An energy of $484 \mathrm{~J}$ is spent in increasing the speed of a flywheel from $60 \mathrm{rpm}$ to $360…

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:
  1. $0.7 \mathrm{~kg}-\mathrm{m}^2$
  2. $3.22 \mathrm{~kg}-\mathrm{m}^2$
  3. $30.8 \mathrm{~kg}-\mathrm{m}^2$
  4. $0.07 \mathrm{~kg}-\mathrm{m}^2$

Solution

Given,
$\begin{aligned}
& \omega_1=60 \mathrm{rpm}=60 \times \frac{2 \pi}{60}=2 \pi \mathrm{rad} / \mathrm{s} \\
& \omega_2=360 \mathrm{rpm}=360 \times \frac{2 \pi}{60}=12 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}$
Change in kinetic energy,
$\begin{aligned}
\Delta \mathrm{K} . \mathrm{E} & =\frac{1}{2} \mathrm{I}\left(\omega_2^2-\omega_1^2\right) \\
484 & =\frac{1}{2} \times \mathrm{I}\left(\omega_2+\omega_1\right)\left(\omega_2-\omega_1\right) \\
\Rightarrow \quad \mathrm{I} & =\frac{484 \times 2}{14 \pi \times 10 \pi}=0.7 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}$
(where, the symbols have their usual meanings)

Asked in: NEET 2022 (Phase 2)

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