Question

the frequency of the rotor of a helicopter is 1800 cycles/min, calculate the period of its rotor in seconds, if the rotor arm is 3 m long, find the tangential speed of a point at the far end of the arm. (Take p to be 3)

Answer 1

Ответ:

The angular speed of rotation of the screw w = 2p * n / t = (2p * 1800/3) = 62.8 s ^ -1

Answer 2

To calculate the period of the rotor, we need to convert the frequency from cycles per minute to cycles per second.

Given: Frequency of the rotor = 1800 cycles/min

To convert cycles/min to cycles/second, we divide by 60 (since there are 60 seconds in a minute): Frequency of the rotor = 1800 cycles/min ÷ 60 = 30 cycles/sec

The period is the reciprocal of the frequency, so we can calculate it as follows: Period = 1 / Frequency = 1 / 30 cycles/sec

To find the tangential speed of a point at the far end of the rotor arm, we can use the formula for tangential speed: Tangential speed = 2πr / T

Given: Rotor arm length (radius), r = 3 m Period, T = 1 / 30 cycles/sec

We can substitute the values into the formula to calculate the tangential speed: Tangential speed = 2π(3) / (1/30) = 60π m/s

Approximating π as 3, we can simplify the expression: Tangential speed ≈ 60(3) m/s = 180 m/s

Therefore, the period of the rotor is approximately 1/30 seconds, and the tangential speed of a point at the far end of the rotor arm is approximately 180 m/s.