How Is Angular Speed Calculated in a Falling Bucket Physics Problem?

[AlphabetTown]

Hi! I really need help answering this question. I am really bad at Physics and have no idea what I'm doing, so if someone could help me out I would really appreciate it! Thank you!

Homework Statement

8. A bucket filled with water has a mass of 23 kg and is attached to a massless rope, which in turn, is wound around a 0.050-m radius cylinder that rotates about a frictionless axle at the top of a well. The bucket is raised to the top of the well and then released to fall back into the well. What is the angular speed of the cylinder at the instant the bucket is moving at 8 m/s?

The answer is 160 rads/s, but how?

Homework Equations

(I think)

Torque = Force * radius
Alpha = [omega (final) - omega (initial)]/time
Alpha = linear speed/angular speed

The Attempt at a Solution

Torque (T) = F (force) * r (radius)
11.5 Nm = (25 kg)(10 m/s^2) * ( .050 m)

I have no idea where to go from here or if it's even relevant to the problem :(

Thanks again in advance.

 

[voko]

Because the rope is wound on the cylinder, you can assume there is no slipping between the rope and the cylinder. So the speed of the rope is always the same as the tangential speed of the cylinder. Knowing that, and the radius of the cylinder, you should be able to determine its angular speed.

 

[utkarshakash]

I think the problem is incomplete. It should also state that there is no slipping between the rope and cylinder. Assuming this to be true, the relative velocity between them is zero. You already know the velocity of rope at that instant. Now think of that equation that relates linear velocity to angular velocity.

 

[SammyS]

I think the problem is incomplete. It should also state that there is no slipping between the rope and cylinder. Assuming this to be true, the relative velocity between them is zero. You already know the velocity of rope at that instant. Now think of that equation that relates linear velocity to angular velocity.

It's complete.

Everything is there to answer the question ... plus some extraneous information.