PYQ NEET- Rotational Motion L-3

Question: An energy of $484 \mathrm{~J}$ is spent in increasing the speed of a flywheel from $60 \mathrm{rpm}$ to $360 \mathrm{rpm}$. The moment of inertia of the flywheel is :#

A) $0.07 \mathrm{~kg}-\mathrm{m}^2$

B) $0.7 \mathrm{~kg}-\mathrm{m}^2$

C) $3.22 \mathrm{~kg}-\mathrm{m}^2$

D) $30.8 \mathrm{~kg}-\mathrm{m}^2$

Answer: $0.7 \mathrm{~kg}-\mathrm{m}^2$#

Solution:#

From work - energy theorem $W=\Delta k$ (change in Kinetic Energy) In rotation, $K E=\frac{1}{2} I \omega^2$ $484=\frac{1}{2} I\left(\omega_f^2-\omega_i^2\right)$ $\Rightarrow 484=\frac{1}{2} I\left[\left(2 \pi \frac{360}{60}\right)^2-\left(2 \pi \times \frac{60}{60}\right)^2\right]$ $\Rightarrow 484=\frac{1}{2} I 4 \pi^2(36-1)$ $\Rightarrow I \simeq 0.7 \mathrm{~kg}-\mathrm{m}^2$