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Inertia, Momentum, Rotation

  1. Nov 7, 2006 #1
    I am stuck on the first part of this question. I thought that I could add the Inertias for child 1 + child 2 + board and that would give me the answer. I used I=mr^2 for the children with the r being L/2 and 1/12*(mass)*(width^2 + length^2). as the I for the board. However I am not getting the right answer. Also I am unsure of how to calculate the angular momentum of the system Any help would be appreciated

    Two children, each with mass m = 10.5 kg, sit on opposite ends of a narrow board with length L = 4.8 m, width W = 0.18 m, and mass M = 6.2 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?

    What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.28 rad/s?

    While the system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed?
     
  2. jcsd
  3. Nov 7, 2006 #2

    radou

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    Well, for the first part, simply use Steiner's theorem on the system consisting of the board and the childern. You must add the mass of each child separately multiplied with its squared distance from the axis of rotation to the moment of inertia of the board, which you can find here: http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html.

    For the second part of the problem, with the children pulling themselves towards the center, here's a hint: there is no external torque - so what is conserved?
     
  4. Nov 7, 2006 #3

    OlderDan

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    You have the right approach for I. If you do not post your answer, there is no way we can check to see if you did the calculation correctly.
     
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