College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

  • radius, r=100 mr=100\ \text{m}
  • angular velocity, ω=0.5 rev/sec×2π rad1 rev=3.1416 rad/sec\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4 kgm=4\ \text{kg}

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c = m a_c which can also be expressed as

Fc=mv2ror Fc=mrω2F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula Fc=mrω2F_c = mr \omega^2. If we substitute the given values, we have

Fc=mrω2Fc=(4 kg)(100 m)(3.1416 rad/sec)2Fc=3947.8602 NFc=4×103 N  (Answer)\begin{align*} F_c & =mr \omega^2 \\ \\ F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\ F_c & = 3947.8602\ \text{N} \\ \\ F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4×103 N4 \times 10^3\ \text{N}.

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