Question

A flywheel is rotating at a rate of 150 rev/minute. If it slows at constant retardation of \( \pi \) rads\(^{-2} \), then the time required for the wheel to come to rest is

• A. \( 2.5 \) s
• B. \( 5 \) s
• C. \( 4 \) s
• D. \( 6 \) s

Hint:

Convert the initial angular velocity from rev/minute to rad/s. Use the first equation of rotational motion, \( \omega = \omega_0 + \alpha t \), where \( \omega \) is the final angular velocity (0 in this case), \( \omega_0 \) is the initial angular velocity, \( \alpha \) is the angular acceleration (retardation is negative acceleration), and \( t \) is the time.

Answer 1

The Correct Option is B

The initial angular velocity of the flywheel is \( \omega_0 = 150 \) rev/minute.
Convert this to rad/s: \( \omega_0 = 150 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{150 \times 2\pi}{60} \text{ rad/s} = 5\pi \text{ rad/s} \) The constant angular retardation (angular acceleration) is \( \alpha = -\pi \) rads\(^{-2} \) (negative because it's retardation).
The final angular velocity is \( \omega = 0 \) rad/s (since the wheel comes to rest).
We use the first equation of rotational motion: \( \omega = \omega_0 + \alpha t \), where \( t \) is the time taken.
Substitute the given values: \( 0 = 5\pi + (-\pi) t \) \( 0 = 5\pi - \pi t \) \( \pi t = 5\pi \) \( t = \frac{5\pi}{\pi} \) \( t = 5 \) s The time required for the wheel to come to rest is 5 seconds.