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Stopping a potter's wheel with a wet rag (angular momentum and friction)

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data
    A potter's wheel having a radius 0.55 m and a moment of inertia 11.6 kg·m2 is rotating freely at 55 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 66 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

    2. Relevant equations
    for a solid wheel I = (1/2)m*R^2

    3. The attempt at a solution
    I started off converting 55rpm into 5.76 rad/s. I then divided 5.76rad/s by 8.00s to get the needed acceleration to stop the wheel and got -.72rad/s^2. Next i solved for the mass of the wheel by dividing the moment of inertia by .5*R^2 and got 7.018kg.

    I figured by setting the force needed to stop the wheel equal to the force applied by the rag times the coefficient of friction i could divide the applied force on both sides and end up with the COF: m*a = F*u u=(m*a)/F u= (7.018kg*-.72rad/s^2)/66N = .07656
     
  2. jcsd
  3. Apr 17, 2010 #2
    if you're going to use tangential acceleration, you have to multiply the rotational acceleration by the radius
    and
    you assumed that the potter's wheel could be approximated by I = .5 MR^2 when you didn't need to.

    This problem is much more easily solved by using torques.

    Recall that torque = moment of inertia x rotational acceleration

    t = I a
    you found rotational acceleration successfully
    Now you need to find the torque from friction

    Recall that torque = radius x force x sin (angle between them)
    t = r x F x sin 90
    Friction = mu x normal force
    = u N
    so the torque t = r u N

    now set torques equal, r u N = I a
    and solve for u

    you do have to memorize a couple of equations for torques, but once you do a lot of problems become a lot easier to figure out
     
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