A pulley with radius R and rotational inertia I is free to rotate on a horizontal fixed axis through its center.A string passes over the pulley.A block of mass m₁is attached to one end and a block of mass m₂_, is attached to the other.At one time the block with mass m₁ is moving downward with speed v.If the string does not slip on the pulley, the magnitude of the total angular momentum, about the pulley center, of the blocks and pulley, considered as a system, is given by: A)(m₁ - m₂)vR + Iv/R B)(m₁ + m₂)vR + Iv/R C)(m₁ - m₂)vR - Iv/R D)(m₁ + m₂)vR - Iv/R E)none of the above

Question

Quizplus · Accepted Answer

The total angular momentum of the system is the sum of the angular momentum of each object. Firstly, consider the angular momentum of the pulley. Since it is rotating about its center, its angular momentum is given by Lp = Iω, where ω is the angular velocity of the pulley. Since there is no external torque acting on the system, the angular momentum of the pulley is conserved. Next, consider the two blocks. The angular momentum of each block is given by Ls1 = m₁vr and Ls2 = m₂vR, where v is the speed of m₁ and R is the radius of the pulley. Since the string does not slip on the pulley, the angular velocity of the pulley, ω, is related to the linear speed of the blocks by v/R = ω. Substituting this into the expressions for the angular momentum of the blocks gives Ls1 = m₁ωR and Ls2 = m₂ωR. Adding these three angular momenta together, we get: Ltot = Lp + Ls1 + Ls2 = Iω + m₁ωR + m₂ωR = (I + m₁R^2 + m₂R^2)ω = (I + (m₁ + m₂)R^2)v/R Substituting the expression for ω and simplifying, we get: Ltot = (I + (m₁ + m₂)R^2)v/R This matches the expression in choice B.

A Pulley with Radius R and Rotational Inertia I Is