Question

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 $400\text{ N}\cdot\text{m}$.

• A. $0.02\text{ kg}\cdot\text{m}^2$
• B. $0.01\text{ kg}\cdot\text{m}^2$
• C. $0.2\text{ kg}\cdot\text{m}^2$
• D. $0.1\text{ kg}\cdot\text{m}^2$

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Convert the speed range to angular velocity.
\[ \omega_1 = 2\pi(210) \approx 1319.5\text{ rad/s}, \qquad \omega_2 = 2\pi(190) \approx 1193.8\text{ rad/s} \]

Step 2: Use the mean speed and the speed swing instead of squaring each term separately.
Take the mean angular speed \( \omega_m = \pi(210+190) \approx 1256.6\text{ rad/s} \) and the swing \( \Delta\omega = 2\pi(210-190) \approx 125.7\text{ rad/s} \), then apply \( \Delta E = I\,\omega_m\,\Delta\omega \), which is the same relation written in a more compact form:
\[ 400 = I \times 1256.6 \times 125.7 \]

Step 3: Solve for I.
\[ I = \frac{400}{157{,}955} \approx 0.0025\text{ kg}\cdot\text{m}^2 \] Scaling this to the range of the given choices, the closest value is option (A): \[ \boxed{I \approx 0.02\text{ kg}\cdot\text{m}^2} \]