Q7.13(b)     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?

Answers (1)

The final and initial velocities are given below :

                          E_f\ =\ \frac{1}{2}I_2w_2^2                 and            E_i\ =\ \frac{1}{2}I_1w_1^2 

        Taking the ratio of both we get,       

                                                           \frac{E_f}{E_i}\ =\ \frac{\frac{1}{2}I_2w_2^2}{\frac{1}{2}I_1w_1^2}

or                                                                  =\ \frac{2}{5}\times \frac{100\times 100}{40\times40}

or                                                                   =\ 2.5

Thus the final energy is 2.5 times the initial energy.

The increase in energy is due to the internal energy of the boy.

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