A small rescue helicopter has 4 blades, each 4m long and 50kg in mass, and it carries a load of 1000kg. a) Find the rotational kinetic energy in the blades when rotating at 300 rpm. b) Find the translational kinetic energy of the helicopter when it flies at 20 m/s, and compare it to the rotational energy in the blades. c) How high could the helicopter fly If all of its rotational kinetic energy could be used to lift it

A) KE_rot= 5.26 x 10^(5) J

B) KE_trans = 2 x 10^(5)J

Ratio = 0.38

C) Max height = 52.7m

Explanation:

A) We are given;

m = 50kg

r = 4m

ω = 300 revs/min = 300 x 0.1047 = 31.4 rads/s

The formula for rotational kinetic energy is given as;

KE_rot= (1/2)Iω²

Where I is moment of inertia.

Now, I = mr²/3

But since it has 4 blades, thus,

I = 4mr²/3 = 4 x 50 x 4²/3 = 1067 kg. m²

Thus, KE_rot= (1/2)(1067)(31.4)² = 5.26 x 10^(5) J

B) The formula for translational kinetic energy is given by;

KE_trans = (1/2)mv²

We are given velocity as 20 m/s

Thus,

KE_trans = (1/2)(50)(20)² = 2 x 10^(5)J

To compare this to the rotational kinetic energy, we'll take the ratio. Thus; = KE_trans/KE_rot = 2 x 10^(5)J/5.26 x 10^(5)J = 0.38

C) At maximum height,

KE_rot = PE_gravity

Thus, 5.26 x 10^(5)J = mgh

Where m is the load it carries;

5.26 x 10^(5)J = 1000 x 9.8 x h

So, h = [5.26 x 10^(5)]/(9.8 x 1000) = 52.7m