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Merry-go-round problem

  1. Jun 19, 2005 #1
    A 5.4m diameter merry-go-round is rotating freely with an angular velocity of 0.730 rad/s. Its total moment of inertia is 1665kg*m2. Four people standing on the ground, each of mass 60.8kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now? Use units of "rad/s".

    r = 2.7 m
    w= .730 rad/s
    I= 1665 kg*m2

    I'm not sure if i'm doing this right but what i did was i set the inertia equal to mk^2. By doing this i was able to find the initial mass to be 228.395 kg. Then i added the weight of the four people to this mass and got 471.595 kg. Then I solved for that moment of inertia to be (471.595)(2.7)^2 which ended up being 3437.93kg*m2. Now i'm not sure how to find the angular velocity...I tried to just set the values equal to one another, hoping the proportion would produce the right value but that didn't work. I have no idea...
  2. jcsd
  3. Jun 19, 2005 #2


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    You do not need to know the total mass. What you need to find is the moment of inertia after the people step on. Moment of inertia is a sum of bits of mass times the distance from the axis squared for each of those bits of mass. Each of the four people add to the moment of inertia of the merry-go-round with their mass times their distance from the axis, which is the radius. What you did worked out correctly because you assumed all the mass of the merry-go-round was at the circumference to calculate its mass. That need not be true. You should have just added 4*60.8kg*radius^2 to the given moment of inertia. The result would have been the same.

    This problem is about conservation of angular mometum. None of the people have any until they get on the merry-go-round. How is angular momentum related to I and w? If you increase I, what must happen to w?
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