1. Physics 106: Mechanics Lecture 06 Wenda Cao NJIT Physics Department

2. February 24, 2011 Conservation of Angular Momentum  Cross Product  Comparison: definitions of single particle torque and angular momentum  Angular Momentum of a system of particles of a rigid object  Conservation of angular momentum  Examples http://www.youtube.com/watch?v=AQLtcEAG9v0

3. February 24, 2011 Cross Product  The cross product of two vectors says something about how perpendicular they are.  Magnitude:  is smaller angle between the vectors Cross product of any parallel vectors = zero Cross product is maximum for perpendicular vectors Cross products of Cartesian unit vectors:  y x z i j k i kj

4. February 24, 2011 Cross Product  Direction: C perpendicular to both A and B (right-hand rule) Place A and B tail to tail Right hand, not left hand Four fingers are pointed along the first vector A “sweep” from first vector A into second vector B through the smaller angle between them Your outstretched thumb points the direction of C  First practice

5. February 24, 2011 More about Cross Product  The quantity ABsin  is the area of the parallelogram formed by A and B  The direction of C is perpendicular to the plane formed by A and B  Cross product is not commutative  The distributive law  The derivative of cross product obeys the chain rule  Calculate cross product

6. February 24, 2011 Torque Angular Momentum Torque and Angular Momentum for a Single Particle

7. February 24, 2011 Angular momentum of a system of particles  Angular momentum of a system of particles angular momenta add as vectors be careful of sign of each angular momentum for this case:

8. February 24, 2011 Angular Momentum of a Rigid Body  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: kgm 2 /s  Calculate L of a 10 kg disc when  = 320 rad/s, R = 9 cm = 0.09 m  L = I  and I = MR 2 /2 for disc  L = 1/2MR 2  = ½(10)(0.09) 2 (320) = 12.96 kgm 2 /s

9. February 24, 2011  A solid sphere and a hollow sphere have the same mass and radius. They are rotating with the same angular speed. Which one has the higher angular momentum? A) the solid sphere B) the hollow sphere C) both have the same angular momentum D) impossible to determine Finding angular momentum

10. February 24, 2011 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

11. February 24, 2011 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

12. February 24, 2011 Demonstration  Start from  Expand using derivative chain rule

13. February 24, 2011 What about SYSTEMS of Rigid Bodies?  i = net torque on particle “i” internal torque pairs are included in sum individual angular momenta L i all about same origin BUT… internal torques in the sum cancel in Newton 3rd law pairs. Only External Torques contribute to L sys 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. Total angular momentum of a system of bodies: net external torque on the system

14. February 24, 2011 Example: A Non-isolated System A sphere mass m 1 and a block of mass m 2 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.

15. February 24, 2011 Masses are connected by a light cord Find the linear acceleration a. Use angular momentum approach No friction between m 2 and table Treat block, pulley and sphere as a non- isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides Constraints: Ignore internal forces, consider external forces only Net external torque on system: Angular momentum of system: (not constant) same result followed from earlier method using 3 FBD’s & 2 nd law I

16. February 24, 2011 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

17. February 24, 2011 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

18. February 24, 2011 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 v i = 2 m/s in a circle of radius r i = 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.

19. February 24, 2011 Angular Momentum Conservation  m = 0.5 kg, v i = 2 m/s, r i = 0.2 m, r f = 0.1 m, v f = ?  Isolated system?  Tension force on m exert zero torque about hole, why?

20. February 24, 2011 Moment of inertia changes Isolated System

21. February 24, 2011 How fast should the student spin? L is constant… while moment of inertia changes The student on a platform is rotating (no friction) with angular speed 1.2 rad/s. His arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the professor, the bricks, and the platform about the central axis is 6.0 kg·m 2. By moving the bricks the student decreases the rotational inertia of the system to 2.0 kg·m 2. (a)what is the resulting angular speed of the platform? (b)what is the ratio of the system’s new kinetic energy to the original kinetic energy?

22. February 24, 2011 KE has increased!! L is constant… while moment of inertia changes, I i = 6 kg-m 2  i = 1.2 rad/s I f = 2 kg-m 2  f = ? rad/s Solution (a): Solution (b):

23. February 24, 2011 Controlling spin () by changing I (moment of inertia) In the air,  net = 0 L is constant Change I by curling up or stretching out - spin rate  must adjust Moment of inertia changes

24. February 24, 2011 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-m 2. There is no friction. Find the angular velocity of the platform after the child has jumped on.

25. February 24, 2011  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. The Merry-Go-Round

26. February 24, 2011 Solution: A merry-go-round problem I = 20 kg.m 2 V T = 4.0 m/s m c = 40 kg r = 2.0 m  0 = 0