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Angular Momentum Conservation of Angular momentum.

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Presentation on theme: "Angular Momentum Conservation of Angular momentum."— Presentation transcript:

1 Angular Momentum Conservation of Angular momentum

2 Angular Momentum L = r x p = r x mvL = r x p = r x mv Magnitude L = rpsinθ = rmvsinθMagnitude L = rpsinθ = rmvsinθ Derivative of angular momentumDerivative of angular momentum dL/dt = (dr/dt x mv) +(r x m dv/dt) = (v x mv) + (r x ma) = r x F = torquedL/dt = (dr/dt x mv) +(r x m dv/dt) = (v x mv) + (r x ma) = r x F = torque

3 Angular Momentum Continued dL/dt = r x F or dL/dt = tau or torquedL/dt = r x F or dL/dt = tau or torque L z = Σ m i r i v i = Σ m i r i 2 ω = IωL z = Σ m i r i v i = Σ m i r i 2 ω = Iω L z = IωL z = Iω Angular Momentum is Conserved!!!!!

4 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.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.

5 Solution KE 1 + U 1 = KE 2 + U 2KE 1 + U 1 = KE 2 + U 2 mgh = ½ mv ½ Iω 2mgh = ½ mv ½ Iω 2 I = ½ MR 2, v = RωI = ½ MR 2, v = Rω mgh = ½ mv 2 +1/2(1/2MR 2 )(v/R) 2mgh = ½ mv 2 +1/2(1/2MR 2 )(v/R) 2 v 2 = 4mgh/(2m + M)v 2 = 4mgh/(2m + M) v= √4mgh/(2m + M)v= √4mgh/(2m + M) TR = ½ MR 2 αTR = ½ MR 2 α α = 2T/MRα = 2T/MR

6 Solution Continued ΣF = maΣF = ma T-mg = -maT-mg = -ma a = 2TR/MRa = 2TR/MR mg –T = m(2T/M)mg –T = m(2T/M) T = mg(M/(M + 2m))T = mg(M/(M + 2m)) a = 2T/M = (2(mg)(M/(M +2m))/Ma = 2T/M = (2(mg)(M/(M +2m))/M a = 2mg/(M +2m)a = 2mg/(M +2m)

7 Solution Continued α = a/R =2mg/(RM +R2m)α = a/R =2mg/(RM +R2m) v 2 = v o 2 +2ahv 2 = v o 2 +2ah v 2 = (2mg/(M + 2m))hv 2 = (2mg/(M + 2m))h v = √(4mgh/(M + 2m))v = √(4mgh/(M + 2m))

8 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?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?

9 Solution L 1 = mvRAngular Momentum of the bulletL 1 = mvRAngular Momentum of the bullet L 2 = mR 2 ω + IωL 2 = mR 2 ω + Iω mvR = mR 2 ω + IωmvR = mR 2 ω + Iω I = 1/2MR 2I = 1/2MR 2 mvR = ω(mR 2 +1/2MR 2 )mvR = ω(mR 2 +1/2MR 2 ) ω = mv/(R(m + 1/2M))ω = mv/(R(m + 1/2M))


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