Angular MomentumAngular Momentum

Conservation of Angular Conservation of Angular momentummomentum

Angular MomentumAngular Momentum

• LL = = rr x x pp = = rr x m x mvv• 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) dL/dt = (dr/dt x mv) +(r x m dv/dt)

= (v x mv) + (r x ma) = r x F = = (v x mv) + (r x ma) = r x F = torquetorque

Angular Momentum Angular Momentum ContinuedContinued

• dL/dt = r x F or dL/dt = r x F or dL/dt = tau or dL/dt = tau or torquetorque

• LLzz = Σ m = Σ miirriivvii = Σ = Σ mmiirrii

22ω = Iωω = Iω

• LLzz = Iω = Iω

• Angular Angular Momentum is Momentum is Conserved!!!!!Conserved!!!!!

ProblemProblem

• A string of negligible weight is wrapped A string of negligible weight is wrapped around a pulley of mass M and radius R around a pulley of mass M and radius R and tied to a mass m. The mass is and tied to a mass m. The mass is released from rest and it drops a released from rest and it drops a distance h to the floor. Use energy distance h to the floor. Use energy principles to determine the speed of the principles to determine the speed of the mass when it hits the floor. Determine mass when it hits the floor. Determine the speed, tension and angular the speed, tension and angular acceleration of the pulley.acceleration of the pulley.

SolutionSolution

• KEKE11 + U + U11 = KE = KE22 + U + U22

• mgh = ½ mvmgh = ½ mv22 + 0 + ½ Iω + 0 + ½ Iω22

• I = ½ MRI = ½ MR22 , v = Rω , v = Rω• mgh = ½ mvmgh = ½ mv22 +1/2(1/2MR +1/2(1/2MR22)(v/R))(v/R)22

• vv22 = 4mgh/(2m + M) = 4mgh/(2m + M)• v= √4mgh/(2m + M)v= √4mgh/(2m + M)• TR = ½ MRTR = ½ MR22αα• α = 2T/MRα = 2T/MR

Solution ContinuedSolution 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)

Solution ContinuedSolution Continued

• α = a/R =2mg/(RM +R2m)α = a/R =2mg/(RM +R2m)

• vv22 = v = voo22 +2ah +2ah

• vv22 = 0 + 2 (2mg/(M + 2m))h = 0 + 2 (2mg/(M + 2m))h• v = √(4mgh/(M + 2m))v = √(4mgh/(M + 2m))

Problem 2Problem 2

• A disk is mounted with its axis A disk is mounted with its axis vertical. It has a radius R and a vertical. It has a radius R and a mass M. It is initially at rest. A mass M. It is initially at rest. A bullet of mass m and a velocity v is bullet of mass m and a velocity v is fired horizontally and tangential to fired horizontally and tangential to the disk. It lodges in the perimeter the disk. It lodges in the perimeter of the disk. What angular velocity of the disk. What angular velocity will the disk acquire?will the disk acquire?

SolutionSolution

• LL11 = mvR = mvR Angular Momentum of Angular Momentum of the bulletthe bullet

• LL22 = mR = mR22ω + Iωω + Iω

• mvR = mRmvR = mR22ω + Iωω + Iω• I = 1/2MRI = 1/2MR22

• mvR = ω(mRmvR = ω(mR22 +1/2MR +1/2MR22))• ω = mv/(R(m + 1/2M))ω = mv/(R(m + 1/2M))