**Answer:**

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)