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NEET PHYSICSEasy

The angular speed of the wheel of a vehicle is increased from 360 rpm360 \text{ rpm} to 1200 rpm1200 \text{ rpm} in 14 seconds14 \text{ seconds}. Its angular acceleration will be:

A

2π rad/s22\pi \text{ rad/s}^2

B

28π rad/s228\pi \text{ rad/s}^2

C

120π rad/s2120\pi \text{ rad/s}^2

D

1 rad/s21 \text{ rad/s}^2

Step-by-Step Solution

Initial angular speed, ω0=360 rpm=360×2π60 rad/s=12π rad/s\omega_0 = 360 \text{ rpm} = 360 \times \frac{2\pi}{60} \text{ rad/s} = 12\pi \text{ rad/s}. Final angular speed, ω=1200 rpm=1200×2π60 rad/s=40π rad/s\omega = 1200 \text{ rpm} = 1200 \times \frac{2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}. Time, t=14 st = 14 \text{ s}. Using the kinematic equation for uniform rotational acceleration, ω=ω0+αt\omega = \omega_0 + \alpha t: α=ω−ω0t\alpha = \frac{\omega - \omega_0}{t} α=40π−12π14\alpha = \frac{40\pi - 12\pi}{14} α=28π14=2π rad/s2\alpha = \frac{28\pi}{14} = 2\pi \text{ rad/s}^2.

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