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Force on Center of Mass

  1. Feb 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A heavy ring of mass m is clamped on the periphery of a light circular disc. A small particle of equal mass is placed at the center of the disc. The system is rotated in such a way that the center moves in a circle of radius r and with a uniform speed v. Find the external force on the system.

    2. Relevant equations
    • Finding the Center of Mass of the system:
    (Considering the center of the plate to be the Reference Point.)

    mr/(m+m) => r/2

    Therefore the center of mass lies at r/2 from the center

    • Finding force on the system:

    Centripetal Force=mv2/r

    Putting r=r/2 (i.e.the center of mass), F=2mv2/r

    3. The attempt at a solution

    It is known to us the position of the center of mass of the system. Therefore the force on the system must be equivalent that on the center of mass. Using this, i formulated the possible expression for the external force on the system. But I am still confused whether my approach is correct.

    Any feedback is deeply valued and appreciated
     
  2. jcsd
  3. Feb 3, 2016 #2

    PeroK

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    How did you get the centre of mass to be r/2? If r is large, then the center of mass would be outside the disk!
     
  4. Feb 3, 2016 #3
    The disk is massless. There are only two masses in the system, the ring and the mass at the center and both are of equal mass. Separation between them is r. So the centre of mass has to be at r/2.
    It must not matter how large r is.
     
  5. Feb 3, 2016 #4

    PeroK

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    r is the radius of the motion. You aren't given the radius of the disk. Where is the centre of mass of a ring?
     
  6. Feb 3, 2016 #5
    Right. I got the catch. I was misinterpreting r to be the radius of the disk.
    On second thought I realise that the center of mass of disk has to be its geometrical center. So the COM of the system is at the center, where the small particle is kept
     
  7. Feb 3, 2016 #6

    PeroK

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    That's right.
     
  8. Feb 3, 2016 #7
    Okay, now the centripetal force can be found.
    Thanks PeroK for your help!
     
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