A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25 m. The other end of the rope is attached to a massless suspended platform, upon which 0.5 kg weights may be placed. While the system is initially in equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling the rope. What is the torque on the pulley when the system is motionless?

The torque on the pulley, when the system is motionless is approximately 9.81 N·m

Explanation:

The given parameters are;

The mass of the object = 2 kg

The friction between the rope and the pulley = 0

The mass of the rope, = 0.5 kg

The mass of the pulley, = 0.01 kg

The radius of the pulley, r = 0.25 m

The torque on the pulley, τ = I·α = F × D

The torque on the pulley, when the system is motionless, τ = F × D

Where;

F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N

D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m

Therefore;

τ = 19.62 N × 0.5 m = 9.81 N·m

The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.

we need to know the time interval during which the force acts. impulse of force = momentum change