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I Centripetal Force for non point particle

  1. Jul 28, 2016 #1
    Say there is a point particle attached to a rope of radius r. It spins with angular velocity w. As a result, the force needed to keep this particle in circular motion is:

    F = mv2/r

    However, now say I have some irregular shaped object, with say some constant surface density for simplicity. It is also attached to a rope of radius r and spins with angular velocity w.

    My question, is the force needed for the rope to exert at the base of the object the area integral of dm*v2/r = dm w2*r over the object? If this is true, what is the intuition? Why can I sum up the forces needed for all the constituent particles and say that is the force applied by the rope at the base of the object?

    Another thought in my head was that you can think of translational forces as being applied directly to the center of mass of the object. Does the math somehow work out that the force necessary is F = mv2/r, where m is the total mass, and r is the center of mass?

    A third possible answer I assume, would be none of the above and would stem from an actual derivation of centripetal force for the general case, and not just for a point particle. I suspect it may be this, since I've never actually derived the force for a point particle personally.
     
  2. jcsd
  3. Jul 28, 2016 #2

    jbriggs444

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    Almost. Remember that force is a vector. The centripetal force for a piece on the leading edge of the object and the centripetal force for a piece on the trailing edge of the object are not parallel. The magnitude of the total force will be less than the sum of the magnitudes of the incremental forces.

    Momentum and force are additive quantities. The rate of change of momentum of the object is the sum of the rates of change of momentum of its pieces. The total force required to produce this total rate of change follows from that.
     
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