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Centripetal Acceleration again

  1. Oct 11, 2005 #1
    Hey all -

    I've got a problem that's rather encompassing...

    The second hand and the minute hand on one type of clock are the same length. What is the period T of the motion for the second hand and for the minute hand?

    It goes on to say the centripetal acceleration is given by a_c = v^2/r where v is speed and r is the radius. How is centripetal acceleration related to the period?

    And then finally it asks for the ratio a_c, second / a_c, minute for the tips of the second hand and the minute hand.

    I'm not even sure where to start really...

    Any help would be great, thanks!
  2. jcsd
  3. Oct 11, 2005 #2
    Basically if you take the parametric equations:

    [tex]x=r\cos{\omega t}[/tex]
    [tex]y=r\sin{\omega t}[/tex]

    ...and note that the second derivatives (acceleration) are:

    [tex]x''=-\omega^2 r\cos{\omega t}[/tex]
    [tex]y''=-\omega^2 r\sin{\omega t}[/tex]

    The magnitude is therefore:

    [tex]\left|\vec{a}\right|=\omega^2 r=\frac{v^2}{r}[/tex]



    Can you use this information and try to answer your question?

  4. Oct 11, 2005 #3
    Sounds complicated.. heh. I'll try and trudge my way thru it and see if I can make more sense of it...

    thanks for the input
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