Let us treat a helicopter rotor blade as a long thin rod, as shown in Fig. 8–49. (a) If each of the three rotor
helicopter blades is 3.75 m long and has a mass of 135 kg,
calculate the moment of inertia of the three rotor blades
about the axis of rotation. (b) How much torque must the
motor apply to bring the blades from rest up to a speed
of 6.0 rev/s in 8.0 s?
Rotor

(a) 1900 kg m²

(b) 8950 Nm

Explanation:

(a) The moment of inertia of a rod about its end is I = ⅓mL².

For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:

I = 3 (⅓ (135 kg) (3.75 m)²)

I = 1900 kg m²

(b) Net torque = moment of inertia × angular acceleration

∑τ = Iα

First, find the angular acceleration.

ω₀ = 0 rad/s

ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s

t = 8.0 s

α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²

∑τ = Iα

∑τ = (1900 kg m²) (4.71 rad/s²)

∑τ = 8950 kg m² / s²

∑τ = 8950 Nm

d

explanation:

i think the answer is d. but im not sure!

ill do it for 100, also this isnt 80 its 40