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Homework Help: Could Someone Check This Answer?

  1. Jun 16, 2005 #1
    Hello!!

    I did this problem and have gotten it wrong-- but my math is right. Maybe I'm missing something? Here's the question:

    We want to support a thin hoop by a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What must the hoop's radius be?


    So this is a physical pendulum, and in order to find the radius, I need to know the period. The period is found by:

    [tex]T=2\pi\sqrt{\displaystyle{\frac{I}{mgd}}}[/tex]
    Where I is the moment of inertia, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of gravity.

    For a hoop, the moment of inertia is: [tex]I=MR^2[/tex]
    And the center of gravity is in the center of the hoop, so d is the radius.

    So with that information, I set up the following:

    [tex]T=2\pi\sqrt{\displaystyle{\frac{MR^2}{MgR}}}[/tex]
    [tex]T=2\pi\sqrt{\displaystyle{\frac{R}{g}}}[/tex]

    And in plugging in the values, I get:

    [tex]2=2\pi\sqrt{\displaystyle{\frac{R}{9.8}}}[/tex]
    Solving for R, I get .9929475

    Which, when I plug into my equation, I get very close to 2 for my period. However, this is incorrect...

    Any ideas as to why?


    Thanks!!!!
     
  2. jcsd
  3. Jun 16, 2005 #2
    That makes sense. How would I go about finding the moment of inertia about the nail?

    ...This is a longshot, but would the moment of inertia be 2MR^2??
    (using the parallel-axis theorem...)
     
  4. Jun 16, 2005 #3
    ...There was a reply here a minute ago... ...That suddenly disappeared...
     
  5. Jun 16, 2005 #4

    StatusX

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    Sorry, that was me. I wasn't sure about my answer because I haven't done this stuff in a while, but you can try it if you want. And yes, that is the correct moment of inertia around the nail.
     
  6. Jun 17, 2005 #5

    CarlB

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    Yes, you're right, you have to adjust the MI by adding the appropriate amount because you're spinning the hoop around a spot on the rim instead of its center.

    If I recall, the formula for adjusting the MI is to take the formula you look up, and add to it [tex]md^2[/tex], where "m" is the mass of the object, and "d" is the distance away from the center of mass. So you get [tex]MR^2+MR^2 = 2MR^2[/tex], like you suspected.

    Carl
     
    Last edited: Jun 17, 2005
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