Question:medium

A flywheel is designed for engine speed variation from $190\text{ rps}$ to $210\text{ rps}$. Calculate the inertia of the wheel if the kinetic energy stored in the flywheel is $400\text{ N}\cdot\text{m}$.

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To simplify calculations involving flywheel energy changes, always use the difference of squares identity: $\omega_{\text{max}}^2 - \omega_{\text{min}}^2 = (\omega_{\text{max}} - \omega_{\text{min}})(\omega_{\text{max}} + \omega_{\text{min}})$. This turns long squaring steps into a quick multiplication.

Updated On: Jul 9, 2026

$0.02\text{ kg}\cdot\text{m}^2$
$0.01\text{ kg}\cdot\text{m}^2$
$0.2\text{ kg}\cdot\text{m}^2$
$0.1\text{ kg}\cdot\text{m}^2$

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The Correct Option is A

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