A long thin rod of mass M = 2:00 kg and length L = 75:0 cm is free to rotate about its
center as shown. Two identical masses (each of mass m = .403 kg) slide
without friction along the rod. The two masses begin at the rod's point of rotation when
the rod is rotating at 10.0 rad/s.
(a) When they have moved halfway to the end of the rod, how fast (rad/s) is the rod
(b) When the masses are halfway to the end of the rod, what is the ratio of the final kinetic energy to the initial kinetic energy (Kf=Ki)?
(c) When they reach the end, how fast is the rod rotating (rad/s)?

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  1. Use conservation of angular momentum.

    I*w = constant

    I is the moment of inertia, which is
    (1/12)Mrod*L^2 + 2m*R^2

    R is the distance of the masses from the center of the rod.

    w = 10.0 when R = 0

    For (a) and (b), R = L/4

    For (c), R = L/2

    Let's do (a)

    Angular momentum with massesw m at R=0:
    = (1/12)*2.00*(0.75)^2*10
    = 9.38*10^-1 kg m^2/s
    (This remains constant).
    When R = L/4 = 0.1875 m,
    I*w = (2/12)*(0.75)^2*w + 2*(0.403)*(0.1875)^2*w
    = (9.38*10^-2 + 2.83*10^-2)w
    = 1.221*10^-1*w = 9.38*10^-1
    w = 7.68 m/s

    For (b), compare initial and final values of (1/2) I w^2

    For (c), repeat the process of (a), but use R = L/2

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