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AP Physics C: Mechanics Chapter 11 Angular Momentum The Vector Product There are instances where the product of two vectors is another vector. Earlier we saw where the product of two vectors was a scalar. This was called the dot product. The vector product of two vectors is called the cross product. Section 11.1 The Vector Product and Torque The torque vector lies in a direction perpendicular to the plane formed by the position vector and the force vector. t = F´r The torque is the vector (or cross) product of the position vector and the force vector. Section 11.1 The Vector Product Defined two vectors, A and B The vector (cross) product of A and defined as a third vector, C = A ´ B . Given C is read as “A cross B”. The B is magnitude of vector C is AB sin q q is the angle between A and. B Section 11.1 More About the Vector Product quantity AB sin q is equal to the area of the parallelogram formed by A and B . The direction of C is perpendicular to the plane formed by A and B . The The best way to determine this direction is to use the right-hand rule. Section 11.1 Properties of the Vector Product The vector product is not commutative. The order in which the vectors are multiplied is important. To account for order, remember A ´ B = -B ´ A If A is parallel to B (q = 0o or 180o), then A ´ B = 0 Therefore A ´ A = 0 If A is perpendicular to B , then A ´ B = AB The vector product obeys the distributive law. A x (B + C) = A x B + A x C Section 11.1 Final Properties of the Vector Product The derivative of the cross product with respect to some variable such as t is ( ) d dA dB A ´B = ´B + A ´ dt dt dt where it is important to preserve the multiplicative order of the vectors. Section 11.1 Vector Products of Unit Vectors ˆi ´ ˆi = ˆj ´ ˆj = kˆ ´ kˆ = 0 ˆi ´ ˆj = - ˆj ´ ˆi = kˆ ˆj ´ kˆ = -kˆ ´ ˆj = ˆi kˆ ´ ˆi = - ˆi ´ kˆ = ˆj Section 11.1 Signs in Cross Products Signs are interchangeable in cross products ( ) A ´ -B = - A ´ B and ˆi ´ (- ˆj)= -ˆi ´ ˆj Section 11.1 Using Determinants The cross product can be expressed as ˆi A ´ B = Ax ˆj Ay Bx By kˆ Ay Az = By Bz Expanding Az ˆi + Ax Bz Bx Az ˆ Ax j+ Bx Bz Ay By kˆ the determinants gives A ´ B = (Ay Bz - AzBy ) ˆi + (Ax Bz - AzBx )ˆj + (Ax By - Ay Bx )kˆ Section 11.1 Vector Product Example Given A = 2ˆi + 3ˆj; B = -ˆi + 2ˆj Find A ´ B Result A ´ B = (2ˆi + 3ˆj) ´ ( -ˆi + 2ˆj) = 2ˆi ´ ( -ˆi ) + 2ˆi ´ 2ˆj + 3ˆj ´ ( -ˆi ) + 3ˆj ´ 2ˆj = 0 + 4kˆ + 3kˆ + 0 = 7kˆ Section 11.1 Torque Vector Example Given the force and location F = (2.00 ˆi + 3.00 ˆj) N r = (4.00 ˆi + 5.00 ˆj) m Find the torque produced t = r ´ F = [(4.00ˆi + 5.00ˆj)N] ´ [(2.00ˆi + 3.00ˆj)m] = [(4.00)(2.00)ˆi ´ ˆi + (4.00)(3.00)ˆi ´ ˆj +(5.00)(2.00)ˆj ´ ˆi + (5.00)(3.00)ˆi ´ ˆj = 2.0 kˆ N × m Section 11.1 Angular Momentum Consider a particle of mass m located at the vector position r and moving with linear momentum p . Find the net torque. r ´ å F = åt = r ´ dp dt dr ´ p (sinceit = 0 ) dt d (r ´ p ) t = å dt Add the term This looks very similar to the equation for the net force in terms of the linear momentum since the torque plays the same role in rotational motion that force plays in translational motion. Section 11.2 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 p mv What if mass of center of object is not moving, but it is rotating? Angular momentum L I May 3, 2017 Angular Momentum I 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: kg m2/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 May 3, 2017 Angular Momentum II Angular momentum of a particle L I mr 2 mv r mvr sin rp sin Angular momentum of a particle L r p m( r v ) r is the particle’s instantaneous position vector p is its instantaneous linear momentum Only tangential momentum component contribute r and p tail to tail form a plane, L is perpendicular to this plane May 3, 2017 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 sinq = mvr sin (90o) = mvr A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path O May 3, 2017 Angular momentum III 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 May 3, 2017 Example: calculating angular momentum for particles Two objects are moving as shown in the figure . What is their total angular momentum about point O? Lnet L1 L2 r1 p1 r2 p2 m2 Lnet r1mv1 sin q1 r2 mv2 sin q 2 r1mv1 r2 mv2 2.8 3.1 3.6 1.5 6.5 2.2 31.25 21.45 9.8 kgm2 / s m1 May 3, 2017 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 May 3, 2017 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 May 3, 2017 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 May 3, 2017 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. May 3, 2017 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 May 3, 2017 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 a M m1 m2 May 3, 2017 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 May 3, 2017 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 May 3, 2017 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. May 3, 2017 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 May 3, 2017 Isolated System τ net 0 L about z - axis L constant I ω I i initial i final f ωf Moment of inertia changes May 3, 2017 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 May 3, 2017 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. May 3, 2017 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. May 3, 2017 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 May 3, 2017