Chapter 11: Rolling, Torque, and Angular Momentum • For an object rolling smoothly, the motion of the center of mass is pure translational. θ θ ω s = Rvcom = ds/dt = d( R)/dt = R ω vcom = R • Rolling viewed as a combination of pure rotation and pure translation • Rolling viewed as pure rotation ω vtop = ( )(2R) = 2 vcom • Different views, same conclusion • Rotational inertia involves not only the mass but also the distribution of mass for continuous masses • Calculating the rotational inertia I = ∫ r2dm The Kinetic Energy of Rolling View the rolling as pure rotation around P, the kinetic energy ω2 K = ½ IP 2 parallel axis theorem: Ip = Icom +MR ω2 2ω2 so K = ½ Icom + ½ MR ω since vcom = R ω2 2 K = ½ Icom + ½ M(vcom) ω2 ½ Icom : due to the object’s rotation about its center of mass 2 ½ M(vcom) : due to the translational motion of its center of mass Sample Problem: A uniform solid cylindrical disk, of mass M = 1.4 kg and radius R = 8.5 cm, rolls smoothly across a horizontal table at a speed of 15 cm/s. What is its kinetic energy K? vc.m. = 0.15m/s 2 2 −3 2 Idisk =1/2MR = (0.5)(1.4kg)(0.085m) = 5.058x10 kg m ω = v/R = (0.15m/s)/0.085m = 1.765 rad/s 1 1 K = K + K = Mv2 + Iω2 trans rot 2 c.m. 2 1 1 − K = K + K = (1.4)(0.15)2 + (5.058x10 3 )(1.765)2 trans rot 2 2 =15.75x10−3 + 7.878x10−3 = 23.63x10−3 J Vector Product (Review) • Vector product of vectors a and b produce a third vector c whose magnitude is c = a b sin φ whose direction follow the right hand rule if a and b parallel => a x b = 0 if a and b perpendicular => a x b =ab i x i = 0 i x j = 1 Note: a x b = - ( b x a ) Torque revisited • For a fixed axis rotation, torque τ = r F sinθ • Expand the definition to apply to a particle that moves along any path relative to a fixed point. r τr = rr × F • direction : right-hand rule • magnitude :τ = rF sin Φ = rF⊥ = r⊥ F Angular momentum • Angular momentum with respect to point O for a particle of mass m and linear momentum p is defined as v r r r r l = r × p = m (r × v) direction: right-hand rule magnitude: l = r psin φ = r mvsin φ • Compare to the linear case p = mv Particles 1, 2, 3, 4, 5 have the same mass and speed as shown in the figure. Particles 1 & 2 move around O in opposite directions. Particles 3, 4, and 5 move towards or away from O as shown. 2r 2r r r Which of the particles has the greatest magnitude angular momentum? Particles 1, 2, 3, 4, 5 have the same mass and speed as shown in the figure. Particles 1 & 2 move around O in opposite directions. Particles 3, 4, and 5 move towards or away from O as shown. 2r 2r r r = φ = φ φ o l r psin r mvsin = 90 only for 1 and 2, but r1= 2r2. Which of the particles has the greatest magnitude angular momentum? 1) 1 2) 2 3) 3 4) 4 5) 5 6) all have the same l Newton’s Second Law in Angular Form r dl τr = net dt r •:τnet the vector sum of all the torques acting on the object r dpr • Comparing to the linear case: F = net dt • Newton’s 2nd law for a system of particles r dL r n r r = total L = l τnet total ∑ i dt i=1 – Net external torque equals to the time rate change of the system’s total angular momentum Torque and Angular Momentum v dl d(vr × mv) vτ = = net dt dt v = v × dv = v ×()v = v × v r m r ma r F dt Torque is the time rate of change of angular momentum. A Quiz Particles 1, 2, 3, 4, 5 have the same mass and speed as shown in the figure. Particles 1 & 2 move around O in opposite directions. Particles 3, 4, and 5 move towards or away from O as shown. 2r 2r r r Which of the particles has the smallest magnitude angular momentum? 1) 1 2) 2 3) 3 4) 4 5) 5 6) all have the same l A Quiz Particles 1, 2, 3, 4, 5 have the same mass and speed as shown in the figure. Particles 1 & 2 move around O in opposite directions. Particles 3, 4, and 5 move towards or away from O as shown. 2r 2r r r l = r psin φ = r mvsin φ φ = 0o for 5. => l = 0 Which of the particles has the smallest magnitude angular momentum? 1) 1 2) 2 3) 3 4) 4 5) 5 6) all have the same l The angular momentum of a rigid body rotating about a fixed axis • Consider a simple case, a mass m rotating z v y about a fixed axis z: r l = r mv sin90o = r m r ω = mr2ω = I ω x • In general, the angular momentum of rigid body rotating about a fixed axis is L = I ω L : angular momentum along the rotation axis I : moment of inertia about the same axis Conservation of Angular Momentum • If the net external torque acting on a system is zero, the angular momentum of the system is conserved. r dL r = ifτr = then L const net dt r r τ = • If net, z = 0 then Li, z = Lf, z L i L f • For a rigid body rotating around a fixed axis, ( L = I w ) the conservation of angular momentum can be written as ω ω Ii i = If f Some examples involving conservation of angular momentum • The spinning volunteer ω ω Ii i = If f Angular momentum is conserved Li is in the spinning wheel Now exert a torque to flip its rotation. − Lf, wheel = Li. Conservation of Angular momentum means that the person must now acquire an angular momentum. +2 Lf, person = Li +2 − so that Lf = Lf, person + Lf, wheel = Li + Li = Li. Problem 11-66 Ring of R1 (=R2/2) and R2 (=0.8m), Mass m2 = 8.00kg. ω i = 8.00 rad/s. Cat m1 = 2kg. Find kinetic energy change when cat walks from outer radius to inner radius. Initial Momentum = + = + ω Li Li,cat Li,ring m1R 2vi I i 1 = m R 2ω + m (R 2 + R 2 )ω 1 2 i 2 2 1 2 i 1 m R 2 = m R 2ω 1+ 2 1 +1 1 2 i 2 2 m1 R 2 Problem 11-66 Ring of R1 (=R2/2) and R2 (=0.8m), Mass m2 = 8.00kg. ω i = 8.00 rad/s. Cat m1 = 2kg. Find kinetic energy change when cat walks from outer radius to inner radius. Problem 11-66 Ring of R1 (=R2/2) and R2 (=0.8m), Mass m2 = 8.00kg. ω i = 8.00 rad/s. Cat m1 = 2kg. Find kinetic energy change when cat walks from outer radius to inner radius. Final Momentum = + = + ω Lf Lf ,cat Lf ,ring m1R1vf I f 1 = m R 2ω + m (R 2 + R 2 )ω 1 1 f 2 2 1 2 f 1 m R 2 = m R 2ω 1+ 2 2 +1 1 1 f 2 2 m1 R1 Problem 11-66 Problem 11-66 Ring of R1 (=R2/2) and R2 (=0.8m), Mass m2 = 8.00kg. ω i = 8.00 rad/s. Cat m1 = 2kg. Find kinetic energy change when cat walks from outer radius to inner radius. Initial Kinetic energy Ki is:.