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What linear speed must a 0.0495-kg hula hoop have if its total kinetic energy is to be 0.113 J? Assume the hoop rolls on the ground without slipping.

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    The kinetic energy of the hoop consists of two parts: (1) translational motion of the center of mass, (1/2) M V^2, and
    (2) rotation about the center of mass,
    (1/2)I w^2 = (1/2)M R^2*(V/R)^2=(1/2)MV^2
    The two parts are equal in this case. The total kinetic energy is M V^2.
    Solve
    (0.0495)V^2 = 0.113
    to obtain V. The answer will be in m/s.

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