Question

The angular speed of an automobile engine is increased at a constant rate from 1260 rev/min to 3460 rev/min in 9.90 s. (a) What is its angular acceleration in revolutions per minute-squared

Answer 1

Answer:

13333.33 rev/min²

Explanation:

Given:

Initial angular speed of the automobile (ω₁) = 1260 rev/min

Final angular speed of the automobile (ω₂) = 3460 rev/min

Time interval for the change in speed (t) = 9.90 s

Angular acceleration of the automobile (α) = ?

Consider the sense of rotation to be positive. So, making use of the equation of motion of rotation, we have:

$\omega_2=\omega_1+\alpha t$

Rewriting in terms of 'α', we get:

$\alpha =\dfrac{\omega_2-\omega_1}{t}$

Converting time 't' from seconds to minutes using conversion factor.

1 sec = $\frac{1}{60}\ min$

So, 9.90 s = $9.9\times \frac{1}{60}=0.165\ min$

Plug in all the given values and solve for 'α'. This gives,

$\alpha =\frac{3460-1260}{0.165}\ rev/min^2\\\\\alpha=\frac{2200}{0.165}=13333.33\ rev/min^2$

Therefore, the angular acceleration in revolutions per minute squared is 13333.33 rev/min².