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Rotational Motion. Apparently, I have no idea what I'm doing.

  1. Feb 1, 2005 #1
    ok... We were given the following question... (I apologize for it's length... please bear with me)

    A thin long cylindrical rod (mass M radius R and length l) is initially rotating about its long axis with angular momentum L. There are no external forces or torques however interestingly Ke is usually not conserved.
    a) Express the initial Ke of the system in terms of M R l and L
    b) Describe how the system can lower its energy while maintaining constant angular momentum.
    c) Calculate the ratio of the final to initial energy
    d) How thin does a rod of radius R have to be at least so that one can consider it as "thin and long"

    Ok. First of all... I have no idea what the formula is for a thin long rod about it's long axis. (The only formulas I can find are for through the cm and perpendicular to the long axis) so I assumed I = 1/2 MR^2 (inertia for a solid cylinder). If that's right, I get Ke = (L^2)/MR^2
    And that's fine. Except that I don't have an l term. Which means that when I go to do part d... I can't find the ratio of l to R necessary.
    Also I have no idea what the "final" energy would be because I have no idea how MUCH Ke is lost based on the question...

    Any help you can give would be greatly appreciated!
    Thanks!
    Nicola
     
  2. jcsd
  3. Feb 1, 2005 #2

    Galileo

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    Find the moment of inertia of the rod about its axis of rotation, then use the formula for the rotational kinetic energy of a spinning rigid body in terms of L and I (the moment of inertia).

    For b): Think of the analogous situation of an iceskater spinning a pirouette. If the arms are outstretched and pulled inwards towards the body the spinning will speed up. Angular momentum is conserved, but the rotational energy increases.
     
  4. Feb 1, 2005 #3

    Andrew Mason

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    That's right. And rotational kinetic energy is:

    [tex]KE = \frac{1}{2}I\omega^2 = \frac{1}{2}L^2/I = \frac{L^2}{MR^2} [/tex]
    so you are right.
    A system can change its energy by changing the distribution of its mass (ie. by increasing the distance of its mass from the axis of rotation, the kinetic energy would be lowered). I don't see how that can be done here. If you wanted to change its orientation, and thereby transfer some of its angular momentum to another axis of rotation, you would have to apply an external torque. (you could lower its temperature, but I think they are asking about kinetic energy).
    This question makes sense only if you can change the moment of inertia. But I don't know how you can do that with a cylindrical rod without applying an external torque.
    R>>l ?? Beats me. The question makes sense if you are talking about electrical charge, but not moment of inertia.

    AM
     
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