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## Homework Statement

A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.1 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.

Solve for:

1) tension in the rope

2) time before the bucket hits the water

3) speed at which the bucket hits the water

4) force exerted on the cylinder by the axle during the fall

## Homework Equations

Torque=R*F*sin(phi)

Torque=I*alpha

I=(M*R^2)/2

## The Attempt at a Solution

M=12.1 kg (mass of cylinder)

R=0.175 (radius of cylinder)

m=14.2 (mass of bucket)

Torque = 0.175 * Tension = 1/2 * 12.1 * 0.175^2 * alpha

Torque = 0.175 * Tension = 0.18528125 * alpha

1.05875 * alpha = Tension

Weight = 14.2 * 9.8 = 139.16

ForceNet (on bucket) = Weight - Tension = m*a

ForceNet = 139.16 - Tension = 14.2 * a

139.6 - 1.05875*alpha = 14.2 * a

Tension = (1/2) * M * a

Tension = 41.57619

But I've been told that this isn't the correct answer. Can somebody set me straight, please?