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Physics 151: Lecture 24 Today’s Agenda

Physics 151: Lecture 24 Today’s Agenda. Topics Angular Momentum Ch. 11.3-5 More fun demos !!! Gyroscopes Ch. 11.6. See text: 11.3. p=mv. Angular Momentum: Definitions & Derivations. We have shown that for a system of particles Momentum is conserved if. Now we also know, .

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Physics 151: Lecture 24 Today’s Agenda

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  1. Physics 151: Lecture 24Today’s Agenda • Topics • Angular Momentum Ch. 11.3-5 • More fun demos !!! • Gyroscopes Ch. 11.6

  2. See text: 11.3 p=mv Angular Momentum:Definitions & Derivations • We have shown that for a system of particles Momentum is conserved if • Now we also know, Angular momentum is conserved if tEXT = 0 Animation • With angular momentum L = r x p

  3. In the absence of external torques See text: 11.5 What does it mean? • where and Total angular momentum is conserved Also, remember that L of a rigid body about a fixed axis is: I  Analogue of p = mv !!

  4. See text: 11.4 Angular momentum of a rigid bodyabout 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 theright hand rule (same as ). • We will omit the ”Z” subscript for simplicity,and write L= I z 

  5. z z Arm Arm A B Demonstration ofConservation of Angular Momentum • Figure Skating : LA = LB IA < IB wA > wB

  6. K1 > K2 w1 w2 disk 1 disk 2 Lecture 23, ACT 2Angular momentum • Two different spinning disks have the same angular momentum, but disk 1 has more kinetic energy than disk 2. • Which one has the biggest moment of inertia ? (a) disk 1 (b) disk 2 (c) not enough info I1 < I2

  7. z z F Example: Two Disks • A disk of mass M and radius R rotates around the z axis with angular velocity 0. 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. What is F ? 0

  8. z 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: 2 1 0

  9. 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 dueto both disks spinning: z 2 1 F

  10. z z F Example: Two Disks • Since LINI = LFIN LINI LFIN 0

  11. z z F Example: Two Disks • Now let’s use conservation of energy principle: EINI = EFIN 1/2 Iw02 = 1/2 (I + I)wF2 wF2 = 1/2 w02 wF = w0 / 21/2 EINI EFIN 0

  12. Example: Two Disks • So we got a different answers ! wF’ = w0 / 21/2 Conservation of energy ! wF = w0 / 2 Conservation of momentum ! Which one is correct ? Because this is an inelastic collision, since E is not conserved (friction) !

  13. Lecture 25, Act 2 • A mass m=0.1kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed wi = 5rad/s in a circle of radius ri = 0.2m. The cord is then slowly pulled from below, and the radius decreases to r = 0.1m. How much work is done moving the mas from ri to r ? (A) 0.15 J (B) 0 J (C) - 0.15 J ri wi animation

  14. Lecture 25, Act 3 • A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction r = i + j. What is its angular momentum (in kg · m2/s) relative to the origin? a. 0 k b. 6 (2)1/2k c. –6 (2)1/2k d. 6 k e. –6 k

  15. See text: Ex. 11.11 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 before after

  16. See text: 11.5 Angular Momentum Conservation • A freely moving particle has a definite angular momentum about any given axis. • If no torques are acting on the particle, its angular momentum will be conserved. • In the example below, the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. y x d m v

  17. See text: Ex. 11.11 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 before

  18. See text: Ex. 11.11 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 after

  19. See text: Ex. 11.11 Example: Bullet hitting stick... • Set LBEFORE = LAFTER using M F D m D/4 v1 v2 before after

  20. See text: Ex. 11.11 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: before after See example 13-9

  21. See text: Ex. 11.11 Example: Throwing ball from stool... • Conserve angular momentum (since there are no external torques acting on the student-stool system): • LBEFORE = 0 • LAFTER = 0 = IF - Mvd M v F d I I top view: before after See example 13-9

  22. See text: 11.6 Gyroscopic Motion: • Suppose you have a spinning gyroscope in the configuration shown below: • If the left support is removed, what will happen ?? pivot support  g

  23. See text: 11.6 Gyroscopic Motion... • Suppose you have a spinning gyroscope in the configuration shown below: • If the left support is removed, what will happen ? • The gyroscope does not fall down ! pivot  g

  24. pivot  See text: 11.6 Gyroscopic Motion... • ... instead it precesses around its pivot axis ! • This rather odd phenomenon can be easily understood using the simple relation between torque and angular momentum we derived in a previous lecture.

  25. See text: 11.6 Gyroscopic Motion... • The magnitude of the torque about the pivot is  = mgd. • The direction of this torque at the instant shown is out of the page (using the right hand rule). • The change in angular momentum at the instant shown must also be out of the page! d L pivot  mg

  26. See text: 11.6 Gyroscopic Motion... • Consider a view looking down on the gyroscope. • The magnitude of the change in angular momentum in a time dt is dL = Ld. • So where is the “precession frequency” top view L(t) dL d pivot L(t+dt)

  27. See text: 11.6 Gyroscopic Motion... • So • In this example  = mgd and L = I: • The direction of precession is given by applying the right hand rule to find the direction of  and hence of dL/dt. d  L pivot  mg

  28. Lecture 24, Act 1Statics • Suppose you have a gyroscope that is supported on a gymbals such that it is free to move in all angles, but the rotation axes all go through the center of mass. As pictured, looking down from the top, which way will the gyroscope precess? (a)clockwise(b)counterclockwise(c) it won’t precess 

  29. Lecture 24, Act 1Statics • Remember that W = t/L. So what is t? t = r x F r in this case is zero. Why? Thus W is zero. It will not precess. At All. Even if you move the base. This is how you make a direction finder for an airplane. Answer (c) it won’t precess

  30. Recap of today’s lecture • Chapter 11, • Conservation of Angular Momentum • Spinning Demos

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