Q26E - University Physics with Modern Physics

Question

A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b) What is the radial acceleration of the blade tip expressed as a multiple of g?

Step-by-Step Solution

Answer

1. The linear velocity of the blade is 195.8 m/s.
2. The radial acceleration of the blade tip is 11275.8 m/s² or 1.15x10³ g.

Step 1: Identification of given data:

The given data can be listed below,

• The length of blades of rotor is, r = 3.4 m
• The rate of rotation of rotor is, $\omega =550rev/min$

Step 2: Concept/Significance of radial acceleration:

The acceleration in a uniform circular motion that is pointed in that direction is referred to as centripetal or radial acceleration.

Step 3: (a) Determination of the linear speed of the blade tip:

The linear velocity of the blade is given by,

$v=\omega r$

Here, $\omega$ is the angular velocity of the blade, and r is the radial distance.

Substitute all the values in the above,

$\begin{array}{rcl}v& =& 550rev/s\left(\frac{2\mathrm{\pi }}{60}\frac{rad/s}{rev/min}\right)\times 3.4m\\ & =& 57.6rev/s\left(3.4 m\right)\\ & =& 195.8m/s\end{array}$

Thus, the linear velocity of the blade is 195.8 m/s.

Step 4: (b) Determination of the radial acceleration of the blade tip expressed as a multiple of g:

The radial acceleration of the object is given by,

$a_r=\frac{v^2}{R}$

Here, v is the velocity of the blade and R is the radius of rotor.

Substitute all the values in the above.

$\begin{array}{rcl}{a}_r& =& \frac{{\left(195.8m/s\right)}^2}{3.4m}\\ & =& 11275.8m/s^2\left(\frac{g}{9.8m/s^2}\right)\\ & =& 1.15\times {10}^{3}g\end{array}$

Thus, the radial acceleration of the blade tip is $11275.8m/s^2 or 1.15\times {10}^{3}g$.