Question

Resuelva los siguientes planteamientos: Debido a su conocimiento de la física y el interés por los medios de transporte, has conseguido un trabajo de verano en la compañía de Estados Unidos Sikorsky, ellos están interesados en investigar el consumo energético de su nuevo helicóptero de tres rotores (Sikorsky S-97 Raider). El aspa de un rotor puede considerarse como una varilla delgada. Si cada una de las tres aspas del rotor del helicóptero mide 3.75 m de largo y tiene una masa de 135 kg, a) Determine la energía necesaria para poner a funcionar los rotores (2) hasta que las aspas alcancen una rapidez de 50.0 rev/s. b) Si se usara gasolina regular para producir esa energía Cuántos galones de la misma se necesitarían? Si el motor de un helicóptero Sikorsky CH-53K proporciona una potencia útil en el eje de 13.5x106 watts,Cuántas revoluciones por minuto deben rotar sus aspas para mantenerlo en el aire por una hora? Explique una forma de reducir el consumo energético de estos helicópteros.

Solve the following approaches: Due to your knowledge of physics and interest in transportation, you have obtained a summer job at the United States company Sikorsky, they are interested in investigating the energy consumption of your new helicopter three rotors (Sikorsky S-97 Raider). The blade of a rotor can affect like a thin rod. If each of the three blades of the helicopter rotor is 3.75 m long and has a mass of 135 kg, a) Determine the energy needed to operate the rotors (2) until the blades reach a speed of 50.0 rev / s. b) If regular gasoline were used to produce that energy, how many gallons of it would be needed? If the engine of a Sikorsky CH-53K helicopter provides a useful shaft power of 13.5x106 watts, how many revolutions per minute should its blades rotate to keep it in the air for an hour? Explain a way to reduce the energy consumption of these helicopters.

Answer 1

(a) The rotational speed, $\inline \omega$= 50.0 rev/s = 50.0 x 2$\inline \pi$ rad/s. The moment of inertia of the rod about the end of the rod will be

$I=\frac{1}{3}ML^{2}=\frac{1}{3}\times 135\times 3.75^{2}= 632.8\ Kg.m^{2}$

The rotational kinetic energy of the blades or energy needed to operate the rotor will be

$E_{k}=\frac{1}{2}I\omega ^{2}\times 3=\frac{3}{2}\times 632.8\times \left ( 50\times 2 \pi \right )^{2}=93.68\times 10^{6}\ J$

(b) Let us consider the energy produced per gallon of gasoline is about 1.3 x 10⁸ J. Then, the amount of gasoline is needed

= (93.68 x 10⁶ J)/(1.3 x 10⁸ J) = 0.72 gallon.

Let us consider the power P and time t (for which it is operating), then energy required will be E = Pt. Therefore, from the above equation you can write (1 h = 3600 s)

$Pt=E_{k}=\frac{1}{2}I\omega ^{2}\times 3$

$\omega =\sqrt{\frac{2Pt}{3I}}=\sqrt{\frac{2\times 13.5\times 10^{6}\times 3600}{3\times 632.8}}=7155.49\ rad/s\approx 68330 \ rev/min$

To reduce energy consumption, you have to reduce the revolution per min. You need only the minimum revolution to operate the helicopter not the maximum one.