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Change in the precession angle

  1. Oct 29, 2006 #1
    Hi. I have: A 3 kg bicycle wheel rotating at a 2484 rev/min angular velocity has its shaft supported on one side, as show in the figure. When viewing from the left (from the positive x-axes), one sees that the wheel is rotating in a clockwise manner. The distance from the center of the wheel to the pivot point is 0.6 m. The wheel is a hoop of radius 0.4 m and its shaft is horizontal.

    Assume all the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9.8 m/s^2.

    Find the change in the precession angle after a 1.5 s time interval.

    In my book I have the equation:

    [tex]d\phi = \frac{(M g h) dt}{L}[/tex]

    so [tex] L = I \omega = M R^2 \omega[/tex]

    therefore [tex]d\phi = \frac{(M g h) dt}{M R^2 \omega} = \frac{(g h) dt}{R^2 \omega}[/tex]

    Now [tex]\omega = \frac{2484 "rev"}{"min"} \frac{1 "min"}{60 "sec"} \frac{2 \pi "rad"}{"rev"} = \frac{260.124 "rad"}{"sec"}[/tex]

    and let [tex]h = 0.6, R = 0.4, g = 9.8, dt = 1.5[/tex]

    Using these I came up with .211918 but the answer was wrong. Any ideas where I went wrong? Thank you.

    [Edit]I tried -.211918 (the bicycle is rotating in a clockwise fashion) but that was wrong as well.
    Last edited: Oct 29, 2006
  2. jcsd
  3. Oct 29, 2006 #2
    I was supposed to answer in degrees but I was putting the answer in radians. This has been a humbling learning experience, but alas, those are the ones that seem to make such a difference. Thanks everyone! :-)
  4. Oct 29, 2006 #3


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    If by everyone, you mean yourself (since nobody actually did anything), then sure.

    Nice to know you got the answer though
  5. Oct 29, 2006 #4
    Thanks! I figured no one responded because they saw I was on the right path and assumed I would figure it out . Hence, by not responding they aided my development. It's happened before! :-)
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