Question

If $2\times 10^{5}J$ of work is done in producing rotational kinetic energy of a wheel having moment of inertia of $10kg.m^{2}$ then the angular velocity (in $rads^{- 1}$ ) acquired by the wheel is equal to

Answer 1

From work-energy theorem,
Work done produces change in kinetic energy,
$\therefore \quad W=\frac{1}{2} I \omega^2 \quad \ldots .\left(\because K . E .=\frac{1}{2} I \omega^2\right)$
$\therefore \omega =\sqrt{\frac{2 W}{I}}=\sqrt{\frac{2 \times 2 \times 10^{5}}{10}}=200rad/s$