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**Conservation of Angular momentum**

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**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

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**Angular Momentum Continued**

dL/dt = r x F or dL/dt = tau or torque Lz = Σ mirivi = Σ miri2ω = Iω Lz = Iω Angular Momentum is Conserved!!!!!

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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.

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**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

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**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)

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**Solution Continued α = a/R =2mg/(RM +R2m) v2 = vo2 +2ah**

v2 = (2mg/(M + 2m))h v = √(4mgh/(M + 2m))

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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?

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**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))

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Phys211C10 p1 Dynamics of Rotational Motion Torque: the rotational analogue of force Torque = force x moment arm = Fl moment arm = perpendicular distance.

Phys211C10 p1 Dynamics of Rotational Motion Torque: the rotational analogue of force Torque = force x moment arm = Fl moment arm = perpendicular distance.

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