Kinetic energy of rotation of a flywheel of radius 2 m , mass 8 kg and angular speed 4 rad s - 1 about an axis perpendicular to its plane and passing through its center is

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Aicharya · Accepted Answer

Given that Mass of wheel $m=8 \mathrm{~kg}$ radius $r=2 \mathrm{~m}$ Angular speed $\omega=4 \mathrm{rad} \mathrm{s}^{-1}$ Rotational kinetic energy of wheel $\begin{aligned} k & =\frac{1}{2} I \omega^{2} \ & \Rightarrow k=\frac{1}{2} \frac{m r^{2}}{2} \omega^{2} \ & \Rightarrow k=\frac{1}{2} \frac{8 \times 4}{2} 4^{2}=128 \mathrm{~J} \end{aligned}$

Kinetic energy of rotation of a flywheel of radius 2 m , mass 8 kg and angular speed 4 rad s - 1 about an…

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