An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

Solution:

The rotation angle \Delta \theta is defined as the ratio of the arc length to the radius of curvature:

\Delta \theta = \frac{\Delta s}{r}

where arc length \Delta s is distance traveled along a circular path and r is the radius of curvature of the circular path.

From the given problem, we are given the following quantities: r=0.260 \ \text{m}, and \Delta s = 80000 \ \text{km}.

\begin{align*}
\Delta \theta & = \frac{\Delta s}{r} \\
\\
\Delta \theta & = \frac{80000 \ \text{km} \times \frac{1000 \ \text{m}}{1 \ \text{km}}}{0.260 \ \text{m}} \\
\\
\Delta \theta & = 307.6923077 \times 10^{6} \ \text{radians} \times\frac{1 \ \text{rev}}{2\pi \  \text{radians}} \\
\\
\Delta \theta & = 48970751.72 \ \text{revolutions}  \\
\\
\Delta \theta & = 4.90 \times 10^{7} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}