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

Answer:

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

(B) $\Theta =2846.30rad$

Explanation:

We have given initial angular speed

$\omega _i=1150rpm=\frac{2\times 3.14\times 1150}{60}=120.366rad/sec$

$\omega _f=2680rpm=\frac{2\times 3.14\times 2680}{60}=280.506rad/sec$

Time t = 14.2 sec

(a) From first equation of motion

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

$280.506=120.366+\alpha \times 14.2$

$\alpha =11.277rad/sec^2$

(b) From third equation of motion

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

$280.506^2=120.366^2+2\times 11.277\times \Theta$

$\Theta =2846.30rad$