Answer:
F = mbg
τ = 1/4×MR³×ω²/h
Explanation:
The force applied to the pulley-axle system is equal to the downward weight of the block of mass mb falling freely.
F = Wb = mbg
The torque on the system is equal to
τ = I×α
But
ω² = ωo² + 2αθ
ωo = 0 rad/s
ω² = 0² + 2αθ
α = ω²/2θ
The block of mass m falls through a height h
So h = θR and θ = h/R
α = ω²/2(h/R) = ω²R/2h
So
τ = I×α = 1/2×MR²×ω²R/2h = 1/4×MR³×ω²/h
τ = 1/4×MR³×ω²/h