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Homework Help: Rotational Kinetic ENERGY! (well laid out)

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    c) Three point masses lie on a rigid, rod of no mass and L = 7.71 m :

    - Two particles,with the same mass m = 3.27 kg, lie on opposite ends.
    - Mass M = 3.24 kg is in the center of the rod.

    the rod lies along the x-axis, and rotates about the y-axis. about a point 1.68 m from one end at constant angular speed ω = 5.82 rad/s.

    Find the kinetic energy of this system:

    2. Relevant equations

    3. The attempt at a solution
    1. First I calculated the center of mass which i found to be at 3.864 m from one end point.
    2. I calculated the KE of rotation from the axis running through the cm.
    3. Then i used the parallel axis theorem. With d being... the distance of the rotation point form rod+the distance of the rod left from the cm
    4. now with d i used the parallel axis theorem and got the wrong answer

    Is my reasoning flawed?
    Thank you for your help

    Attached Files:

  2. jcsd
  3. Nov 2, 2008 #2
    The center of mass is right in the middle (7.71/2=3.855 from either end).
    You don't need the moment of inertia.
    You have point masses at different distances from the axis.
    KE=1/2 m*omega^2*r^2 for each mass.
    r1=L-1.68m, r2=L/2+1.68m, r3=1.68m
  4. Nov 2, 2008 #3
    Isnt the center of mass Dependant of all 3 point masses?
    so wont it be a little off the center since the middle mass is a little lighter?

    and can you also explain how you got the three r's
    Thank you very much
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