Physics 1501: Lecture 23 Today ’ s Agenda

1. Physics 1501: Lecture 23Today’s Agenda • Announcements • HW#8: due Oct. 28 • Honors’ students • see me Wednesday at 2:30 in P-114 • Topics • Review of • rolling motion • Angular momentum • Gyroscope …

2. 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

3. 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

4. Example: Throwing ball from stool... • Conserve angular momentum (since there are no external torques acting on the student-stool system): • LBEFORE = 0 , Lstool= IF • LAFTER = 0 = Lstool - Lball , Lball = Iballball = Md2 (v/d) = M d v • 0 = IF - M d v F =M v d / I M v F d I I top view: before after

5. Summary of rotation:Comparison between Rotation and Linear Motion Angular Linear x  = x /R  = v / R v a  = a /R

6. ComparisonKinematics Angular Linear

7. Comparison: Dynamics m I = Si mi ri2 Angular Linear F = a m t = r x F = a I L = r xp = I w p = mv W = F •x W =  D K = WNET K = WNET

8. Lecture 23, Act 1 • 1. You are working with Marriott’s Great America and are given the task of designing their new roller coaster. Unfortunately, the cars have already been purchased, but you do get to design the track. The idea is to pull the cars up to the top of a hill, then release them from a standstill to go barreling down the hill, around a vertical loop, and on to the next thrill. You decide that this will all work best if the car falls completely under the influence of gravity and that the riders feel completely weightless at the top of the loop. The cars consist of a box for the passengers that when completely loaded have a mass of 1000kg. They also have four wheels, each of which is a solid cylinder of radius 0.25 m and a mass of 125 kg. The county regulations say that the maximum height of the entire rollercoaster can only be 30 meters. So how big is the loop you can make?

9. z z F Reconsider: 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. The disks eventually they rotate together with angular velocity F. 0

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

11. Example: Two Disks • Remember that using conservation of angular momentum: LINI = LFIN we got a different answer ! wF = ½ w0 ½M R2w0 = M R2 wF wF’ = w0 / 21/2 Conservation of energy ! Conservation of momentum ! wF = w0 / 2 wF’ > wF Which one is correct ?

12. z F Example: Two Disks • Is the system conservative ? • Are there any non-conservative forces involved ? • In order for top disc to turn when in contact with the bottom one there has to be friction ! (non-conservative force !) • So, we can not use the conservation of energy here. • correct answer: wF = w0/2 • We can calculate work being done due to this friction ! W = DE = 1/2 I w02 - 1/2 (I+I)(w0/2)2 = 1/2 I w02 (1 - 2/4) = 1/4 Iw02 = 1/8 MR2 w02 This is 1/2 of initial Energy !

13. Lecture 23, 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 mass from ri to r ? (A) 0.15 J (B) 0 J (C) - 0.15 J ri wi

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

15. 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

16. 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. pivot 

17. 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

18. 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” L(t) dL d pivot L(t+dt) top view

19. 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

20. Lecture 23, Act 3Statics • 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 