Answer:
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
