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=(\omega_f-\omega_i)/(t)=(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)=(1)/(2)\alpha t^2 =(1)/(2)(2.5 rev/s^2)(2 s)^2=5 rev`

so, 5 revolutions.