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Homework Help: 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
    t=Ia
    a=w/t

    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

    hage567

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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

    hage567

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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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