Posted by Emily on Sunday, November 1, 2009 at 10:41pm.
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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