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Moment of inertia of a flywheel

  1. Feb 24, 2008 #1
    [SOLVED] moment of inertia

    1. The problem statement, all variables and given/known data
    The flywheel of a gasoline engine is required to give up 750 J of kinetic energy while its angular velocity decreases from 870 rev/min to 410 rev/min.

    What moment of inertia is required?

    2. Relevant equations

    [tex]I=\frac{2K}{\omega^{2}}[/tex]

    3. The attempt at a solution

    I'm not sure how to derive the rad/sec used in this equation. I've converted both 870 rev/min and 410 rev/min to radians before attempting the problem (91.1 rad/sec and 42.9 rad/sec respectively). I've tried taking the difference between the two, then attempting the average, then just using the final radial velocity and ignoring the initial. I'm not sure what to do at the moment. Can anyone help?
     
  2. jcsd
  3. Feb 24, 2008 #2
    Well they're saying "give up" 750 joules of KE. So you don't actually know any absolute kinetic energies. You know there is a Ki, and that Kf=Ki-750J, you know Ki corresponds with 870rpm and Kf corresponds with 410 rpm, and that is all you're given to work with
     
  4. Feb 24, 2008 #3

    D H

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    You started on the right track but got off on the wrong foot.

    [tex]
    \begin{aligned}
    K_i &= \frac 1 2 I \omega_i^2 \\[4pt]
    K_f &= \frac 1 2 I \omega_f^2 \\
    &\Rightarrow \\
    K_i-K_f &= \frac 1 2 I (\omega_i^2 - \omega_f^2)
    \end{aligned}[/tex]

    You can take it from here.
     
  5. Feb 24, 2008 #4
    I did something, but I'm still not sure I understand it, but it seemed to work.

    If I assume [tex]K_{f}[/tex] is 0, then I can set up the problem like so:

    [tex]K_{i}=\frac{1}{2}I(\omega_{i}^{2}-\omega_{f}^{2})[/tex]

    Solving for I, I get


    [tex]\frac{2K_{i}}{(\omega_{i}^{2}-\omega_{f}^{2})}=I[/tex]

    Plug and chug and got .232 kg*m^2, which is the correct answer.

    What I don't understand is why [tex]K_{f}[/tex] is 0.
     
  6. Feb 24, 2008 #5

    D H

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    You cannot assume [itex]K_f[/itex] is zero. You are given the change in angular kinetic energy. The problem statement says exactly what [itex]K_i-K_f[/itex] is: 750 joules.
     
  7. Feb 24, 2008 #6
    So in other words. What K initial and K final are is not important, because the change in kinetic energy is already given in the problem?
     
  8. Feb 24, 2008 #7

    cepheid

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  9. Feb 24, 2008 #8
    ^_^; man I feel stupid now. Well, anyway thanks!
     
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