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Rotational Inertia of a Ball-Rod system

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A cylindrical rod 32.2 cm long has mass
    0.919 kg and radius 1.54 cm. A 8.89 kg
    ball of diameter 11.3 cm is attached to one
    end. The arrangement is originally vertical
    with the ball at the top and is free to pivot
    about the other end.
    The acceleration of gravity is 9.8 m/s2 .
    After the ball-rod system falls a quarter
    turn, what is its rotational kinetic energy?
    Answer in units of J.
    At the same point, what is its angular speed?


    2. Relevant equations
    K=.5Iw^2


    3. The attempt at a solution
    I=1/3mL^2+2/5mR^2


    Alright, I'm on the second part of the problem, having gotten the energy part. I cannot, however, for the stupid life of me, define this Ball-Rod system's moment of inertia in a fashion that gets me a reasonable answer. I'm wondering if I'm just missing something. If somebody could help me see it I can get the rest no problem.

    I also set it up in a way that it was just the rod with a big mass on the end of it, and that failed miserably. And I searched the forum for a similar problem, but the guy doing it had already found moment of inertia and didn't want to share.
     
    Last edited: Nov 14, 2008
  2. jcsd
  3. Nov 14, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    Isn't the sphere a point mass at the end of the rotating rod? It is not actually rotating about the center of the sphere is it?

    Won't its distance be at 43.5 cm?
     
  4. Nov 17, 2008 #3
    Sorry, but I got it. I believe the solution I used was 1/3mL^2+2/5mR^2 + MD where M is mass of the sphere and D is the distance from it to the center of the sphere. It does seem counter intuitive, but the way the system's set up is that it APPEARS the sphere is rotating as its moving with the rod. If you hold a ball and put a big X on the side facing you, and then move it in a way that its face never changes direction, when it's to another side, it has 'rotated' away from you.

    The concept's very abstract and I don't even see the validity in it, but the equation worked on a hunch. Blarg.
     
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