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Physics 106: Mechanics Lecture 06 Wenda Cao NJIT Physics Department 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 February 24, 2011 Cross Product C A B B B sin The cross product of two vectors says something about how perpendicular they are. Magnitude: A A sin C A B AB sin 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: iˆ ˆj kˆ; iˆ kˆ ˆj; ˆj kˆ iˆ iˆ iˆ 0; ˆj ˆj 0; kˆ kˆ 0 y j i x k z i j k 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 A B B A ? A B - B A A B B A ? 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 A B - B A The distributive law A (B C) A B A C The derivative of cross product obeys the chain rule Calculate cross product d dA dB A B B A dt dt dt A B ( Ay Bz Az By )iˆ ( Az Bx Ax Bz ) ˆj ( Ax By Ay Bx )kˆ February 24, 2011 Torque and Angular Momentum for a Single Particle Torque r F rF rF sin() r F rF sin() Angular Momentum L r p L rp rp sin() p mv L rp rp sin() p mv February 24, 2011 Angular momentum of a system of particles Angular momentum of a system of particles Lnet L 1 L 2 ... L n Li all i ri p i all i angular momenta add as vectors be careful of sign of each angular momentum for this case: Lnet L1 L2 r1 p1 r2 p2 | Lnet | r1 p1 - r2 p2 February 24, 2011 Angular Momentum of a Rigid Body Angular momentum of a rotating rigid object L I L 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: kgm2/s Calculate L of a 10 kg disc when = 320 rad/s, R = 9 cm = 0.09 m L = I and I = MR2/2 for disc L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s February 24, 2011 Finding angular momentum 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 L I 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 dv dp Fnet F ma m dt dt p mv IL Fnet t p t 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 dp dL Fnet F net dt dt and L to be measured about the same origin The origin should not be accelerating, should be an inertial frame February 24, 2011 Demonstration dp dL Fnet F net dt dt Start from dL d (r p ) m d (r v ) dt dt dt Expand using derivative chain rule dL d dr dv m (r v ) m v r mv v r a dt dt dt dt dL mv v r a mr a r (ma ) r Fnet net dt February 24, 2011 What about SYSTEMS of Rigid Bodies? dL i Rotational 2 law for a single body : i dt • individual Total angular momentum angular momenta Li Lsys L i • all about same of a system of bodies: origin dLsys • i = net torque on particle “i” dLi i • internal torque pairs are dt dt included in sum i nd BUT… internal torques in the sum cancel in Newton 3rd law pairs. Only External Torques contribute to Lsys dLsys dt i ,ext net net external torque on the system i 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. February 24, 2011 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. a a ext m1 gR February 24, 2011 a 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 nonisolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides • Constraints: Equal v' s and a' s for block and sphere v ωR for pulley I a α d / dt a αR dv/dt • Ignore internal forces, consider external forces only • Net external torque on system: m gR net • Angular momentum of system: (not constant) 1 about center of wheel Lsys m1vR m2vR Iω m1vR m2vR MR 2ω dLsys m1aR m2 aR MR 2α (m1 R m2 R MR)a τ net m1 gR dt m1 g same result followed from earlier a method using 3 FBD’s & 2nd law M m1 m2 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 dLtot ext 0 dt Ltot constant or Li L f February 24, 2011 Angular Momentum Conservation Ltot constant or Li L f 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, Ltot Ln L1 L2 L3 constant For an isolated system is deformable I ii I f f constant 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 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. February 24, 2011 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? Li L f L r p r (mv ) Li mri vi sin 90 mri vi L f mrf v f sin 90 mrf v f ri 0.2 v f vi 2 4m/s rf 0.1 v 2f 42 T mac m 0.5 80 N rf 0.1 February 24, 2011 Isolated System τ net 0 L about z - axis L constant I ω I i initial i final f ωf Moment of inertia changes February 24, 2011 How fast should the student spin? 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·m2. By moving the bricks the student decreases the rotational inertia of the system to 2.0 kg·m2. (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? L is constant… while moment of inertia changes February 24, 2011 Ii = 6 kg-m2 If = 2 kg-m2 i = 1.2 rad/s f = ? rad/s L is constant… while moment of inertia changes, Zero external torque ... about a fixed axis Solution (a): Solution (b): f Kf Ki L final L initial L L I i i I f f Ii 6 i 1.2 3.6 rad/s If 2 1 2 1 2 I f 2f I f f 2 I f Ii 2 Ii ( ) ( ) 3 2 Ii f Ii I f If I ii KE has increased!! February 24, 2011 Controlling spin () by changing I (moment of inertia) In the air, net = 0 L is constant L I ii I f f Change I by curling up or stretching out - spin rate must adjust Moment of inertia changes 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-goround 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. February 24, 2011 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. February 24, 2011 Solution: A merry-go-round problem Ltot I i ωi I f f Li I ii I0 mc vT r mc vT r L f I f ω f ( I mc r 2 )ω f I = 20 kg.m2 VT = 4.0 m/s mc = 40 kg r = 2.0 m 0 = 0 ( I mc r 2 )ω f mc vT r ωf mc vT r 40 4 2 1.78 rad/s 2 2 I mc r 10 40 2 February 24, 2011