Question

The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 0.99 m, respectively. The respective rotational speeds are 449 rev/min and 4,149rev/min. Calculate the speeds of the tips of both rotors.

main rotor	m/s
tail rotor	m/s

Compare these speeds with the speed of sound, 343 m/s.

vmain rotor / vsound	=
vtail rotor / vsound	=

Answer 1

diameter of main rotor = d_m = 7.67 m

radius of main rotor = r_m= d_m/2 = 7.67/2 = 3.835 m

diameter of tail rotor = d_t = 0.99 m

radius of tail rotor = r_t= d_m/2 = 0.99/2 = 0.495 m

Rotational speed of main rotor = W_m = 449 rev/min

W_m = 449 *2(3.14)/60 rad/sec = 47 rad/sec                            ( since 1 rev = 2 pi    , 1 min = 60 sec)

speed of tip of main rotor can be given as ::

V_m = r_m W_m

V_m = (3.835 m) ( 47 rad/sec)

V_m = 180.25 m/s

Rotational speed of tail rotor = W_t = 4149 rev/min

W_t = 4149 *2(3.14)/60 rad/sec = 434.3 rad/sec                            ( since 1 rev = 2 pi    , 1 min = 60 sec)

speed of tip of tail rotor can be given as ::

V_t = r_t W_t

V_t = (0.495 m) ( 434.3 rad/sec)

V_t = 214.98 m/s

V_mainrotor /V_sound = 180.25 m/s / 343 m/s = 0.53

V_tailrotor /V_sound = 214.98 m/s / 343 m/s = 0.63