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Interesting problem - solution

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  • Interesting problem - solution

    Thanks to all those who responded to my little problem. An
    interesting point is that not everyone got the right answer!! but I am not going
    to publish who did and who did not. Here is my interpretation.

    (a) The answer to the first part of the question is quite straight
    forward. Since angular momentum remains constant (no external
    torques),then the ratio of the final and initial kinetic energies is
    given by:

    I1 w1 = I2 w2

    1.5 I1 = 5 I2 or I1 = 3.33 I2

    (1/2 I1 w1^2) / (1/2 I2 w2^2) (ratio of initial to final KE)

    after substitution the ratio is 1:3.33

    (b) To increase the energy of the system, work must be done. The only
    force that you can apply is the centripetal one and you can clearly
    see that it must do positive work since the masses move inward.
    Here is my attempt at the explanation. I hope that you can all
    follow. If not, I will send you a hard copy done in an equation editor.

    Imagine that you model the system as a point mass located at a
    distance r from the centre axle of the turntable (this distance could
    actually be the radius of gyration). If you let the velocity of this mass
    be v and the centripetal force equal F the the work done by
    this force is,

    Fs = (mv^2)/r dr

    By using the work-energy principle, we can say that the change in
    energy is equal to the work done. That is

    Fs = (mv^2)/r dr = dE = d(1/2 m v^2)

    -(mv^2)/r dr = m v dv (I have included the minus sign because r is

    -dr/r v = dv

    0 = (v + r) dv/dr

    0 = d/dr(vr) - which says that the rate of change of the vr (which
    is effectively the moment of the linear momentum) is zero.

    Thus, the whole thing works out nicely. The increase in energy can be
    accounted for and it turns out that the angular momentum (moment of the linear momentum) remains constant.

    Cheers and best wishes.


    Robert Neal, PhD
    Department of Human Movement Studies
    The University of Queensland

    ph 61 7 365 6240
    FAX 61 7 365 6877