Question

A wheel of angular speed 600 rev/min is made to slow down at a rate of \( 2 \) rad/s\(^2\). The number of revolutions made by the wheel before coming to rest is

• A. 157
• B. 314
• C. 177
• D. 117

Hint:

Always convert angular speed to rad/s before applying rotational kinematics equations.

Answer 1

The Correct Option is A

Using the kinematic equation for rotational motion: \[ \omega^2 = \omega_0^2 + 2 \alpha \theta \] Given: Initial angular speed, \( \omega_0 = 600 \) rev/min \( = 600 \times \frac{2\pi}{60} = 20\pi \) rad/s Final angular speed, \( \omega = 0 \) rad/s Angular acceleration, \( \alpha = -2 \) rad/s\(^2\) Solving for \( \theta \): \[ 0 = (20\pi)^2 + 2(-2) \theta \] \[ 400\pi^2 = 4\theta \] \[ \theta = \frac{400\pi^2}{4} = 100\pi^2 \] Since 1 revolution corresponds to \( 2\pi \) radians, \[ \text{Revolutions} = \frac{100\pi^2}{2\pi} = 157 \] Thus, the total number of revolutions is 157.