1. Physics 111: Mechanics Lecture 11 Dale Gary NJIT Physics Department

2. August 3, 2015 Angular Momentum  Vectors  Cross Product  Torque using vectors  Angular Momentum

3. August 3, 2015 Vector Basics  We will be using vectors a lot in this course.  Remember that vectors have both magnitude and direction e.g.  You should know how to find the components of a vector from its magnitude and direction  You should know how to find a vector’s magnitude and direction from its components Ways of writing vector notation  y x axax ayay

4. August 3, 2015  y x z   Projection of a Vector in Three Dimensions  Any vector in three dimensions can be projected onto the x - y plane.  The vector projection then makes an angle  from the x axis.  Now project the vector onto the z axis, along the direction of the earlier projection.  The original vector a makes an angle  from the z axis. y x z 

5. August 3, 2015 Vector Basics (Spherical Coords)  You should know how to generalize the case of a 2-d vector to three dimensions, e.g. 1 magnitude and 2 directions  Conversion to x, y, z components  Conversion from x, y, z components  Unit vector notation:  y x z 

6. August 3, 2015 A Note About Right-Hand Coordinate Systems  A three-dimensional coordinate system MUST obey the right-hand rule.  Curl the fingers of your RIGHT HAND so they go from x to y. Your thumb will point in the z direction. y x z

7. August 3, 2015 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

8. August 3, 2015 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

9. August 3, 2015 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

10. August 3, 2015 Derivation  How do we show that ?  Start with  Then  But  So

11. August 3, 2015  The torque is the cross product of a force vector with the position vector to its point of application  The torque vector is perpendicular to the plane formed by the position vector and the force vector (e.g., imagine drawing them tail-to-tail)  Right Hand Rule: curl fingers from r to F, thumb points along torque. Torque as a Cross Product Superposition :  Can have multiple forces applied at multiple points.  Direction of  net is angular acceleration axis

12. August 3, 2015 Calculating Cross Products Solution: i kj Calculate torque given a force and its location Solution: Find: Where:

13. August 3, 2015 i kj Net torque example: multiple forces at a single point x y z  3 forces applied at point r : Find the net torque about the origin: Here all forces were applied at the same point. For forces applied at different points, first calculate the individual torques, then add them as vectors, i.e., use: oblique rotation axis through origin

14. August 3, 2015 Angular Momentum  Same basic techniques that were used in linear motion can be applied to rotational motion. F becomes  m becomes I a becomes  v becomes ω x becomes θ  Linear momentum defined as  What if mass of center of object is not moving, but it is rotating?  Angular momentum

15. August 3, 2015 Angular Momentum I  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: kg-m 2 /s Calculate L of a 10 kg disk when  = 320 rad/s, R = 9 cm = 0.09 m L = I  and I = MR 2 /2 for disk L = 1/2MR 2  = ½(10)(0.09) 2 (320) = 12.96 kgm 2 /s * When rotation is about a principal axis

16. August 3, 2015 Angular Momentum II  Angular momentum of a particle r is the particle’s instantaneous position vector p is its instantaneous linear momentum Only tangential momentum component contribute Mentally place r and p tail to tail form a plane, L is perpendicular to this plane

17. August 3, 2015 Angular Momentum of a Particle in Uniform Circular Motion  The angular momentum vector points out of the diagram  The magnitude is L = rp sin  = mvr sin(90 o ) = mvr  A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path O Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v.

18. August 3, 2015 Angular momentum III  Angular momentum of a system of particles angular momenta add as vectors be careful of sign of each angular momentum for this case:

19. August 3, 2015 Example: calculating angular momentum for particles Two objects are moving as shown in the figure. What is their total angular momentum about point O? Direction of L is out of screen. m2m2 m1m1

20. August 3, 2015  What would the angular momentum about point “P” be if the car leaves the track at “A” and ends up at point “B” with the same velocity ? Angular Momentum for a Car A)5.0  10 2 B)5.0  10 6 C)2.5  10 4 D)2.5  10 6 E)5.0  10 3 P A B

21. August 3, 2015 Recall: 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

22. August 3, 2015 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 are to be measured about the same origin  The origin must not be accelerating (must be an inertial frame)

23. August 3, 2015 Demonstration  Start from  Expand using derivative chain rule

24. August 3, 2015 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

25. August 3, 2015 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.

26. August 3, 2015 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

27. August 3, 2015 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

28. August 3, 2015 Angular Momentum Conservation  Here i denotes initial state, f is the final state  L is conserved separately for x, y, z direction  For an isolated system consisting of particles,  For an isolated system that is deformable

29. August 3, 2015 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.

30. August 3, 2015 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?

31. August 3, 2015 Moment of inertia changes Isolated System

32. August 3, 2015 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

33. August 3, 2015 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.

34. August 3, 2015  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

35. August 3, 2015 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