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Rotational motion/Torque

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A potter's wheel - A thick stone disk of radius 0.5m and mass 100kg is freely rotating at 50rev/min. The potter can stop the wheel in 6s by pressing a rag against the rim and exerting a radially inward force of 70N. Find the effective coefficient of kinetic friction between the wheel and the rag.

    2. Relevant equations

    torque = RF
    Torque = I(moment of inertia) x radial acceleration

    3. The attempt at a solution

    Sum of forces on the x axis: F(applied) - F(normal) = 0
    F(applied) = F(normal)
    The froce of friction is on the y axis and is opposite to the direction the wheel turns, and F(friction) = (coefficient of friction)F(normal)

    Sum of forces on the y axis: I'm not sure what other force opposes the frictional force. Normally I would have thought F(friction) = ma, but it is not a particle that we are looking at, so i'm a little confused.

    And for the sum of the forces for torque, i know Torque = I(angular acceleration). Fir this i think I= 1/2 MR^2, and I can find angular acceleration with angular speed(final) = angular speed(initial) + angular acceleration X t.

    I'm also not sure what forces affect torque.

    Any help is appreciated.
  2. jcsd
  3. Jan 15, 2007 #2


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    Your nearly there. Basically, the friction from the rag is the torque, you know how long it takes the wheel to stop and its initial angular velocity and therefore you can work out the angular acceleration. From this you can work out the applied torque, and hence the frictional force; from which you can calculate the coefficient of friction.
  4. Jan 15, 2007 #3
    So, i find the angular acceleration, using:
    final angular speed = initial angular speed + (angular acceleration)t

    Then, I use the angular acceleration to find the frictional force by:
    torque = RF(friction) and torque = I x (angular acceleration), therefore,
    RF(friction) = I x angular acceleration.

    Finally I use F(friction) = coefficient x F(normal), where F(normal) = F(applied), and I solve for the coefficient of friction.

    Is this what I need to do?
    Thanks a lot for all your help.
  5. Jan 15, 2007 #4


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    That's exactly what you need to do :approve:
    Twas a pleasure
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