# Angular Acceleration of a washer

1. Dec 6, 2003

### ryan838

I dont really know what to do on this problem. So if someone could get me pointed in the right direction I would appreciate it.

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 13.0 s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 14.0 s. Through how many revolutions does the tub turn during this 27 s interval? Assume constant angular acceleration while it is starting and stopping.

2. Dec 6, 2003

### Staff: Mentor

You have to apply the kinematic equations for constant acceleration:
x = x0 + v0t + 1/2 a t2
vf2 = vi2 + 2a&Delta;x

Of course, instead of distance, velocity, and acceleration, you must use angular displacement, angular velocity, and angular acceleration. Make sense?

3. Dec 6, 2003

### ryan838

No that doesnt really make to much sense. Here is what I did so maybe you can tell me where I went wrong.

I took the 7 rev/s and divided it by 13s to get the acceleration to be .54 rev/s. Then to get the deacceleration I took 7 rev/s divided by 14s to get -.5 rev/s. So for the first 13s it is speeding up by .54 rev/s right? Then the last 14s it is deacclerating at a rate of -.5 rev/s? From there I dont get how to put it into the equation you posted. Thanks for your help though.

4. Dec 6, 2003

### Staff: Mentor

Right. The units should be rev/s2. (I should have given you the equation &Delta;v = at .)