A motorcyclist rides on the inside wall of a vertical…

A motorcyclist rides on the inside wall of a vertical cylinder ("wall of death") of radius $R = 15\ \mathrm{m}$. The coefficient of friction between the tyres and the wall is $\mu = 0.4$. What is the smallest speed at which he does not slide down?

Explanation of Method

Minimum speed on the wall of death.

Step 1: Forces

The wall's normal force points to the axis and is the centripetal force; friction along the wall must hold the weight: $N = \frac{mv^2}{R}$.

\[ f = mg \le \mu N. \]

Step 2: Minimum speed

\[ mg \le \mu\,\frac{mv^2}{R}, \]

\[ v \ge \sqrt{\frac{gR}{\mu}} = \sqrt{\frac{9.81 \cdot 15}{0.4}} = \sqrt{367.9}. \]

\[v_{\min} \approx 19.2\ \mathrm{m/s} \approx 69\ \mathrm{km/h}\]

$N = \tfrac{mv^2}{R}$ $f \le \mu N$ $mg$ $R = 15\ \mathrm{m}$ $\mu\,\frac{mv^2}{R} \ge mg$ $v \ge 19.2\ \mathrm{m/s}$

The wall pushes the rider toward the axis; that push is the centripetal force. The friction it allows, at most $\mu N$, must carry the weight, so the rider needs enough speed to be pressed hard enough.

Result

\[\boxed{\,v_{\min} \approx 19.2\ \mathrm{m/s} \approx 69\ \mathrm{km/h}\,}\]

Remark

Check. The original result is correct. The mass cancels, so the same speed holds for any rider and machine; a rougher wall (larger $\mu$) allows a slower ride.