Question

A helicopter blade starts to rotate from rest with a constant angular acceleration of \alpha=\:α = 0.98 radians/second2. How many revolutions will the blade make before it is rotating at 500 revolutions per minute (RPM)?

Answer 1

Answer:

n = 223 revolutions

Explanation:

It is given that,

The angular acceleration of a helicopter blade, $\alpha =0.98\ rad/s^2$

Initial speed of the helicopter blade, $\omega_i=0$

The final speed of the blade, $\omega_f=500\ rpm=500\times \dfrac{2\pi}{60}\ rad/s=52.35\ rad/s$

We need to find the number of revolutions. Firstly we will find the angle turned by the blade. Let the angle is $\theta$. So,

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2}{2\alpha}$

$\theta=1398.22\ rad$

Let there are n number of revolutions made by the blade. So,

$n=\dfrac{\theta}{2\pi}\\\\n=\dfrac{1398.22}{2\pi}\\\\n=222.53\ rev$

or

n = 223 rev

So, there are 223 revolutions.