Asked by Emily
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.
Answers
Answered by
drwls
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.
(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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