A bucket hangs from a rope that is attached to a cylindrical, massive pulley wit

Question

A bucket hangs from a rope that is attached to a cylindrical, massive pulley with a radius of 0.420 m and a mass of 3.25 kg. The pulley spins as the bucket falls, undergoing an angular The bucket starts at rest and falls a distance 4.95 m before hitting the water. During this time the pulley undergoes an angular acceleration of 3.67 rad/s2, which means the bucket is falling with a linear acceleration of 1.54 m/s2 (a=r).
What is the moment of inertia of the pulley? (remember, this is a solid cylinder.)

What is the magnitude of the torque that the rope exerts on the pulley? (Think about the rotational version of Newton's Second Law.)

Since you know the torque, what is the force that the rope is exerting on the pulley?

How much time elapses between when the bucket is released and when it hits the water?

Explanation / Answer

a.

the moment of inertia of the pulley is,

I = 0.5 MR2 = 0.5 * 3.25 * 0.42*0.42 = 0.28665 kg.m2 = 0.28 kg.m2

b.

the magnitude of the torque is,

torque = I *angular acceleration = 0.28665 * 3.67 = 1.052 N.m

c.

The force is calculated as follows:

F = Torque / R = 1.052 / 0.420 = 2.50 N

d.

The time is calculated as follows:

t = sqrt [2d / a] =sqrt [2 * 4.95 / 1.54] = 2.54 s

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