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Homework Help: Mechanics question - circular motion

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A yo-yo of mass m , on a string of length [itex]\ell[/itex] is held horizontally and then released, as shown in the diagram. Assume that there is no friction and that the yo-yo swings freely in a vertical plane. Give your answers in terms of m, [itex]\ell[/itex], [itex]\theta[/itex], and g.

    Diagram:

    [PLAIN]http://img43.imageshack.us/img43/6735/physprob2.png [Broken]

    The part I'm stuck on: What is the angular speed [itex]\omega = \dot{\theta}[/itex] as a function of [itex]\theta[/itex]?

    2. Relevant equations

    Kinetic energy: [tex](1/2)mv^2 = (1/2)m\omega^2\ell^2[/tex]

    3. The attempt at a solution

    The way I solved it was using conservation of energy. If we take the bottom of the arc to be zero, the yo-yo was dropped from a height [itex]\ell[/itex]. Then at some angle [itex]\theta[/itex], the yo-yo would be at a height of [itex]\ell\cos\theta[/itex] (and have a corresponding potential energy of [itex]mg\ell\cos\theta[/itex]). So the equation becomes

    [tex](1/2)m\omega^2\ell^2 + mg\ell\cos\theta = mg\ell[/tex]

    After some algebra, the answer comes out to be

    [tex]\omega = \sqrt{ 2g(1-\cos\theta) /\ell}[/tex].

    (for some reason, the LaTeX coding is being silly with the square root; however, the entire expression is supposed to be under the square root)

    My professor does it differently. In his solution, he uses conservation of energy with the equation

    [tex](1/2)m\omega^2\ell^2 = mg\ell\cos\theta[/tex]

    and solves from there to get the same thing I got except with just [itex]\cos\theta[/itex] where I have [itex]1-\cos\theta[/itex]. What is wrong with my reasoning here?

    Thanks for your help!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 27, 2010 #2

    gabbagabbahey

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    Looking at your diagram, when [itex]\theta=0[/itex], the mass will be at the bottom of the arc and should have zero potential energy (or at least less than it had at the top of the arc)....what does your formula for the potential energy give at [itex]\theta=0[/itex]?
     
    Last edited by a moderator: May 4, 2017
  4. Apr 27, 2010 #3
    Then you'd get an angular velocity of zero, which now that you mention it, does not make sense. However, where does the mistake in my reasoning come in?
     
  5. Apr 27, 2010 #4

    gabbagabbahey

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    Your expression for the potential energy is clearly incorrect...how did you derive it?
     
  6. Apr 27, 2010 #5
    I just derived it from the expression for gravitational potential energy, using [itex]\ell\cos\theta[/itex] as the height.
     
  7. Apr 27, 2010 #6

    gabbagabbahey

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    [itex]\ell\cos\theta[/itex] is the distance below the pivot...not the height:wink:
     
  8. Apr 27, 2010 #7
    Ohhh haha, thank you! :)
     
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