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Question
Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution
`"Final Kinetic Energy of rotation"/"Initial Kinetic Energy of rotation" = (1/2 I_2omega_2^2)/(1/2I_1omega_1^2) = (1/2 I_2(2piv_2)^2)/(1/2I_1(2piv_1)^2) = (I_2v_2^2)/(I_1v_1^2) = (2/5I_1xx(100)^2)/(2/5I_1xx(40)^2) = 2.5`
Clearly, final (K.E) becomes more because the child used his internal energy when he folds his hands to increase the kinetic energy
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