The angular speed of the wheel of a vehicle is increased from $360 ext{ rpm}$ to $1200 ext{ rpm}$ in $14 \

Question

The angular speed of the wheel of a vehicle is increased from $360 	ext{ rpm}$ to $1200 	ext{ rpm}$ in $14 	ext{ seconds}$. Its angular acceleration will be:

Accepted Answer

Initial angular speed, $\omega_0 = 360 	ext{ rpm} = 360 	imes \frac{2\pi}{60} 	ext{ rad/s} = 12\pi 	ext{ rad/s}$.
Final angular speed, $\omega = 1200 	ext{ rpm} = 1200 	imes \frac{2\pi}{60} 	ext{ rad/s} = 40\pi 	ext{ rad/s}$.
Time, $t = 14 	ext{ s}$.
Using the kinematic equation for uniform rotational acceleration, $\omega = \omega_0 + \alpha t$:
$\alpha = \frac{\omega - \omega_0}{t}$
$\alpha = \frac{40\pi - 12\pi}{14}$
$\alpha = \frac{28\pi}{14} = 2\pi 	ext{ rad/s}^2$.

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