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Stuck in a Physics Project

  1. Nov 6, 2004 #1
    I am currently working on a physics project. The idea is a measure rate of rotation of different gymastics flips (i.e tuck, twist, layout, pike). I calculated the moment of inertia for each of the different positions. I then taped a gymnast preforming the different flips on a trapoline. I then found how long it takes her to do one full rotation for each position. This was done repeatley for each position. I then took the ratio the different positions. I am trying to get my math calcs to support my experimental data. I tried taking the ratio of the moment of inertia of the positions. I thought that would match up with the experimental ratio, but it didn't. They were in the same order but the values were off. IF you have any ideas about what I should do, or if you think it is because of experimental error please tell me. I am open to any advice. Thanks for you time.
  2. jcsd
  3. Nov 6, 2004 #2
    Have you considered that the non-uniform construction of the body could be affecting your center of mass and moment of inertia calculations? Though I don't know how big of a factor that'd be since I don't know much about anatomy.
  4. Nov 6, 2004 #3
    of course there is experimental error, you do not know the exact moment of inertia, moreover the moment of inertia will not be constant during the flips because the body is not stiff but changes shape (at least a little) during the flips and you will not measure the timing precise, but well that is obvious...

    Another thing is that the torques are probably not equal when the gymnast does different flips, which affects the rotation times:

    [tex]\tau = I\alpha \Leftrightarrow \alpha = \frac{\tau}{I}[/tex]
    [tex]\omega = \alpha{\Delta{t}}[/tex]

    so the duration of one rotation is:

    [tex]t = \frac{2\pi}{\omega} \Leftrightarrow
    t = \frac{2\pi}{\alpha\Delta{t}} \Leftrightarrow
    t = \frac{2\pi{I}}{\tau\Delta{t}} [/tex]

    One thing that you could do is using your approximation of the moment of inertia, the rotation times you measured and an estimate of how long the torque was applied (perhaps you can get this from the videotapes) to determine what the torques where:
    [tex]\tau = \frac{2\pi{I}}{t\Delta{t}}[/tex]

    EDIT: I think it may even be quite interesting if you can show that the gymnast uses different torques for the different flips.
    Last edited: Nov 6, 2004
  5. Nov 6, 2004 #4


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    You probably ment
    [tex]\omega = \frac{\alpha}{\Delta t}[/tex]

    So the final formula states:
    [tex] t = \frac{2\pi}{\omega} \Leftrightarrowt = \frac{2\pi\Delta t}{\alpha}\Leftrightarrow t = \frac{2\pi I\Delta t}{\tau} [/tex]
  6. Nov 6, 2004 #5
    No, [itex]\omega[/itex] is the angular velocity and [itex]\alpha[/itex] is the angular acceleration, so [itex]\omega=\alpha\Delta{t}[/itex]
  7. Nov 6, 2004 #6
    Air friction could play a small role here as well, and it would have a variable effect due to the varying positions of the gymnast. This should not have too large an effect on results, however.
  8. Nov 8, 2004 #7


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    For me,[tex] \alpha,\beta,\theta,\vartheta [/tex] and so on and so forth are (in classical context) notations for angles.When taught mechanics in school,it was [tex] \epsilon [/tex] the Greek letter that designated angular acceleration.Let's call it a misinterpretation of the notation used.I agree with it.
    Still,you probably meant [itex]\Delta\omega=\alpha\Delta{t}[/itex][/QUOTE],right?
  9. Nov 8, 2004 #8
    No you are wrong, just face it please!!
  10. Nov 8, 2004 #9
    No, acceleration is how much velocity changes per time unit. In this case the angular acceleration is applied for a short time [itex]\Delta{t}[/itex], so the angular velocity ([itex]\omega[/itex]) after this time will be [itex]\alpha\Delta{t}[/itex].
  11. Nov 8, 2004 #10
    I truly appreciate everone who wrote, it was very interesting. My method of calculating the moment of inertia did divded the body in eight sections. I compared that moment of inertia with the simplified sphere or rod inertia equations. The results were comparable. I believe, as stated by gerben, that most of the error is coming from torque issues. How would I measure how long the torque was applied? One more thought: I know that ratios do not have units, but is it a mistake to compare moment of inertia ratios to experimental ratios(i.e seconds per flip). Once again your time is truly appreciated. Thanks
  12. Nov 9, 2004 #11
    The gymnast can only apply force when in contact with the trampoline. You do not really know whether torque is applied during all the time contact is made, but I would just use this time as the best possible approximation (assuming that in all flips torque is applied during a similar percentage of the contact time).

    that would be ok if the torques (and the time the toques are applied) are equal, from the formula below you can see that the time of a rotation is proportional to the moment of inertia, they differ by a factor [itex]\frac{2\pi}{\tau\Delta t}[/itex], so if this factor is constant it will have no effect on the ratios.

    [tex]t = I\times\frac{2\pi}{\tau\Delta t}[/tex]
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