An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many times around the edge of the pulley and the free end attached to a block of mass mbmb , which is held at rest. When the block is released, the block falls downward. Consider clockwise to be the positive direction of rotation, frictional effects from the axle are negligible, and the string wrapped around the disk never fully unwinds. The rotational inertia of the pulley is 12MR212MR2 about its center of mass. The block falls for a time t0t0, but the string does not completely unwind. What is the change in angular momentum of the pulley-block system from the instant that the block is released from rest until time t0t0

ΔL = MmRgt / (2m + M)

Explanation:

The pulley starts at rest, so the change in angular momentum is equal to the final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To find the angular velocity ω, first draw a free body diagram for each the pulley and the block.

For the block, there are two forces: weight force mg pulling down, and tension force T pulling up.

For the pulley, there three forces: weight force Mg pulling down, reaction force pulling up, and tension force T pulling down.

Sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

Sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substitute:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

The angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)