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Where does ω=sqrt(MgL/I) come from?

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the frequency of a hoop pivoting on an axle. The hoop has mass M and radius R.


    2. Relevant equations



    3. The attempt at a solution
    Obviously f = ω / 2pi

    But apparently ω=sqrt(MgL/I), which you plug into the above. I don't understand where this part comes from, though. L is the center of gravity of the hoop (which is just R).

    Any help?
     
  2. jcsd
  3. Mar 10, 2010 #2

    Andrew Mason

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    You will have to explain the problem better. It sounds like the hoop is oscillating like a pendulum but it is unclear why it would do this if the axle is through the centre of the hoop. Is there a diagram?

    AM
     
  4. Mar 10, 2010 #3
    Whoops sorry, the hoop pivots through its edge. There was no diagram given, but if the hoop is like a unit circle, then imagine an axle passing through the hoop at the 90 degree position.
     
  5. Mar 11, 2010 #4

    Andrew Mason

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    Ok. It is similar to a single bob pendulum with the bob located at the centre of mass (ie the centre of the hoop) BUT with a slight difference.

    Analyse the forces the same as you would a simple pendulum, BUT you have to let:

    [tex]\vec{F}\cdot{\vec{R}} = \vec{\tau} = I\vec{\alpha}[/tex]

    Here is a little more help:

    For small angles,

    [tex]\vec{\tau} = -mgR\sin\theta \approx -mgR\theta = I\alpha[/tex]

    which reduces to:

    [tex]\alpha = \ddot\theta = -\frac{mgR}{I}\theta [/tex]

    [Hint: Find the general solution to that differential equation ]

    AM
     
    Last edited: Mar 11, 2010
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