A model of a helicopter rotor has four blades, each of length 3.30 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 590 rev/min. a. What is the linear speed of the blade tip?

b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity?

Answer: a) 195.82m/a

b) 1149.7g

Explanation: parameters given are:

angular velocity, ω

= 550 rpm

Convert rev / min to rad/s

= 550 * 2π / 60

= 55π/3 radians/s

radius, r = 3.4 m

A.

Using linear speed formula

Linear speed, v

= r ω m/s

= 3.4 * 55 π/3 m/s

= 195.83 m/s

B.

Let use acceleration derives from centripetal acceleration

a = v^2 / r

This lead to

radial acceleration

= r ω2

= 3.4 (55π/3) 2 m/s/s

= 11279 m/s/s

Acceleration due to gravity g = 9.81m/s^2

Expressing as a multiple of g gives

= 11279 / 9.81 g

= 1149.7 g

a) Linear Speed; v = 590 x 2π/60 x 3.3 = 203.91 m/s

b) We get the radial acceleration of the tip of the blade as

a = w²r or v²/r

a = (203.91 m/s) ² / 3.3

= 12600.50 m/s²

Acceleration due to gravity of the earth = 9.8m/s²

ratio of radial acceleration to g = a-rad/g

= 12600.50 / 9.8 = 1285.76g