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Final angular velocity of a merry-go-round?

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A child (mc = 36 kg) is playing on a merry-go-round (mm = 225 kg, R = 2.9m) that is initially at rest. The child then jumps off in a direction tangent to the edge of the merry-go-round. The child has a speed of 5.0 m/s just before she lands on the ground. What is the magnitude of the final angular velocity of the merry-go-round?

    2. Relevant equations
    L = I*ω
    Li,c + Li,m = Lf,c + Lf,m
    I = (1/2)mmR2 for the merry-go-round.
    I = mcR2 for the child.

    3. The attempt at a solution
    The initial angular momentum of the system is 0 so Li,c + Li,m = 0.
    So Lf,c + Lf,m = 0
    (1/2)mmR2 * ωf + mcR2 * ωf = 0
    (1/2)mmR2 * ωf = - mcR2 * ωf

    However when I plug the values for mass of the merry-go-round and the child as well as the radius for the merry-go-round I cannot seem to find a way to isolate ωf to one side of the equation, I just keep getting 0 rad/s.

     
  2. jcsd
  3. May 6, 2015 #2

    jbriggs444

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    The angular velocity of the child is not very well specified since the child is not in circular motion around the selected axis of rotation. But that turns out not to matter. The angular momentum of the child can be calculated more directly. Angular momentum for a non-rotating body moving in a straight line is the product of momentum times the perpendicular offset from the selected axis of rotation.
     
  4. May 6, 2015 #3
    So I can find the final angular momentum for the child, and then plug it into the equation for the conservation of momentum to solve for the final angular velocity of the merry-go-round? So the momentum for him would be mass*velocity (momentum) and then I'm not sure what you mean by the perpendicular offset for the selected axis of rotation. The axis of rotation for the merry-go-round goes straight through the middle, and the child jumps off the edge in a direction tangent to that edge.
     
  5. May 7, 2015 #4

    PeroK

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    You could use conservation of angular momentum, but you can also think of the child applying an impulse (torque) when it jumps off. As the child is not rotating or in a circular orbit, I think it's confusing to consider its angular velocity. There is, of course, a more fundamental definition/equation for angular momentum that it would be better to use.

    To understand the angular momentum in this case, you could consider the following:

    A particle moves at a constant velocity ##v## from the point (0, 1) along the line ##y=1##. Calculate its angular momentum about the origin and check that this is conserved as it follows its linear path.
     
  6. May 7, 2015 #5

    jbriggs444

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    The "perpendicular offset" is the distance from the child's path to the center of rotation. More precisely, you would look at the center of mass of the child and its distance from the center of rotation. Consider this distance as having two components, one in the direction of the child's momentum and one perpendicular to the child's momentum. The component that matters is the perpendicular one.

    Or take the approach that PeroK has suggested. You can calculate the child's momentum. You know what linear impulse had to be applied to produce that momentum. You know the distance from the axis of rotation where that impulse was applied. So you can compute the associated change in angular momentum.

    It ends up being the same calculation with either approach.
     
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