Question

A typical helicopter has four blades that rotate at 334 rpm and have a total kinetic energy of 4.55 105 J. What is the magnitude of the total angular momentum of the blades about the axis of rotation?

Answer 1

Answer:

the magnitude of the total angular momentum of the blades is 743.71 kg·m²

Explanation:

Converting the angular speed into radians per second:

ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)

ω = 34.98 rad/s

The rotational kinetic energy of the blades is given by:

EK = 1/2 I ω²

where

• I is the moment of inertia
• ω is the angular speed

Therefore, rearranging the above equation, we get:

1/2 I ω² = EK

I ω² = 2 EK

I = 2(EK) / ω²

I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²

I = 743.71 kg·m²

Therefore, the magnitude of the total angular momentum of the blades is 743.71 kg·m².