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C.O.M. issues?

  1. Aug 7, 2005 #1

    I am rather perplexed by the solution to the following problem.

    Particles of mass M1 and M3 are fastened to the ends of a light rod having a length l. A bead of mass M2 is free to slide along the rod between M1 and M3. Point p is the center of mass of M1 and M3, not including M2.

    I is the moment of inertia of the M1, M3 rod arrangement about an axis perpendicular to the rod and passing through p . All motion is considered in a plane. :yuck:

    Now the translational aspect of the KE is very simple. But it was the rotational part that had me confused.

    In my humble opinion we cannot use I and p coz as we add M2 the center of mass changes.

    However in the answer for the rot. KE they have

    [tex]\frac{1}{2}[/tex] I @theta/@dt + [tex]\frac{1}{2}[/tex] M2[...............................................]

    The thing that bothers me is that surely with the addition of M2, the value of p and hence I would change. So how can they treat the other masses independently from the third?????


    I hope I have been fairly clear and am not being too silly.

    Thanks guys!
  2. jcsd
  3. Aug 8, 2005 #2
    hmmm, I see this has been moved.

    In fairness to me I should point out that this is not my homework and I am not at college.

    The problem comes from a textbook on Lagrangian Dynamics I thought I would work through for interest.

    :redface: :redface: :redface: :redface: :redface: :redface:
  4. Aug 8, 2005 #3
    Adding M2 does change the total CoM, but you should recall the superposition principle. The center of mass of the whole system relies on where M2 is placed, but 'p' is defined to be the CM of M1 and M3, which are not moving. The axis of rotation is through this point and is also fixed. Adding M2 to a point on the beam does change the total center of mass, but it does not affect 'p' by definition.
  5. Aug 8, 2005 #4


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    Yes you are right. They can't use same I and P if they consider all the three masses as a single system, but I think they are calculating the rotational kinetic energy of M1, M3 system and M2 separately and then adding them, and therefore no question for I of the system with M1,M2 and M3.
  6. Aug 8, 2005 #5
    Thanks Whozum,

    Seeing it put in other words makes things a lot clearer!

    I did not realise that the axis was a constraint on the system.

    cheers to u too Mukundpa,

    Last edited: Aug 8, 2005
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