2Angular Momentum L = r x p = r x mv Magnitude L = rpsinθ = rmvsinθ Derivative of angular momentumdL/dt = (dr/dt x mv) +(r x m dv/dt) = (v x mv) + (r x ma) = r x F = torque
3Angular Momentum Continued dL/dt = r x F or dL/dt = tau or torqueLz = Σ mirivi = Σ miri2ω = IωLz = IωAngular Momentum is Conserved!!!!!
4ProblemA string of negligible weight is wrapped around a pulley of mass M and radius R and tied to a mass m. The mass is released from rest and it drops a distance h to the floor. Use energy principles to determine the speed of the mass when it hits the floor. Determine the speed, tension and angular acceleration of the pulley.
5Solution KE1 + U1 = KE2 + U2 mgh = ½ mv2 + 0 + ½ Iω2 I = ½ MR2 , v = Rωmgh = ½ mv2 +1/2(1/2MR2)(v/R)2v2 = 4mgh/(2m + M)v= √4mgh/(2m + M)TR = ½ MR2αα = 2T/MR
6Solution Continued ΣF = ma T-mg = -ma a = 2TR/MR mg –T = m(2T/M) T = mg(M/(M + 2m))a = 2T/M = (2(mg)(M/(M +2m))/Ma = 2mg/(M +2m)
8Problem 2A disk is mounted with its axis vertical. It has a radius R and a mass M. It is initially at rest. A bullet of mass m and a velocity v is fired horizontally and tangential to the disk. It lodges in the perimeter of the disk. What angular velocity will the disk acquire?