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

n = 223 revolutions

Step-by-step 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* (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=(\omega_f^2)/(2\alpha)`

`\theta=1398.22\ rad`

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

`n=(\theta)/(2\pi)\\\\n=(1398.22)/(2\pi)\\\\n=222.53\ rev`

or

n = 223 rev

So, there are 223 revolutions.