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Angular change in velocity (momentum)

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A merry-go-round has a radius of 3.0 m and a rotational inertia of 600 kg m2. The merry-go-round is initially at rest. A 20 kg child is running at 5.0 m/s along a line tangent to the rim. Find the angular velocity of the merry-go-round after the child jumps on.

    a. 0.38 rad/s
    b. 0.45 rad/s
    c. 0.71 rad/s
    d. 0.56 rad/s
    e. 1.2 rad/s


    2. Relevant equations

    p(angular)=Iω
    p=mv
    p(angular)=p*r


    3. The attempt at a solution

    i tried to use conservation of momentum by saying:

    r*mv+0=(I+mr)ω

    and i got ω=0.45 but the answer key says the answer is a. 0.38 rad/s
     
  2. jcsd
  3. Oct 28, 2011 #2

    Doc Al

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    Staff: Mentor

    Right idea, but you used the wrong formula for the rotational inertia of the child. What's the rotational inertia of a point mass (we can treat the child as a point mass) that is some distance from the axis?

    (Always check units. What are the units of I? Does mr have those units?)
     
  4. Oct 28, 2011 #3
    Ohh so it will be mr^2 right? :)
    How is it derived anyways?
     
  5. Oct 28, 2011 #4
  6. Oct 28, 2011 #5

    Doc Al

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    Staff: Mentor

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