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Rotational Kinetic Energy of a grinding wheel

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A grinding wheel in the shape of a solid disk is 0.240m in diameter and has a mass of 2.50kg. The wheel is rotating at 2050rpm about an axis through its center.

    What is its kinetic energy?

    How far would it have to drop in free fall to acquire the same amount of kinetic energy?

    2. Relevant equations
    K = 1/2Iw^2
    U(grav) = mgh


    3. The attempt at a solution

    If I can find the first answer I can equate this number to mgh to find the kinetic energy. (I think).

    Now for the first question, I have no idea what I am doing wrong. First off, the radius is .12m. We can then find the moment of inertia by doing:

    I = mr^2 => 2.5(.12^2) = .036kg*m^2

    Then we can find angular velocity by doing:

    w = 2050/60 * 2pi = 214.7rad/s

    Plugging these values into the kinetic energy equation we get:

    K = 1/2(.036)(214.7^2) => 830 J

    This isn't right so any help is appreciated! Thanks!
     
  2. jcsd
  3. Apr 10, 2009 #2

    hage567

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    Homework Helper

    Double-check the formula you are using for I. You want the one for a solid disk.
     
  4. Apr 10, 2009 #3
    Ah...so I = 1/2MR^2. Thanks a lot!
     
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