Physics 1 11 : Mechanics Lecture 11

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

2. Angular Momentum • Vectors • Cross Product • Torque using vectors • Angular Momentum

3. y ay q x ax Vector Basics Ways of writing vector notation • 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

4. z q q f f y x 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 f 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 q from the z axis. z y x

5. z q f y x 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:

6. z y x 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.

7. y q j i x i k z j k 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:

8. 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. 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. Derivation • How do we show that ? • Start with • Then • But • So

11. Torque as a Cross Product • 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. Superposition: • Can have multiple forces applied at multiple points. • Direction of tnetis angular acceleration axis

12. i j k Calculating Cross Products Find: Where: Solution: Calculate torque given a force and its location Solution:

13. z q y x i j k Net torque example: multiple forces at a single point 3 forces applied at point r : Find the net torque about the origin: oblique rotation axis through 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:

14. 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. 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-m2/s Calculate L of a 10 kg disk when  = 320 rad/s, R = 9 cm = 0.09 m L = I and I = MR2/2 for disk L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s *When rotation is about a principal axis

16. Angular Momentum II • Angular momentum of a particle • 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. O Angular Momentum of a Particle in Uniform Circular Motion 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. • The angular momentum vector points out of the diagram • The magnitude is L = rp sin = mvr sin(90o) = mvr • A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path

18. 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. 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. m2 m1

20. A B P Angular Momentum for a Car • 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 ? • 5.0  102 • 5.0  106 • 2.5  104 • 2.5  106 E) 5.0  103

21. 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. 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. Demonstration • Start from • Expand using derivative chain rule

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

25. Example: A Non-isolated System A sphere mass m1 and a block of mass m2 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. Masses are connected by a light cord. Find the linear acceleration a. • Use angular momentum approach • No friction between m2 and table • Treat block, pulley and sphere as a non- isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides • Constraints: I • 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 & 2nd law

27. 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. 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. 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 vi = 2 m/s in a circle of radius ri = 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. Angular Momentum Conservation • m = 0.5 kg, vi = 2 m/s, ri = 0.2 m, rf = 0.1 m, vf = ? • Isolated system? • Tension force on m exert zero torque about hole, why?

31. Isolated System Moment of inertia changes

32. Controlling spin (w) by changing I (moment of inertia) In the air, tnet= 0 L is constant Change I by curling up or stretching out - spin rate w must adjust Moment of inertia changes

33. 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-m2. There is no friction. Find the angular velocity of the platform after the child has jumped on.

34. The Merry-Go-Round • 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.

35. Solution: A merry-go-round problem I = 20 kg.m2 VT = 4.0 m/s mc = 40 kg r = 2.0 m w0 = 0