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Two children stand on the edge of a merry-go-round

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Two children stand on the edge of a merry-go-round while their friend accelerates them to an angular speed of 3 revolutions per second. The merry-go-round has a radius 10 m and a mass 100 kg and both kids weigh 35 kg. Once they reach this speed one of the kids jumps off.
    What is the angular speed of the merry-go-round right after the child jumps for his life?


    2. Relevant equations


    By conservation of angular momentum

    3. The attempt at a solution

    Moment of Inertia before the Jump = (35 * 10^2 ) * 2 = 70 * 10^2 kg-m^2
    Moment of inertia after one of them jumps = 35*10^2 kg-m^2


    70 * 10^2 * 3 = ωf * 35 * 10^2
    => ωf = 2*3 = 6 rad/sec
     
  2. jcsd
  3. Dec 19, 2011 #2

    Simon Bridge

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    Re: Merry-go-round

    Doesn't the angular momentum of the merry-go-round play a part here?
    What is the relationship between angular speed and revolutions per second?
    Did you have a specific question of your own?

    Tip: when you relate your working, show your reasoning as well.
    It can be hard to follow just a bunch of numbers like that.
    Try manipulating the equations first and showing that.
     
  4. Dec 19, 2011 #3
    Re: Merry-go-round

    I believe the angular momentum should play a part, yes.
    w=2*pi*f
    I'm just hoping someone spots an error in my work b/c I don't believe I'm right.
     
  5. Dec 19, 2011 #4

    Simon Bridge

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    Re: Merry-go-round

    Relate the answers you just gave me to your attempt at a solution.

    When doing conservation problems - start with the totals before and after.

    eg.
    Before:

    [itex]L_B = 2L_{child}+L_{mgr}[/itex]

    you just did the two kids.

    Now try it for the "after" situation.
     
  6. Dec 19, 2011 #5
    Re: Merry-go-round

    I'm lost, L=Iw so I=1/2MR^2?
     
  7. Dec 19, 2011 #6

    PeterO

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    Re: Merry-go-round

    The answer depends on the way the child "jumps off" the merry-go-round.
    Which direction does the child push on the merry-go-round as they jumped off.

    The child may have pushed directly towards the centre, and thus made no difference to the speed of the ride as they left. This method will mean that the child will reach the ground traveling at the speed they were moving at before they left the ride. That might make it difficult to not fall over when you land on the ground.
    The child may have "pushed" the merry-go-round forward, so that the child's momentum is transferred to the ride. That method has the advantage of the child landing on the ground with little or no momentum, so not tending to fall over.
    [This method of jumping off is frequently used by skateboard riders when they decide to stop]
    The child might actually push back back on the merry-go-round causing the ride to slow slightly - but meaning the child will hit the ground travelling even faster that they wer moving before leaving the ride.

    Given that the outer edge of the merry-go-round is travelling at almost 21 m/s [over 70 km/hr or about 40 mph - the speed limit for cars in the suburbs is way less than that] I am not sure that the child would fancy "jumping off" at all!! A child of mass 35 kg I can understand - a merry-go-round which is 20m in diameter and only has a mass of 100 kg is an interesting concept!!

    To give you a straight line example of what I meant above:
    Suppose a 60kg boy is riding on a 5 kg skateboard, while carrying a 10kg Shotput.

    What happens to the speed of the boy and board if he:
    (a) drops the shotput
    (b) throws the shotput forward, in the direction they are already traveling.
    (c) throws the shotput backwards, in the direction opposed to the direction they are already travelling.
     
  8. Dec 19, 2011 #7

    Simon Bridge

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    Re: Merry-go-round

    Yes - it's a ******** question.
    They are normally more carefully phrased.
    I think we understand the intent of the question - if Cisneros778 want's to go for bonus marks, then critiquing the question after answering it would be one way to go.

    @Cisneros778: don't worry about what each of the L's are just yet, just write them out.
    This makes sure that you have included all the angular momenta you need.

    Take it one step at a time.
    1st step: [itex]L_A = \ldots[/itex]
     
  9. Dec 19, 2011 #8

    PeterO

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    Re: Merry-go-round

    The difficulty is that I believe the intent of the question is that when one child jumps off - straight out from the merry-go-round - then that child takes with him/her, all that part of the angular momentum they had while in contact with the ride, and thus the ride neither slows down nor speeds up.

    There was a similar question on this forum recently about a person sitting on a rotating seat, with two masses held in hands out-stetched sideways, with the question being "what happens when the person releases the masses". Of course the person and seat do not speed up, nor slow down in that case.
     
  10. Dec 20, 2011 #9

    Simon Bridge

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    Re: Merry-go-round

    Hmm-hmmm ... so the [itex]L_A[/itex] will still have a contribution due to the child that drops off. Were we disagreeing? [rethinking: ah - I misunderstood your motivation for that discussion on how the child jumps off - apologies.]

    I suppose now I think about it there are problems about here that have a mass dropped onto a turntable and the person who set the problem may think they are setting the reverse of this process (just removing a mass). But if we go around wondering if the problem-setter has made a mistake right at the start we won't get anywhere.

    BTW: way to give away the answer :)
     
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