Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What is its angular acceleration in revolutions per minute-squared

Answer 1

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 $\alpha = 6261 \ rev/minutes^2$

b

 $\theta = 613 \ revolutions$

Explanation:

From the question we are told that

   The initial  angular speed is $w_i = 1120 \ rev/minutes$

    The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$  is $w_f = 2560 \ rev/minutes$

    The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$  

 Generally the angular acceleration is mathematically represented as

         $\alpha = \frac{w_f - w_i }{t}$

=>      $\alpha = \frac{2560 - 1120 }{0.23}$  

=>      $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$  is mathematically represented as

           $\theta = \frac{1}{2} * (w_i + w_f)* t$

=>      $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=>      $\theta = 613 \ revolutions$