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Rotational mechanics, angular momentum revision

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is attached as an image. Note this is from a past exam.

    2. Relevant equations

    Conservation of angular momentum.
    Rotational kinetic energy.

    3. The attempt at a solution

    a) The moment of inertia of the man and stool is given as 3 kg m^2, and the dumbells can be considered point masses, so we can just add them all up:

    [tex]I_\mathrm{initial} = I_\mathrm{man} + 2I_\mathrm{dumbell~~~ away} = 3 + 2(3 \times 1^2) = 9[/tex]

    [tex]I_\mathrm{final} = I_\mathrm{man} + 2I_\mathrm{dumbell ~~~pulled~~~ in} = 3 + 2(3 \times 0.25^2) = 3.375[/tex]

    b) The initial angular velocity of the man is [tex]1.5[/tex], and its initial moment of inertia is [tex]9[/tex], so the system's angular momentum is [tex]L = 1.5 \times 9 = 13.5[/tex]. From conservation of angular momentum, the system must have the same angular momentum after the dumbells have been pulled in, so [tex]L = I_\mathrm{final} \omega_\mathrm{final}[/tex]. So [tex]\omega_\mathrm{final} = \frac{L}{I_\mathrm{final}} = \frac{13.5}{3.375} = 4[/tex].

    c) Using the rotational kinetic energy formula:

    [tex]K_\mathrm{initial} = \frac{1}{2} I_\mathrm{initial} \omega^2_\mathrm{initial} = \frac{1}{2} \times 9 \times 1.5^2 = 10.125 J[/tex]

    [tex]K_\mathrm{final} = \frac{1}{2} I_\mathrm{final} \omega^2_\mathrm{final} = \frac{1}{2} \times 3.375 \times 4^2 = 27 J[/tex]

    I am note sure I got it right, shouldn't kinetic energy be conserved? Or does some of it go into potential energy because of the increased radius? I mean clearly the guy is going to spin faster, so the extra kinetic energy must be coming from somewhere.

    PS: imagine the correct units are in there, I'm just too lazy to type them up in LaTeX :tongue:
     

    Attached Files:

  2. jcsd
  3. May 10, 2012 #2

    gneill

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    Staff: Mentor

    Your results look fine.

    Where did the energy come from to move the dumbbells? :wink:
     
  4. May 10, 2012 #3
    Does it come from the work done by the man to pull the dumbells in closer? (chemical energy stored in the man's muscles I suppose) - or maybe I'm overthinking it?
     
  5. May 10, 2012 #4

    gneill

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    Staff: Mentor

    No, not overthinking; That's correct. So, since mechanical energy is being added by a system-internal source, the kinetic energy will not be conserved.
     
  6. May 10, 2012 #5
    Right, that makes sense! Thanks a lot!
     
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