College Physics Problem: The centripetal acceleration at the tip of a helicopter blade

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

We are given the following values: $r=4.00\ \text{m}$ , and $\omega = 300 \ \text{rev/min}$ .

Let us convert the angular velocity to unit of radians per second.

\omega = 300 \ \frac{\text{rev}}{\text{min}} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}}\times \frac{1\ \text{min}}{60 \ \text{sec}} = 31.4159 \ \text{rad/sec}

The centripetal acceleration at the tip of the helicopter blade can be computed using the formula

a_{c} = r \omega ^2

If we substitute the given values into the formula, we have

\begin{align*} a_{c} & = r \omega^2 \\ \\ a_{c} & = \left( 4.00\ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 3947.8351 \ \text{m/s}^2 \\ \\ a_{c} & = 3.95 \times10^3 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We are asked to solve for the linear velocity of the blade’s tip. We are going to use the formula

v=r \omega

We just needed to substitute the given values into the formula.

\begin{align*} v & = r \omega \\ \\ v & = \left( 4.00 \ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right) \\ \\ v & = 125.6636 \ \text{m/s} \\ \\ v & = 126 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Let us compare this with the speed of light which is 340 m/s.

\frac{125.6636 \ \text{m/s}}{340\ \text{m/s}} \times 100 \%= 36.9599 \% =37.0\%

The linear velocity of the blades tip is 37.0% of the speed of light.