The ideal banking angle of a curve on a highway
Problem:
What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?
Solution:
The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:
\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)
We are given the following values:
- radius of curvature, \displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
- linear velocity, \displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}
- acceleration due to gravity, \displaystyle g = 9.81\ \text{m/s}^2
If we substitute these values into our formula, we come up with
\begin{align*} \theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 4.1333 ^\circ \\ \\ \theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
The ideal banking angle for the given highway is about 4.13 ^\circ.
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