Question

An automobile engine slows down from 4500 rpm to 1200 rpm in 2.5 seconds. (A) Calculate its angular acceleration. (Give your answer in radians per second squared written as rad/s^2)

Answer 1

Answer:

-138.23 $rad/s^2$

Explanation:

The initial angular velocity = 4500 rpm

The final angular velocity = 1200 rpm

1 revolution = 2π radians

1 minute = 60 seconds

So,

$Initial\ velocity=\frac {4500\times 2\pi}{60}\ radians/sec=471.2389\ radians/sec$

$Final\ velocity=\frac {1200\times 2\pi}{60}\ radians/sec=125.6637\ radians/sec$

Time = 2.5 seconds

Angular acceleration is:

$\alpha=\frac {\omega_f-\omega_i}{t}$

$\alpha=\frac {125.6637-471.2389}{2.5}\ rad/s^2$

Angular acceleration =- 138.23 $rad/s^2$