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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?

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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.

Posted by

Devendra Khairwa

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