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Homework Help: Rotating Disk

  1. May 3, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin, 50.0 g disk with a diameter of 8.00 cm rotates about an axis through its center with 0.190 J of kinetic energy. What is the speed of a point on the rim?


    2. Relevant equations
    [tex]C = \frac{1}{2}MR^2[/tex]
    [tex]K_{rot}=\frac{1}{2}I\omega^2[/tex]


    3. The attempt at a solution
    Since the formula for kinetic rotational energy is [tex]K_{rot}=\frac{1}{2}I\omega^2[/tex], and the constant for the moment of inertia for a solid disk with the axis of rotation about it's diameter is [tex]C = \frac{1}{2}MR^2[/tex], I substituted the I in the second equation with the first equation, resulting in the following:

    [tex]K_{rot}=\frac{1}{2}(\frac{1}{2}MR^2)\omega^2[/tex]

    Substituting the values that I was supplied with in the problem statement, I came up with this equation:

    [tex]0.190=\frac{1}{2}(\frac{1}{2}(0.05)(0.04)^2)\omega^2[/tex]

    Solving for [tex]\omega[/tex], I ended up with [tex]\omega \approx \pm 97.5[/tex], which was determined to be incorrect.

    Any help would be extremely appreciated.
     
  2. jcsd
  3. May 3, 2010 #2
    It asks for the SPEED of a point on the rim. It does not ask for the angular speed.
     
  4. May 3, 2010 #3
    I figured it was something like that, so shortly after posting the initial topic, I tried [tex]v=r\omega[/tex] as [tex]v=(0.04)(97.5)[/tex] which results in the correct answer of 3.9. Originally I recieved an answer that was something like 360 (I must have entered a wrong number or something into the calculator) so I ignored it. Then after you posted and backed up my thoughts, I tried again and got the correct answer. Thanks!
     
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