total moment inertia = 2* single rod I
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
(a) To find the total moment of inertia of the two blades about the axis of rotation, we can use the parallel-axis theorem. The moment of inertia of a thin rod about its center axis is given by the formula:
I = (1/12) * m * L^2
where I is the moment of inertia, m is the mass of each blade, and L is the length of each blade.
Since there are two blades, we need to calculate the moment of inertia for each blade separately and then add them together. Therefore, the moment of inertia of one blade is:
I1 = (1/12) * m * L^2
= (1/12) * 220 kg * (6.7 m)^2
To find the total moment of inertia, we multiply the moment of inertia of one blade by 2 since there are two blades:
Total moment of inertia = 2 * I1
Now we can substitute the values to find the moment of inertia.
Total moment of inertia = 2 * (1/12) * 220 kg * (6.7 m)^2
(b) To calculate the rotational kinetic energy of the spinning blades, we use the formula:
Rotational kinetic energy = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular speed of the rotation.
We can now substitute the value of the moment of inertia and the angular speed to calculate the rotational kinetic energy:
Rotational kinetic energy = (1/2) * Total moment of inertia * (angular speed)^2
Substitute the values to find the rotational kinetic energy.
Rotational kinetic energy = (1/2) * Total moment of inertia * (45 rad/s)^2