Question

For a flywheel, if \( I \) is the mass moment of inertia of the flywheel, \( \omega_{\text{av}} \) is the average rotational speed, and \( K_s \) is the coefficient of fluctuation of speed given by \( K_s = \frac{\omega_{\max} - \omega_{\min}}{\omega_{\text{av}}} \), then find the maximum fluctuation of energy during a cycle.

• A. \( I K_s (\omega_{\max}^2 - \omega_{\min}^2) \)
• B. \( 0.5 \omega_{\text{av}} (\omega_{\max} - \omega_{\min}) K_s \)
• C. \( \frac{1}{2} I K_s \omega_{\text{av}}^2 \)
• D. \( I K_s \omega_{\text{av}}^2 \)

Hint:

- Use fluctuation of energy formula for flywheels: \( \frac{1}{2} I K_s \omega_{\text{av}}^2 \). - Higher \( K_s \) leads to greater energy variation.

Answer 1

The Correct Option is C

- Maximum fluctuation of energy for a flywheel is: \[ \Delta E = \frac{1}{2} I K_s \omega_{\text{av}}^2 \] - It depends on the mass moment of inertia and the fluctuation of speed.