Question

A wheel of the moment of inertia $10kg-m^2$ is rotating at $10$ rotations per minute. Find the work done in increasing its speed to $5$ times of its initial value.

Answer 1

Given Moment of inertia , $I=10kg-m^2$ , speed $N=10rpm$

$\Rightarrow {\omega }_1=\frac{2\pi N}{60}=\frac{2\pi \times 10}{60}=\frac{\pi }{3}$

$\Rightarrow {\omega }_2=5{\omega }_1=\frac{5\pi }{3}$

$\therefore$ Work done $=\frac{1}{2}I({\omega }_2^2-{\omega }_1^2)=\frac{1}{2}\times 10\times (\frac{25{\pi }^2}{9}-\frac{{\pi }^2}{9})=131.6J$