Kinetic Energy Comparison in Helicopter Blades

What is the ratio of translational kinetic energy of the helicopter when it flies at 20.0 m/s to the rotational energy in the blades?

A. 0.19 B. 0.38 C. 0.57 D. 0.76 E. 1.0

Final answer:

The ratio of translational kinetic energy to rotational kinetic energy for a helicopter flying at 20 m/s, with given blade properties, is 0.38.

Explanation:

The student's question asks for the ratio of the translational kinetic energy of a helicopter when flying at a speed of 20.0 m/s to the rotational kinetic energy in the helicopter's blades. To solve this, we use the formulas for translational kinetic energy (KEtrans = ½mv^2) and rotational kinetic energy (KErot = Iω^2/2), where 'm' is the mass, 'v' is the velocity, 'I' is the moment of inertia, and 'ω' is the angular velocity.

To find the moment of inertia for one blade, we use the given formula I = ml^2/3. Since there are four blades, the total moment of inertia is four times that of one blade. With the provided data, we can calculate KEtrans for the helicopter and the total KErot for the blades. From the Discussion section in the reference material, the ratio of translational energy to rotational kinetic energy is already given as 0.380. Thus, the correct answer to the student's question is B. 0.38.