What Distance Does the Bucket Cover in 3.5 Seconds?

[tsnikpoh11]

[SOLVED] How far does the bucket fall?

Homework Statement

A cylindrical pulley with a mass of M = 5.7 kg and a radius of r = 0.66 m is used to lower a
bucket with a mass of m = 2 kg into a well. The bucket starts from rest and falls for 3.5 s.
The acceleration of gravity is 9:8 m/s^2 :

Homework Equations

Moment of Inertia
Newtons second law
W(f)=W(i)+(angular acceleration)(Time)

The Attempt at a Solution

I = 1/2(5.7)(.66^2)=1.241

W(f)=0+(9.8?)(3.5) Is this the right equation to figure out how far the bucket dropped and does moment of Inertia have anything to effect it? would you do W(f) = 34.3, but do I multiply that by the moment of inertia?

 

[dynamicsolo]

I had to read this again: you are calling the angular speed of the pulley W(t), rather than \omega(t). That's fine: we'll call the angular acceleration of the pulley A.

If you use an approach to this problem involving forces and torques, you are going to need to work out the torque on the pulley in order to find its angular acceleration. What force provides the torque on the pulley and how much is that torque? You have the pulley's moment of inertia, so what is the angular acceleration A for the pulley?

Since you want the distance the bucket drops, you don't actually care about W(t). Instead, what would be the equation describing the angle the pulley rotates by as a function of time. (Hint: it is analogous to the equation for the linear distance an object moves in time t, if it has constant linear acceleration.) If you know the angle the pulley rotates by in 3.5 seconds, you can find how much rope will pay out in that time (which assumes it is all wrapped around at a constant radius from the axle), which is how far the bucket could drop.