Question

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

A) Calculate the centripetal acceleration at the tip of a 3.6 m long helicopter blade that rotates at 290 rev/min in m/s².

B) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s). Give your answer in percent of the speed of sound.

Answer 1

The acceleration of blades of helicopter moving in with a velocity in a circular path will be as below

a = v^2/R

Where v is the linear velocity of blades and r is the length of blades or can be considered as a radius of circular path

From Angular velocity w and linear velocity v relationship

V = wr

W can be calculated by simply converting 290rev/Min to rad/s and that will be

w = 30rad/s

We have r = 3.6 m

V =wr = 30*3.6= 108 rad/s

Now acceleration

a = v^2/r = (wr)^2/r = (108)^2/3.6

a =3240 rad/s^2

Part(2)

V=wr= 108 rad/s this is the velocity of blades

Now velocity of blades in terms of speed of sound will be as using following formula

V/C = 108/3*20^8 = 0.3176 m/s

In percentage 31%

Therefore linear velocity will be equal to 31% of sound speed.