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Rotational kinetic energy / moment of inertia

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A thin, cylindrical rod = 27.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.

    kd71qs.gif

    After the combination rotates through 90 degrees, what is its rotational kinetic energy?

    What is the angular speed of the rod and ball?

    What is the linear speed of the center of mass of the ball?

    How does it compare with the speed had the ball fallen freely through the same distance of 32.0 cm?

    2. Relevant equations

    KE = 1/2 I ω^2

    v = ωr



    3. The attempt at a solution

    I tried using conservation of energy for the first part but i don't see how if i get mgh = 1/2 I ω^2...where the height is unknown and also i don't understand how to even get I to be quite honest.
     
  2. jcsd
  3. Oct 28, 2012 #2

    Ibix

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    Science Advisor

    You can look up the moment of inertia of a rod and a sphere about their respective centers, or derive from first principles. You then need to work out their moments of inertia about an axis through the hinge, not through he center of mass (hint: that axis is parallel to the axis through the center of mass). Then note that the moment of inertia about some axis of a rigid system made of several parts is the sum of the moments of inertia of the component parts about that axis.

    Your conservation of energy approach is correct. The height is the change in altitude of the center of mass. This is so because the center of mass is the mean position of the mass in the object. Its change in height, therefore, is the mean change in height of all the mass in the object.

    Does that help?
     
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