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Angular Velocity of Rotating Mass

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid sphere of mass m rotates uniformly in a horizontal circle suspended from a fixed
    point on a cord of length L and negligible mass. The cord
    makes an angle theta from the vertical. What is the angular
    velocity?

    2. Relevant equations

    Torque= (Rotational Inertia)x(angular acceleration)


    3. The attempt at a solution

    Tried to mess around with a few things but ultimately got no-where so any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 14, 2011 #2
    This is the Diagram for the problem
     

    Attached Files:

  4. Sep 14, 2011 #3

    Hootenanny

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    Welcome to Physics Forums.

    I would suggest that you draw a free body diagram, identifying all the forces acting on the mass.
     
  5. Sep 14, 2011 #4
    thanks for the welcome,

    so there would be a tension force through the cord, a gravitational force acting straight down and a centripetal force. am i missing anything?
     
  6. Sep 14, 2011 #5

    Hootenanny

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    No, that's spot on. So, what can you say about the horizontal and vertical components of the forces acting on the mass?
     
  7. Sep 14, 2011 #6
    the vertical components sum to zero.
    does the horizontal component of the tension= the centripetal force?
     
  8. Sep 14, 2011 #7

    Hootenanny

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    Correct!
     
  9. Sep 14, 2011 #8
    ok so i got to
    tan(theta)=(v^2)/rg where r= Lsin(theta)

    i was thinking solve for v and then use Angular velocity=v(r)

    is this heading in the right direction?
     
  10. Sep 14, 2011 #9

    Hootenanny

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    Looks okay to me.
     
  11. Sep 14, 2011 #10
    ok so i got angular velocity= g/Lcos(theta)

    but the answers tell me that it should be the sqrt of that.
    im a bit lost as to where i have made my mistake :S
     
  12. Sep 14, 2011 #11
    oh wait i forgot to square both sides of the equation!

    thanks for your help!
     
  13. Sep 14, 2011 #12

    Hootenanny

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    So you have

    [tex]\frac{v^2}{r} = g\tan\theta[/tex]

    And you know that [itex]v = \omega r[/itex]. Thus,

    [tex]\omega^2 r = g\tan\theta[/tex]

    Do you follow and can you take it from here?
     
  14. Sep 14, 2011 #13

    Hootenanny

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    Ahh! Never-mind!

    No problem, it was a pleasure!
     
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