The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is: ____.

The problem gives the initial angular speed ω₁ = 600 rpm and the final angular speed ω₂ = 1200 rpm, with the change occurring over a time interval t = 10 s. We assume the angular acceleration is uniform.

First convert the speeds from revolutions per minute (rpm) to revolutions per second (rps):

$\mathrm{\omega ₁}=600\mathrm{rpm}=\frac{600}{60}\mathrm{rps}=10\mathrm{rps}$

$\mathrm{\omega ₂}=1200\mathrm{rpm}=\frac{1200}{60}\mathrm{rps}=20\mathrm{rps}$

With constant angular acceleration, the average angular speed during the interval is the arithmetic mean of the initial and final speeds:

$\mathrm{\omega \_\{avg\}}=\frac{(}{\mathrm{\omega ₁}}=\frac{(}{10}=15\mathrm{rps}$

The total number of revolutions N made in the 10 s is the average speed multiplied by the time:

$N=\mathrm{\omega \_\{avg\}}·t=15·10=150$

Therefore, the flywheel completes 150 revolutions during the acceleration period.