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Conservation of angular momentum on a piano stool

  1. Oct 30, 2006 #1
    A student on a piano stool rotates freely with an angular speed of 3.33 rev/s. The student holds a 1.44 kg mass in each outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 4.34 kgm2, a value that remains constant.
    (a) As the student pulls his arms inward, his angular speed increases to 3.54 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?
    (b) Calculate the initial and final kinetic energy of the system.

    For the first part, I tried doing mr^2w(initial)=mr^2w(final) and it said that my answer was within 10% of the correct answer. I tried doing different variations, playing around with the given moment of inertia, but I couldn't figure out how to factor it in along with the varying radius.
     
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  3. Oct 30, 2006 #2

    Doc Al

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    You realize that angular momentum is conserved. You need to figure out how the rotational inertia of the total system (student+stool+masses) changes as the masses are brought closer to the center. What's the contribution of the masses to the total rotational inertia?
     
  4. Oct 30, 2006 #3
    Well, the masses have to be factored in twice since there are two of them but I'm not sure how to combine it with the initial moment of inertia given in the problem...
     
  5. Oct 30, 2006 #4

    Doc Al

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    What's the moment of inertia of a point mass M at a distance R from an axis?
     
  6. Oct 30, 2006 #5
    moment of inertia I=mr^2
     
  7. Oct 30, 2006 #6

    Doc Al

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    Exactly. So what's the total moment of inertia of the system at those two points (arms outstretched; arms in)?
     
  8. Oct 30, 2006 #7
    arms outstretched: mr^2=(2)(1.44)(0.759)^2
    arms in: mr^2=(2)(1.44)r^2
     
  9. Oct 30, 2006 #8

    Doc Al

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    That's just the contribution of the two masses. Now add that to the moment of inertia of the "student+stool", which is given as a constant.
     
  10. Oct 30, 2006 #9
    ...and so to calculate the kinetic energies I use the equation KE=1/2 Iw^2=1/2 [I+2mr^2](w^2) ?
     
  11. Oct 30, 2006 #10

    OlderDan

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    That should do it.
     
  12. Oct 30, 2006 #11
    I tried using that equation and plugging in my numbers, but apparently it isn't correct...
     
  13. Oct 30, 2006 #12

    OlderDan

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    OK.. You have two different I involved. I assumed from what you did earlier you had them distinguished. So let's be more explicit.

    KE=1/2 (I_total)w^2=1/2 [(I_ss)+2mr^2](w^2)

    where I_ss is the assumed constant moment of inertia of the student and stool combination. You have two calculations to do for the two different positions of the masses with their corresponding angular velocities. If you got the correct r earlier, you should be getting the correct KE.
     
    Last edited: Oct 30, 2006
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