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Torque on a cylindrical rod

  1. Jul 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A cylindrical rod 36.4 cm long has mass 0.655kg and radius 1.1cm. A 18.5kg ball of diameter 11.4cm is attached to one end. The arrangement is originally vertical with the ball at the top and is free to pivot about the other end. After the ball-rod system falls a quarter turn, what is its rotational kinetic energy?

    2. Relevant equations

    KEr = 1/2 Iw^2.
    mgh = 1/2Iw^2 + 1/2 mv^2.


    3. The attempt at a solution

    I found the moment of inertia of the ball-rod system, but then I don't know what to do next.
     
  2. jcsd
  3. Jul 12, 2007 #2

    Doc Al

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    How are they defining "rotational KE"? Do they mean rotational KE about the center of mass? (The object can be considered to be in pure rotation about the end point, and thus all its KE is rotational KE about that point.)

    If they mean rotation KE about the center of mass, then use [itex]v = \omega r_{cm}[/itex] to relate rotational and translational speeds.
     
  4. Jul 12, 2007 #3

    Dick

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    During the quarter turn rotation what is the vertical displacement of the center of mass? Hence what can you say about the change in gravitational potential energy of the system? Where does that energy go? Looks pretty clear to me that they mean you to consider rotational energy about the pivot point, so yes, consider all motion rotational.
     
  5. Jul 12, 2007 #4

    Doc Al

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    If that's the case, then no need to find the moment of inertia. :wink: (To me it's still a bit ambiguous.)
     
  6. Jul 12, 2007 #5

    Dick

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    Good point. Please ignore things that "are pretty clear to me" without foundation.
     
  7. Jul 12, 2007 #6
    Thank you so much! I got it.
     
  8. Jul 12, 2007 #7
    So if I were to find the angular speed of the system at that point, would I just do
    Rotational Kinetic energy = 1/2 Iw^2, where I = 1/3ML^2 + 2/5MR^2 + (L+R)^2?
     
  9. Jul 12, 2007 #8

    Dick

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    So the rotational axis they are referring to IS through the pivot point?? No translational KE after all?
     
  10. Jul 12, 2007 #9
    yes just rotational KE
     
  11. Jul 12, 2007 #10

    Dick

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    Then, yes, just use moment of inertia to solve for angular velocity. I notice you've included the R of the sphere to the parallel axis part for the sphere. So you have decided that the ball is glued on to the end of the rod? (BTW you only have one M symbol in your moment expression and none at all on the (L+R)^2 part. Just typos, right?).
     
  12. Jul 12, 2007 #11
    lol yeah :)
     
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