Question

A high-speed drill rotating counterclockwise takes 2.5 s to speed up to 2400 rpm. (a) What is the drill’s angular acceleration?

Answer 1

Answer:

(a) Angular acceleration will be $100.53rad/sec^2$

(b) 50 revolution

Explanation:

We have given time t = 2.5 sec

Initial speed of the drill $\omega _0=0rad/sec$

Speed after 2.5 sec $=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec$

From first equation of motion we know that

$\omega =\omega _0+\alpha t$

$251.327 =0+\alpha \times 2.5$

$\alpha =100.53rad/sec^2$

(b) From second equation of motion we know that

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2$

$\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times 2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution$