College Physics by Openstax Chapter 6 Problem 25

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:

θ=tan⁡−1(v2rg)\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

  • radius of curvature, r=1.20 km×1000 m1 km=1200 m\displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
  • linear velocity, v=105 km/h×1000 m1 km×1 h3600 s=29.1667 m/s\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, g=9.81 m/s2\displaystyle g = 9.81\ \text{m/s}^2

If we substitute these values into our formula, we come up with

θ=tan⁡−1(v2rg)θ=tan⁡−1[(29.1667 m/s)2(1200 m)(9.81 m/s2)]θ=4.1333∘θ=4.13∘  (Answer)\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∘ 4.13 ^\circ.

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