The angular speed of a motor wheel is increased from $1200 ext{ rpm}$ to $3120 ext{ rpm}$ in $16

Question

The angular speed of a motor wheel is increased from $1200	ext{ rpm}$ to $3120	ext{ rpm}$ in $16	ext{ s}$. Assuming uniform angular acceleration, how many revolutions does the engine make during this time?

Professor PC Thomas Classes · Answer

Initial angular speed $\omega_0 = \frac{1200 	imes 2\pi}{60} = 40\pi	ext{ rad/s}$. Final angular speed $\omega = \frac{3120 	imes 2\pi}{60} = 104\pi	ext{ rad/s}$. Angular acceleration $\alpha = \frac{104\pi - 40\pi}{16} = 4\pi	ext{ rad/s}^2$. Angular displacement $	heta = \omega_0t + \frac{1}{2}\alpha t^2 = 40\pi(16) + \frac{1}{2}(4\pi)(16)^2 = 1152\pi	ext{ rad}$. Number of revolutions $= \frac{1152\pi}{2\pi} = 576$.