Question

A model of a helicopter rotor has four blades, each of length 3.0 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 560 rev/min.part A :What is the linear speed of the bladetip?part B :What is the radial acceleration of the bladetip expressed as a multiple of the acceleration of gravity,g?

Answer 1

Answer with Explanation:

We are given that

r=3 m

Angular frequency=$\omega$=560rev/min

A.1 revolution=$2\pi$ radian

560 revolutions=$560\times 2\pi$ rad

Angular frequency=$2\times 3.14\times \frac{560}{60}$rad/s

1 min=60 s

$\pi=3.14$

Angular frequency=$\omega=58.6rad/s$

Linear speed=$\omega r$

Using the formula

Linear speed=$58.6\times 3=175.8m/s$

Hence, the linear speed of the blade tip=175.8m/s

B.Radial acceleration=$a_{rad}=\frac{v^2}{r}$

By using the formula

Radial acceleration=$a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2$

Radial acceleration=$\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g$

Where $g=9.8m/s^2$

Hence, the radial acceleration$=1.05\times 10^3 g$