An energy of 484,J is spent in increasing the | vedclass

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(A) The initial angular velocity is $\omega_i = 60\,rpm = 60 	imes \frac{2\pi}{60} = 2\pi\,rad/s$.The final angular velocity is $\omega_f = 360\,rpm

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(A) The initial angular velocity is $\omega_i = 60\,rpm = 60 	imes \frac{2\pi}{60} = 2\pi\,rad/s$.The final angular velocity is $\omega_f = 360\,rpm = 360 	imes \frac{2\pi}{60} = 12\pi\,rad/s$.The change in rotational kinetic energy is given by $\Delta K.E. = \frac{1}{2} I (\omega_f^2 - \omega_i^2) = 484\,J$.Substituting the values: $\frac{1}{2} I ((12\pi)^2 - (2\pi)^2) = 484$.$\frac{1}{2} I (144\pi^2 - 4\pi^2) = 484$.$\frac{1}{2} I (140\pi^2) = 484$.$70\pi^2 I = 484$.Using $\pi^2 \approx 9.86$,we get $70 	imes 9.86 	imes I = 484$.$690.2 I = 484$.$I = \frac{484}{690.2} \approx 0.701\,kg\cdot m^2$.

An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $.............\,kg\cdot m^2$.

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