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Angular Acceleration of a washer

  1. Dec 6, 2003 #1
    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. jcsd
  3. Dec 6, 2003 #2

    Doc Al

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    Staff: Mentor

    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?
     
  4. Dec 6, 2003 #3
    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.
     
  5. Dec 6, 2003 #4

    Doc Al

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    Staff: Mentor

    Right. The units should be rev/s2. (I should have given you the equation Δv = at .)
    Right. Same comment about units.
    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!)
     
  6. Dec 7, 2003 #5
    Thank you for your help. I got the right answer which turned out to be 94.63 revolutions.
     
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