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Angular momentum merry go round problem

  1. Dec 1, 2013 #1
    Fred, mass 36 kg, stands at the center of a merry-go-round that has mass 50.0 kg and radius 2.0m and that rotates one full revolution every 4.5 seconds. Treat the merry go round as a solid disc, and Fred as a point object. As it turns, he walks to the edge of the merry-go-round, then jumps off and runs (in a straight line, tangent to the edge of the merry-go-round) with a velocity of 1.5 m/s.
    (a) What is the angular momentum of the merry-go-round and Fred, together, when he is at the center?
    (b) What is the angular velocity (in radians per second) of the merry-go-round when Fred reaches its edge?
    (c) What is the angular velocity of the merry-go-round after he jumps off?
    (d) Did the kinetic energy of the merry-go-round increase, decrease, or remain the same when he jumped off? If it changed, by how much, and where did the energy go or come from?
    (e) Did the total kinetic energy of the merry-go-round and Fred combined round increase, decrease, or remain the same when he jumped off? If it changed, by how much, and where did the energy go or come from?



    I know I changes from 1/MR^2 =100 to 1/2MR^2 +MR^2 =244 and angular momentum is conserved so for part a I got 0.572rad/s
    For b I don't understand how the principle of momentum applies when fred is no longer apart of the merry go round system- will he be pushing off and thus exerting a torque causing the merry go round to slow? Or because his angular moment when running will be in the opposite direction to the angular momentum of the roundabout the roundabout will speed up?

    Thanks
     
  2. jcsd
  3. Dec 1, 2013 #2

    Andrew Mason

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    ?? I think you meant I = MR^2/2 = 100. Your answer is correct

    You are to assume that he does not push off in any horizontal direction. If he just jumps straight up he will leave the merry-go-round. Assuming he does that, what is his horizontal velocity when he jumps up and leaves the merry-go-round? When he does that, is there any horizontal force acting on him?

    AM
     
  4. Dec 1, 2013 #3
    Ok thanks. So his horizontal velocity will be his angular velocity times the radius? but this would only be when he is is the air then he runs at 1.5m/s. I don't understand how him running will affect the angular velocity of the merry go round, or does it not affect it?
     
  5. Dec 1, 2013 #4

    CWatters

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    I wondered why they give his velocity? My reading of the problem is that he jumps off intending to land on the ground at a velocity of 1.5m/s (wrt the ground). If that's faster than the roundabout is going he would have to push off in a direction that slows the roundabout. If it's slower then the roundabout then he pushes off in the opposite direction speeding up the roundabout.
     
  6. Dec 1, 2013 #5
    sorry I still dont get it. so is the angular momentum of just the merry go round conserved or the merry go round and fred running?
     
  7. Dec 1, 2013 #6

    haruspex

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    That's how I read it too.
    It's not so much Fred running as Fred the instant he's lost contact with the merry-go-round. As CWatters says, you should take it that he does so with a tangential velocity of 1.5m/s. In the act of jumping, there may therefore be a tangential force between Fred and the merry-go-round. So for conservation of angular momentum through this event you need to take the whole system.
    What is Fred's angular momentum about the centre of the merry-go-round after he has jumped?
     
  8. Dec 1, 2013 #7
    ok i think I get it so when he has jumped his angular momentum will be mvr=36x1.5x2. and this plus the angular momentum of the merry go round will equal the initial angular momentum of the merry go round and fred before he jumps? for an answer I got 0.3157rad/s so the merry go round slows down?
     
  9. Dec 1, 2013 #8

    haruspex

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    Yes.
     
  10. Dec 2, 2013 #9

    Andrew Mason

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    The problem is not clear whether he leaves the merry-go-round with a velocity of 1.5 m/sec or 1.14 m/sec., but it does say he runs in a straight line tangential to the merry-go-round from the point he jumps off. It says that he runs at 1.5 m/sec. It does not say he jumps off at that speed.

    It would be difficult jump off the merry-go-round travelling at 1.5 m/sec in a line tangential to the merry-go-round because the acceleration is not instantaneous. While he is jumping his velocity increasing and is also changing direction, so the jump would have to be carefully done so that the acceleration of the jump plus the centripetal acceleration resulted in acceleration that was always in a direction tangential to the merry-go-round at the point he leaves it.

    AM
     
  11. Dec 2, 2013 #10

    gneill

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    Even if we assume that Fred leaves the merry-go-round traveling at 1.5 m/s tangentially and that he somehow accomplishes this feat with an instantaneous leap, the problem does not state whether Fred jumps with or against the motion of rim... tangent lines have two "sides".

    Assume that Fred is very clever. Or very lucky! I suspect that Fred's middle name is "Ideal" :smile:
     
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