Motion in a circle of helicopter rotor

[Edwardo_Elric]

Homework Statement

A model of a helicopter rotor has four blades, each 3.20m in length from the central shaft to the blade tip. The model is rotated in a wind tunnel at 600rev/min. a.) What is the linear speed of the blade in m/s?
b.) What is the radial acceleration of the blade expressed as a multiple of the acceleration due to gravity g?

Homework Equations

a_{rad} = \frac{V^2}{R}
a_{rad} = \frac{4{\pi}^2R}{T^2}

The Attempt at a Solution

a.) convert: 600rev / min ( 1 min / 60secs) = 10 rev / s
Multiplied 3.20m by 10 rev / s = 32.0m/s?

b.) a_{rad} = \frac{V^2}{R}
a_{rad} = (32.0m/s)^2 / (3.20m) = 320m/s^2

i don't understand the radial acceleration of blade expressed as a multiple of g...
is my answers correct?

 

[Doc Al]

a.) convert: 600rev / min ( 1 min / 60secs) = 10 rev / s
Multiplied 3.20m by 10 rev / s = 32.0m/s?

Careful. To convert between linear and angular speed, use v = \omega r, where \omega is in radians/sec, not rev/s.

Alternatively, realize that 1 revolution covers 1 circumference, which equals 2 \pi r.

b.) a_{rad} = \frac{V^2}{R}
a_{rad} = (32.0m/s)^2 / (3.20m) = 320m/s^2

Redo this with the correct speed.

i don't understand the radial acceleration of blade expressed as a multiple of g...

Since g = 9.8 m/s^2, just divide your answer by that value to get an answer as a multiple of g.