, Act 1 Solution Just before release, the velocity of the ball is tangent to the circle it is moving in. w

4 Lecture 22, Act 1 Solution After release it keeps going in the same direction since there are no forces acting on it to change this direction. w

5 Angular Momentum: Definitions & Derivations
p = mv Angular Momentum: Definitions & Derivations We have shown that for a system of particles Momentum is conserved if What is the rotational version of this?? The rotational analogue of force F is torque  Define the rotational analogue of momentum p to be angular momentum

6 Definitions & Derivations...
First consider the rate of change of L: So (so what...?)

7 Definitions & Derivations...
Recall that  EXT Which finally gives us: Analogue of !!

8 What does it mean? where and In the absence of external torques
Total angular momentum is conserved

9 Angular momentum of a rigid body about a fixed axis:
Rolling chain Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle: (since ri and vi are perpendicular) v1 We see that L is in the z direction. m2 j Using vi =  ri , we get r2 r1 m1 i v2 r3 m3 v3 I Analogue of p = mv!!

10 Angular momentum of a rigid body about a fixed axis:
In general, for an object rotating about a fixed (z) axis we can write LZ = I  The direction of LZ is given by the right hand rule (same as ). We will omit the Z subscript for simplicity, and write L = I  z

11 Example: Two Disks i f
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 f. z f z i

12 Example: Two Disks First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! Initially, the total angular momentum is due only to the disk on the bottom: z 2 1 0

13 Example: Two Disks First realize that there are no external torques acting on the two-disk system. Angular momentum will be conserved! Finally, the total angular momentum is due to both disks spinning: z 2 1 f

14 Example: Two Disks Wheel rim drop Since Li = Lf
An inelastic collision, since E is not conserved (friction)! z f z Li Lf f

15 Example: Rotating Table
A student sits on a rotating stool with his arms extended and a weight in each hand. The total moment of inertia is Ii, and he is rotating with angular speed i. He then pulls his hands in toward his body so that the moment of inertia reduces to If. What is his final angular speed f? i f Ii If

16 Example: Rotating Table...
Student on stool Drop mass from stool Again, there are no external torques acting on the student-stool system, so angular momentum will be conserved. Initially: Li = Iii Finally: Lf = If f i f Ii If Li Lf

17 Lecture 22, Act 2 Angular Momentum
A student sits on a freely turning stool and rotates with constant angular velocity w1. She pulls her arms in, and due to angular momentum conservation her angular velocity increases to w2. In doing this her kinetic energy: (a) increases (b) decreases (c) stays the same w1 w2 I1 I2 L L

18 Lecture 22, Act 2 Solution (using L = I) L is conserved: I2 < I1
K2 > K K increases! w1 w2 I2 I1 L

19 Lecture 22, Act 2 Solution Since the student has to force her arms to move toward her body, she must be doing positive work! The work/kinetic energy theorem states that this will increase the kinetic energy of the system! w1 w2 I1 I2 L L

20 Angular Momentum of a Freely Moving Particle
We have defined the angular momentum of a particle about the origin as This does not demand that the particle is moving in a circle! We will show that this particle has a constant angular momentum! y x v

21 Angular Momentum of a Freely Moving Particle...
Consider a particle of mass m moving with speed v along the line y = -d. What is its angular momentum as measured from the origin (0,0)? y x d m v

22 Angular Momentum of a Freely Moving Particle...
We need to figure out The magnitude of the angular momentum is: Since r and p are both in the x-y plane, L will be in the z direction (right hand rule): y x r d p=mv

23 Angular Momentum of a Freely Moving Particle...
So we see that the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. L is clearly conserved since d is constant (the distance of closest approach of the particle to the origin) and p is constant (momentum conservation). y x d p

24 Example: Bullet hitting 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. The initial speed of the bullet is v1, and the final speed is v2. What is the angular speed F of the stick after the collision? (Ignore gravity) M F D m D/4 v1 v2 initial final

25 Example: Bullet hitting stick...
Conserve angular momentum around the pivot (z) axis! The total angular momentum before the collision is due only to the bullet (since the stick is not rotating yet). M D m D/4 v1 initial

26 Example: Bullet hitting stick...
Conserve angular momentum around the pivot (z) axis! The total angular momentum after the collision has contributions from both the bullet and the stick. where I is the moment of inertia of the stick about the pivot. F D/4 v2 final

27 Example: Bullet hitting stick...
Set Li = Lf using M F D m D/4 v1 v2 initial final

28 Example: Throwing ball from stool
A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation. What is the angular speed F of the student-stool system after she throws the ball? M v F d I I top view: initial final

29 Example: Throwing ball from stool...
Conserve angular momentum (since there are no external torques acting on the student-stool system): Li = 0 Lf = 0 = IF - Mvd M v F d I I top view: initial final

30 Lecture 22, Act 3 Angular Momentum
A student is riding on the outside edge of a merry-go-round rotating about a frictionless pivot. She holds a heavy ball at rest in her hand. If she releases the ball, the angular velocity of the merry-go-round will: (a) increase (b) decrease (c) stay the same w2 w1

31 Lecture 22, Act 3 Solution The angular momentum is due to the girl, the merry-go-round and the ball. LNET = LMGR + LGIRL + LBALL Initial: w v R m same Final: w v m

32 Lecture 22, Act 3 Solution Since LBALL is the same before & after, w must stay the same to keep “the rest” of LNET unchanged. w w

33 Lecture 22, Act 3 Conceptual answer
Since dropping the ball does not cause any forces to act on the merry-go-round, there is no way that this can change the angular velocity. Just like dropping a weight from a level coasting car does not affect the speed of the car. w2 w

34 Recap of today’s lecture
Angular Momentum: (Text: 10-2, 10-4) Definitions & Derivations What does it mean? Rotation about a fixed axis (Text: 10-2, 10-4) L = I Example: Two disks Student on rotating stool Angular momentum of a freely moving particle (Text: 10-2, 10-4) Bullet hitting stick Student throwing ball Look at textbook problems Chapter 10: # 21, 29, 35, 41, 47, 49


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