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Crazy dude jumping on a Merry-Go-Round

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data

    A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.686 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

    (a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round?
    increase
    stay the same
    decrease

    (b) Calculate the initial and final kinetic energies for this system.
    Ki = kJ
    Kf = kJ



    2. Relevant equations
    KErot= 1/2*Iw^2
    I=1/2mr^2


    3. The attempt at a solution
    a) I would think it should increase but im probably wrong, lol.

    b) The merry-go-round initially is just spinning so I trued finding the KE using the 1/2*I*w^2 formula so I got 1/4*m*r^2*w^2and I came up with 4.98 kJ and thats the wrong answer... is there something I did wrong?

    I don't have any idea how to find the KEf. Can I assume that its the rotational KE of the merry-goround plus the KE of the guy jumping on it?
     
  2. jcsd
  3. Mar 19, 2007 #2
    Lets start with the initial KE. You have a rotating disc included in your system; is there anything else that has KE?
     
  4. Mar 19, 2007 #3

    Astronuc

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    Staff: Mentor

    Try comparing the rim's tangential speed (where the person jumps on) with the speed of the individual, and keep in mind, when the person jumps on, they must have the 'same' tangential speed.

    If the person is traveling faster, the person slows down, but the merry-go-round must increase its rev/s.
     
  5. Mar 19, 2007 #4
    No the disk is the only thing that has KE? Maybe there is something I am just not understanding about this disk.
     
  6. Mar 19, 2007 #5
    Sorry, maybe I do not understand what your system is?
    The person is running.
     
  7. Mar 19, 2007 #6
    the edge of the merry go round is moving at 11.336 m/s much fast then the man. So the merry-go-round slows down. I did conservation of momentum.
    Initally...
    Lmgr= I*w= 2310.58
    Lp=I*(v/r)=623.326
    Ltot= 2933.9

    using this I found the angular speed of the merry-go-round fonal
    2933.9/(Imgr + Ip)=w=3.09832 rad/sec

    But I can not see how this helps me find the initial or final conditions.
     
  8. Mar 19, 2007 #7
    So I must include the guy running although hes not on the merry-go-round yet?
     
  9. Mar 19, 2007 #8
    So I did KEi= .5*Imgr*Wmgr^2 + .5*Ip*Wp^2 and got 5.4524 kJ
    then i did KEf= .5*(M1+M2)*r^2*.5*Wf^2 and got 3.559kJ

    Am I even close?
     
  10. Mar 20, 2007 #9
    I guess Im not because thats still the wrong answer
     
  11. Mar 20, 2007 #10

    Mentz114

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    Gold Member

    Robb has given the answer.

    The initial KE = KE ( disc) + KE (man)

    Final KE = initial KE = KE(disc)

    Remember that when the man steps on the rim, he increases the MoI of the roundabout by m*r^2.
     
    Last edited: Mar 20, 2007
  12. Mar 20, 2007 #11
    True Mentz144 but the Final KE does not equal the Initial KE. I finally figured it out. my KEi was right but I added masses when finding the new MoI not m*r^2 to the MoI of the disk.
     
  13. Mar 20, 2007 #12

    Mentz114

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    Gold Member

    Phyz, The question asks is energy being lost or gained by the >>system<< during the process. Taking the man and turntable together as the system, I venture that the energy is the same before and after.
     
  14. Mar 20, 2007 #13
    Energy is being lost because energy is required to make the mans speed the same speed as the merry-go-round. But thank you for your input.
     
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