Question

An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular acceleration is constant, how many revolutions does the propeller undergo during this time?

Answer 1

First of all, we need to convert the angular speed from rev/min into rev/s:

$\omega_f=300 rev/min=5 rev/s$

The angular acceleration is the variation of angular speed divided by the time:

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2$

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

$\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev$

so, 5 revolutions.