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Rotational Inertia

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    a. A merry go round is rotating in the counter-clockwise direction. Initially, a 50 kg child is sitting on the edge of the merry-go-around, which rotates at 0.5 radians per second. The moment of inertia of the merry-go-round is 2150 kg-m2. The radius is 2.50 meters. The child can be treated as a point mass. What is the total angular momentum?

    b. If the child crawls to the center of the merry-go-round, will the speed at which it rotates change? If so, what is the new rotation speed?


    2. Relevant equations
    [tex]\vec{L}[/tex] = I[tex]\vec{w}[/tex]
    I = m1r12 + m2r22

    3. The attempt at a solution

    The moment of inertia of the merry-go-round is 2150 kg-m2. We need to add to that value the moment of inertia of the child.

    Ichild = (2.50 m) * (50 kg) = 125 kg-m2
    Itotal = 2150 + 125 = 2275 kg-m2

    [tex]\vec{L}[/tex] = I[tex]\vec{w}[/tex] = (2275)(0.5) = 1137.5
     
  2. jcsd
  3. May 19, 2009 #2

    rock.freak667

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    Homework Helper

    If the child can be considered a point mass, then the child's moment of inertia is given by mr2. Recalculate this, then the total moment of inertia and then the total angular momentum.

    For part b, if the child moves closer to the centre of rotation, what property of the child is changing?
     
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