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Potter's Wheel problem

  1. Dec 2, 2003 #1
    Problem 3.
    A potter's wheel of radius 0.19m and mas 82.10 kg is freely rotating at 46.0 rev/min. The potter can stop the wheel in 8.4 s by pressing a wet rag against the rim
    a. What is the angular acceleration of the wheel? In rad/s^2.
    Note: Is the formula to use in this problem t=I*a?Where I is moment of inertia and a = angular acceleration. How would the data in the problem be set up in the formula?
  2. jcsd
  3. Dec 2, 2003 #2


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    "Is the formula to use in this problem t=I*a?Where I is moment of inertia and a = angular acceleration."
    What is t? Normally t means time but that doesn't make sense. I*a = force.

    Acceleration is change in speed divided by the time required for the change (the problem says "In rad/s^2"- that's rad/s divided by s). Here you are told that the initial speed is 42 rev/ min (how many revolutions is that per second) and that it slows to 0 in 46 seconds. That's all the information you need.
  4. Dec 2, 2003 #3
    t= torque.
    This equation is Newton's second law of rotation. I was wondering if I= 12.5kg*m^2, then to find torque it is force times distance. I was wondering what would be the a to find the force? In addition if it is rev./min for a, do i just leave it in there?
  5. Dec 2, 2003 #4
    I don't see how you're getting that value for I.
    For this object
    Icm = (1/2)mr2

    But anyway, it's irrelevant here.

    As HallsofIvy already told you, you don't need torque or moment of inertia here. You have been given the angular velocity in rev/min, and you are told how long it takes to stop it.

    For linear motion:
    aavg = Δv/Δt, right?

    Just do the equivalent for angular motion (after converting your given angular velocity to the appropriate units).
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