Question

An electric motor is turned off, and its angular velocity decreases uniformly from 1000 revolutions per minute(rpm) to 400 revolutions per minute in 5 s.

Find the angular acceleration and the number of revolutions made by the motor in the 5-s interval.

Answer 1

$\vec\omega(t)=\vec\omega_0+\vec\alpha t$

$\vec\alpha =\frac{\vec\omega(t)-\vec\omega_0}{t}$

$t =5s = \frac{1}{12}min$

$\omega_0=1000rpm =\frac{2\pi}{60}1000= 100 rad/s$

$\omega(t)=400rpm =\frac{2\pi}{60}4000= 40 rad/s$

$\vec\alpha=\frac{400-1000}{t}=-7200rpm/min$

$\theta_t-\theta_0=\frac{\vec\omega_0+\vec\omega_t}{2}*t;$

$\Delta\theta=\frac{1000+400}{2}*\frac{1}{12}=58.3\text{ turnover}$

$\text{Answer:}-7200rpm/min;58.3\text{ turnover}$