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 (rev)/(min) \cdot (2 \pi rad/rev)/(60 s/m)=0.157 rad/s`

the angular acceleration is given by

`\alpha = (\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 = (\omega _f)/(\alpha)= (0.157 rad/s)/(64.3 rad/s^2)=2.44 s`