Question

How much time does it take a power drill accelerating at 64.3 rad/s2 to achieve 1500. rotations per minute(RPM)?

Answer 1

First of all, we need to convert the final angular velocity from rpm (revolutions per minute) to rad/s. Keeping in mind that
$1 rev = 2\pi rad$
$1 min = 60 s$
We have
$\omega _f = 1500 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/m}=0.157 rad/s$

the angular acceleration is given by
$\alpha = \frac{\omega _f - \omega _i}{\Delta t}$
where $\omega _i$ is the initial velocity (in this case, zero), and $\Delta t$ is the time needed to accelerate the drill to its final velocity. Using $\alpha = 64.3 rad/s^2$, we can calculate $\Delta t$:
$\Delta t = \frac{\omega _f}{\alpha}= \frac{0.157 rad/s}{64.3 rad/s^2}=2.44 s$