1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Rotational Kinetic ENERGY! (well laid out)