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Can anyone solve this?

  1. Feb 4, 2005 #1
    A uniform rod has a length of 2m. It is hinged to a wall at the left end and held horizontally by a vertical massless string at the right end. What is the angular acceleration of the rod the moment the string is released?
  2. jcsd
  3. Feb 4, 2005 #2


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    angular accel = tangential accel (in this case gravity) / radius
  4. Feb 4, 2005 #3


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    Easiest way : [tex]\tau = I\alpha[/tex], which means torque = moment of inertia*angular acceleration.

    What is the moment of inertia of a rod of uniform length about the end ?

    What is the torque exerted by the weight of the rod about the end ?
  5. Feb 4, 2005 #4
    I think that [tex]I=1/3(ML^2)[/tex]
  6. Feb 4, 2005 #5


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    Correct. You don't have to know the derivation, but it's fairly simple using a little calculus.

    What's the torque about the end ?
  7. Feb 4, 2005 #6
    Ok I got it, thanks, [tex]\tau = I\alpha[/tex] when [tex]\tau =1/2mgL[/tex] and therefore [tex]\frac{1}{3}(ML^2)\alpha=1/2mgL[/tex] and angular acceleration ends up being 7.35.

    You see, the problems I post here are the ones I am studying for the next physics science league, is a competition between schools here in NJ.
    Last edited: Feb 4, 2005
  8. Feb 4, 2005 #7


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    Excellent, you got it (except both your m's need to be the same case) ! :smile:
  9. Feb 5, 2005 #8
    Please forward me your request to my email address <ken@mv.ru>
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