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)
