## Physics (10th Edition)

$\omega=6.1\times10^5rev/min$
The flywheel is a solid disk having the rotation axis at its center, so $$I=\frac{1}{2}MR^2=\frac{1}{2}(13kg)(0.3m)^2=0.585kg.m^2$$ The speed the flywheel will have to rotate to store an energy $KE=1.2\times10^9J$ is $$KE=\frac{1}{2}I\omega^2$$ $$\omega=\sqrt{\frac{2KE}{I}}=6.4\times10^4rad/s\Big(\frac{1rev}{2\pi rad}\Big)\Big(\frac{60s}{1min}\Big)$$ $$\omega=6.1\times10^5rev/min$$