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?

(b) How many revolutions does it make as it reaches top speed?

Answer 1

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

(b) 50 revolution

Step-by-step explanation:

We have given time t = 2.5 sec

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

Speed after 2.5 sec
`=2400rpm=(2\pi * 2400)/(60)=251.327rad/sec`

From first equation of motion we know that

`\omega =\omega _0+\alpha t`

`251.327 =0+\alpha * 2.5`

`\alpha =100.53rad/sec^2`

(b) From second equation of motion we know that

`\Theta =\omega _0t+(1)/(2)\alpha t^2`

`\Theta =0* 2.5+(1)/(2)* 100.53*  2.5^2=314.16rad=(314.16)/(2\pi )=50revolution`