Question

The angular speed of an automobile engine is increased at a constant rate from 1150 rev/min to 2680 rev/min in 14.2 s. (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 14.2 s interval?

Answer 1

(A)
`\alpha =11.277rad/sec^2`

(B)
`\Theta =2846.30rad`

Step-by-step explanation:

We have given initial angular speed

`\omega _i=1150rpm=(2* 3.14* 1150)/(60)=120.366rad/sec`

`\omega _f=2680rpm=(2* 3.14* 2680)/(60)=280.506rad/sec`

Time t = 14.2 sec

(a) From first equation of motion

`\omega _f=\omega _i+\alpha t`

`280.506=120.366+\alpha * 14.2`

`\alpha =11.277rad/sec^2`

(b) From third equation of motion

`\omega _f^2=\omega _i^2+2\alpha \Theta`

`280.506^2=120.366^2+2* 11.277* \Theta`

`\Theta =2846.30rad`