Inertia question

  1. 1. The problem statement, all variables and given/known data
    Which is harder to rotate?
    A rod with a mass (m1) in the center, or a rod with two masses (m2)at the ends.
    Assume mass m1 = 2*m2, so the total masses of rod+masses are equal.
    I am rotating the masses at the center.

    My professor said that the one with the mass in the center is harder to rotate, but I dont understand how he explained it. I actually thought the one with the mass in the center was easier to rotate.

    Can someone explain why the one with the mass in the center is harder to rotate?
    Is it because if we were to rotate each to the same angular velocity, it would take more work to get the first rod to that angular velocity?

    Relevant equations.
    I think I'd use Kr = I [tex]\omega^{2}[/tex]
    and if they go up to the same angular velocities, it requires more work?
    It doesn't seem right, because if the I of the first rod is lower, so the Kr would be lower, so that means the work required to get it to that speed was lower?
    I don't see why the one with the mass in the center is easier to rotate.
     
  2. jcsd
  3. mjsd

    mjsd 860
    Homework Helper

    assuming that you are rotating about an axis that goes thru the center then it would harder to rotate the m2 case because it has larger moment of inertia. unless i have misunderstood the setup.
    torque = moment of inertia x angular acceleration
    you may have mis-heard your prof.. or there are something else in the system?
     
  4. Shooting Star

    Shooting Star 1,979
    Homework Helper

    It wouldn't take any energy to rotate the rod with m1 in the centre, because the velo of the point mass at the centre would be zero.

    Perhaps you misunderstood your professor, as mjsd says?
     
  5. Yeah, I thought what u said was the answer too, I'll have to ask my professor again, maybe i switched up what he said.

    Thanks for helping
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook