Question

A school has installed a modestly-sized wind turbine. The three blades are 4.0 m long; each blade has a mass of 45 kg. You can assume that the blades are uniform along their lengths. Part A When the blades spin at 240 rpm, what is the kinetic energy of the blade assembly?

Answer 1

To solve the exercise it is necessary to apply the concepts related to kinetic energy by rotation and the moment of rotational inertia.

Rotational energy is defined as

$KE = \frac{1}{2}I\omega^2$

Where,

I = Inertia moment

$\omega =$ Angular velocity

While the Rotational inertia of each blade is given as

$I = \frac{1}{3}ml^2$

Where,

m= mass

l = length

We have also that the assembly of the motor has three blade, then the total rotational inertia is

$I_m = 3*\frac{1}{3}ml^2$

$I_m = ml^2$

Replacing with our values

$I_m = 45*4^2$

$I_m = 720Kgm^2$

We have the angular velocity in rev per minute then in rad per second is

$\omega= 240rpm (\frac{2\pi rad}{1rev})(\frac{1min}{60s})$

$\omega=25.13rad/s$

Then the total Kinetic Energy at the system is

$KE = \frac{1}{2}720*(25.13)^2$

$KE= 2.27*10^5J$

Therefore the total Kinetic Energy at the system is $2.27*10^5J$