Question

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 mb , 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 12MR2 about its center of mass. How many forces are applied to the pulley-axle system, and how many torques are applied to the pulley about its center when the block is released from rest

Answer 1

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