The angular speed of a flywheel moving with uniform angular acceleration changes from $1200 ext{ rpm}$ to $3

Question

The angular speed of a flywheel moving with uniform angular acceleration changes from $1200 	ext{ rpm}$ to $3120 	ext{ rpm}$ in $16 	ext{ s}$. The angular acceleration in $	ext{rad/s}^2$ is:

Accepted Answer

Given:
Initial angular speed, $\omega_0 = 1200 	ext{ rpm} = \frac{1200 	imes 2\pi}{60} 	ext{ rad/s} = 40\pi 	ext{ rad/s}$.
Final angular speed, $\omega = 3120 	ext{ rpm} = \frac{3120 	imes 2\pi}{60} 	ext{ rad/s} = 104\pi 	ext{ rad/s}$.
Time taken, $t = 16 	ext{ s}$.

Using the first equation of rotational kinematics:
$\omega = \omega_0 + \alpha t$
$\alpha = \frac{\omega - \omega_0}{t}$
$\alpha = \frac{104\pi - 40\pi}{16}$
$\alpha = \frac{64\pi}{16} = 4\pi 	ext{ rad/s}^2$.

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