Angular Acceleration of a washer

AI Thread Summary
The discussion revolves around calculating the angular displacement of a washer tub during its spin-dry cycle, which involves a period of acceleration and deceleration. The tub accelerates to an angular speed of 7.0 rev/s in 13 seconds and then decelerates to rest in 14 seconds. Participants clarify the need to use angular kinematic equations, emphasizing the correct units for angular acceleration. After applying the equations, the correct total angular displacement is determined to be 94.63 revolutions. This highlights the importance of understanding angular motion and the application of kinematic principles.
ryan838
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I don't 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.
 
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You have to apply the kinematic equations for constant acceleration:
x = x0 + v0t + 1/2 a t2
vf2 = vi2 + 2aΔx

Of course, instead of distance, velocity, and acceleration, you must use angular displacement, angular velocity, and angular acceleration. Make sense?
 
No that doesn't 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 don't get how to put it into the equation you posted. Thanks for your help though.
 
Originally posted by ryan838
I took the 7 rev/s and divided it by 13s to get the acceleration to be .54 rev/s.
Right. The units should be rev/s2. (I should have given you the equation Δv = at .)
Then to get the deacceleration I took 7 rev/s divided by 14s to get -.5 rev/s.
Right. Same comment about units.
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 don't get how to put it into the equation you posted.
You know the times and the accelerations. Now you need to find the angle (distance in revs). Which of the two equations I gave give the distance? (The other one you don't need!)
 
Thank you for your help. I got the right answer which turned out to be 94.63 revolutions.
 
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