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Angular Momentum. Conservation of Angular momentum. Angular Momentum. L = r x p = r x m v Magnitude L = rpsinθ = rmvsinθ Derivative of angular momentum dL/dt = (dr/dt x mv) +(r x m dv/dt) = (v x mv) + (r x ma) = r x F = torque. dL/dt = r x F or dL/dt = tau or torque
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Angular Momentum Conservation of Angular momentum
Angular Momentum • L = r x p = r x mv • Magnitude L = rpsinθ = rmvsinθ • Derivative of angular momentum • dL/dt = (dr/dt x mv) +(r x m dv/dt) = (v x mv) + (r x ma) = r x F = torque
dL/dt = r x F or dL/dt = tau or torque Lz = Σ mirivi = Σ miri2ω = Iω Lz = Iω Angular Momentum is Conserved!!!!! Angular Momentum Continued
Problem • A 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.
Solution • KE1 + U1 = KE2 + U2 • mgh = ½ mv2 + 0 + ½ Iω2 • I = ½ MR2 , v = Rω • mgh = ½ mv2 +1/2(1/2MR2)(v/R)2 • v2 = 4mgh/(2m + M) • v= √4mgh/(2m + M) • TR = ½ MR2α • α = 2T/MR
Solution 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))/M • a = 2mg/(M +2m)
Solution Continued • α = a/R =2mg/(RM +R2m) • v2 = vo2 +2ah • v2 = 0 + 2 (2mg/(M + 2m))h • v = √(4mgh/(M + 2m))
Problem 2 • A 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?
Solution • L1 = mvR Angular Momentum of the bullet • L2 = mR2ω + Iω • mvR = mR2ω + Iω • I = 1/2MR2 • mvR = ω(mR2 +1/2MR2) • ω = mv/(R(m + 1/2M))