Answer to Question #185644 in Classical Mechanics for Jethro

Question

A grinding wheel of 0.2 m in diameter, of mass 3kg, is rotating at 3600 revolutions per minute about an axis through its center.

What is its kinetic energy?

Answer 1

$E=\frac{I\omega^2}{2},$

$I=\frac{mr^2}{2}=\frac{md^2}{8},$

$\omega=\frac{2 \pi \nu}{60}=\frac{\pi \nu}{30},$

$E=\frac 12 \cdot \frac{md^2}{8}\cdot \frac{\pi^2\nu^2}{900}=\frac{md^2\pi^2 \nu^2}{14400}=1066~J.$