Question

A wheel is rotating at 1800 rpm about its own axis. When the power is switched off, it comes to rest in 2 minutes. Then the angular retardation in $\text{rad s}^{-2}$ is}

• A. $2\pi$
• B. $\pi$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$
• E. $\frac{\pi}{6}$

Hint:

Always perform unit conversions first! 1800 rpm is a frequency; you must convert it to $\omega$ before plugging it into kinematic formulas.

Answer 1

The Correct Option is C

Concept: This problem involves Rotational Kinematics with uniform angular acceleration (retardation).
• Angular Speed ($\omega$): Must be in $\text{rad s}^{-1}$. $1 \text{ rpm} = \frac{2\pi}{60} \text{ rad s}^{-1}$.
• Kinematic Equation: $\omega_f = \omega_i + \alpha t$, where $\alpha$ is angular acceleration.
• Retardation: A negative angular acceleration that reduces angular speed over time.

Step 1: Convert units to SI (radians and seconds). Initial frequency $f_i = 1800 \text{ rpm}$. \[ \omega_i = 1800 \times \frac{2\pi}{60} = 30 \times 2\pi = 60\pi \text{ rad s}^{-1} \] Final angular speed $\omega_f = 0$ (comes to rest). Time $t = 2 \text{ minutes} = 2 \times 60 = 120 \text{ seconds}$.

Step 2: Calculate the angular retardation ($\alpha$). Using the first equation of rotational motion: \[ \omega_f = \omega_i - \alpha t \text{ (using minus sign for retardation)} \] \[ 0 = 60\pi - \alpha(120) \] \[ 120\alpha = 60\pi \] \[ \alpha = \frac{60\pi}{120} = \frac{\pi}{2} \text{ rad s}^{-2} \]