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Torque on potter's wheel

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A potter's wheel of radius 0.50 m and mass 100.0 kg is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.0 seconds by pressing a wet rag against the rim.
    a. What is the angular acceleration of the wheel?
    b. How much torque does the potter apply to the wheel?

    2. Relevant equations

    3. The attempt at a solution
    a. (50/60*2*pi)=5.235988 rad/sec
    5.235988/6=.87 rad/sec^2
    b. I didnt understand this portion of the question. I tried .5^2*100*5.235988 but the answer was not 11 Nm which I know is the answer. Please help! Any help is greatly appreciated. Thank you.
  2. jcsd
  3. Apr 18, 2007 #2


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    Homework Helper

    You want

    torque = I * alpha

    not I * omega, that is angular momentum. Check your calculation.

    Also, if you are assuming the wheel is a solid cylindrical disk, then I = 0.5MR^2, so your above I value is not correct if that's the case.
    Last edited: Apr 18, 2007
  4. Apr 19, 2007 #3
    So...in this case, i should go .5*100*.5^2=12.5?
    But the answer given is 11 Nm. I dont understand. :(
  5. Apr 19, 2007 #4


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    Homework Helper

    You didn't multiply by alpha!! torque = I*alpha. If you do that you will get the right answer. You found the right value of alpha in part (a).
  6. Apr 19, 2007 #5
    Oh, ok. My fault. Ok. Alpha.
    so.... .5*100*.5^2*.87=10.9=11Nm!! ok. Thank you!!
    Sorry, I must have overlooked that alpha part. Thank you.
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