Question

The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective rotational speeds are 453 rev/min and 4137 rev/min. Calculate the speeds of the tips of both rotors.

Answer 1

Answer:

Explanation:

Given

Diameter of main rotor $d_1=7.56 m$

Tail rotor $d_2=1 m$

$N_1=453 rev/min$

$N_2=4137 rev/min$

$\omega =\frac{2\pi N}{60}$

$\omega _1=\frac{2\pi 453}{60}=47.44 rad/s$

$\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s$

Speed of the tip of main rotor$=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s$

Speed of tail rotor$=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s$