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Merry-Go-Round Velocity

  1. Apr 13, 2008 #1
    Problem:
    In a plyground there is a small merry-go-round of radius 3.91 m and rotational inertia 5.68e+03 kg m^2. A child of mass 199 kg runs at a speed of 3.09 m/sec tangent to the rim of the merry-go-round when it is at rest and then jumps on. Assume no friction in the bearing of the merry-go-round. What is the angular velocity of the merry-go-round and child?

    Besides the child be a heavy example, I wasn't able to come up with the right answer of 0.276 rad/sec. What might I be doing wrong?

    My Work:
    I searched the archives and found two ways of solving the problem.
    [tex]method \ 1 \ \Rightarrow m*v*r = (I_{merry-go-round} + I_{child}) \omega \ ; solving \ for \ \omega[/tex]

    [tex]method \ 2 \ \Rightarrow I_{merry-go-round} * \omega \ + \ m*v*r = (I_{merry-go-round} + I_{child}) \omega^{prime} \ ; \ solving \ for \ \omega^{prime}[/tex]

    What I came up with:

    [tex]method \ 1 \ \Rightarrow (199 kg \ + (\frac{5680 kg*m^{2}}{3.91m^{2}}))*(3.09 \frac{m}{sec})*(3.91m) = ((5680 kg*m^{2}) + (199 kg * ((3.91m)^{2}))) \omega \ ; solving \ for \ \omega[/tex]

    [tex]\omega \ = \ 2.87[/tex]

    [tex]method \ 2 \ \Rightarrow ((5680 kg*m^{2}) * (\frac{3.09 \frac{m}{sec}}{3.91 m} * 2 \pi)) + (199 kg \ + (\frac{5680 kg*m^{2}}{(3.91m)^{2}})*(3.09 \frac{m}{sec})*(3.91m) = (5680 kg*m^{2} + (199 kg * (3.91m^{2})) \omega^{prime} \ ;[/tex]
    [tex]solving \ for \ \omega^{prime}[/tex]

    [tex]\omega^{prime} \ = \ 5.52[/tex]
     
  2. jcsd
  3. Apr 13, 2008 #2

    Doc Al

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    Staff: Mentor

    This is a conservation of angular momentum problem, so method #1 makes sense. (You could use method #2 by setting [itex]\omega = 0[/itex]--since it starts at rest, but then you're back to method #1.) But you didn't plug in the numbers properly. Hint: What's "m" stand for?
     
  4. Apr 13, 2008 #3
    Am I to assume that 'm' is total mass of the system (merry-go-round + child) or merry-go-round for interia of the merry-go-round?
     
  5. Apr 13, 2008 #4

    Doc Al

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    Staff: Mentor

    That would be incorrect. What's the total angular momentum of the system initially?
     
  6. Apr 13, 2008 #5
    [tex]L = r *p[/tex]

    [tex]L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child[/tex]
     
  7. Apr 13, 2008 #6

    Doc Al

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    This is correct.

    But this isn't quite right. What's 371.53 kg the mass of? What does "p" stand for?

    Hint: What's the only thing moving initially?
     
  8. Apr 13, 2008 #7
    I thought 'p' was for momentum of the merry-go-round. The only object moving initially is the child. If I apply the child attached to the merry-go-round initially with the starting speed, if I understand this correctly, would be

    [tex]L = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}[/tex]
     
    Last edited: Apr 13, 2008
  9. Apr 13, 2008 #8

    Doc Al

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    You forgot the speed of the child. (Yes, "p" stands for momentum, not just mass.)
     
  10. Apr 13, 2008 #9
    I had a lapse of thought there. So I shouldn't include the mass of the merry-go-round?
     
  11. Apr 13, 2008 #10

    Doc Al

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    Staff: Mentor

    Until the kid jumps on it, the merry-go-round isn't moving and thus has no angular momentum. Only the kid has angular momentum (with respect to the center of the merry-go-round).
     
  12. Apr 13, 2008 #11
    So if I revisit method one...

    [tex]L_{child} = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec}[/tex]

    [tex]L_{merry-go-round} = (3.91 m) * (371.531 kg) * (0 \frac{m}{sec}) = 0 \frac{kg*m^{2}}{sec}[/tex]

    [tex]I_{child} = (199 kg) * (3.91 m)^{2} = 3042.33 kg*m^{2}[/tex]

    [tex]method \ 1 \ \Rightarrow 2404.3 \frac{kg*m^{2}}{sec} = (5680 kg*m^{2} + 3042.33kg*m^{2} ) \omega \ ; solving \ for \ \omega \ = 0.2756 \frac{rad}{sec} = \ 0.276 \frac{rad}{sec}[/tex]

    Thanks!
     
  13. Apr 13, 2008 #12

    Doc Al

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    Staff: Mentor

    Looks good!
     
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