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Merry-go-round angular speed

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data


    A merry-go-round rotates at the rate of
    0.12 rev/s with an 86 kg man standing at
    a point 1.4 m from the axis of rotation.
    What is the new angular speed when the
    man walks to a point 0 m from the center?
    Consider the merry-go-round is a solid 45 kg
    cylinder of radius of 1.4 m.
    Answer in units of rad/s

    theres a follow up question:
    What is the change in kinetic energy due to
    this movement?
    Answer in units of J

    but don't really worry about that right now.
    2. Relevant equations
    v = rw ?


    3. The attempt at a solution
    Is it just me being dumb or is there a lot of excess information/ kind of a trick question? The angular speed of the merry go round would still be 0.12 rev/sec (need to convert to rad). just the linear velocity changes with the radius.... I think? maybe some of that other information has to do with the change in energy question?

    I just thought about the follow up question for a minute, maybe I know more than I think, but I'm not confident/have been getting these wrong.
    >would you use V= rw to get the linear velocity and then plug that into 1/2mv^2 , and when the radius is 0 the energy goes to 0 since v=rw?
     
  2. jcsd
  3. Nov 18, 2012 #2
    Angular momentum is conserved.

    As the man walks to the centre the total moment of inertia about the centre is changing. Hence, by the principle of conservation of angular momentum, the angular velocity will also change.
     
  4. Nov 18, 2012 #3
    so if L= Iw and I = 1/2 MR^2
    then you get L = (0.5) (131) (1.4) (0.7539822).
    but wouldn't w need to go to infinite if R is 0?
     
  5. Nov 18, 2012 #4
    Note that the system consists of the merry-g-round wheel TOGETHER with the man!

    Hence altough the moment of inertia of the wheel remains the same value , yet the moment of inertia of the man is changing.

    Now I am at a loss what to do!

    Because if the man were to be considered as a particle (!) his new moment of inertia about the centre of the wheel would be 0. But can one consider the man as a particle?
     
  6. Nov 18, 2012 #5

    rcgldr

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    Homework Helper

    The total angular momentum is the sum of the angular momentum of the wheel and the man. If the man walks to the center of the wheel and can be considered a particle (or a point mass), then all of the angular momentum is in the wheel. The total angular momentum continues to remain constant.
     
    Last edited: Nov 18, 2012
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