How far can a cyclist lean (tilt the bicycle from the…

How far can a cyclist lean (tilt the bicycle from the vertical) when riding through a curve without the risk of falling? The coefficient of friction between the tyres and the road is $\mu$.

Explanation of Method

How far a cyclist can lean in a curve.

Step 1: Forces from the road

The road pushes up with $N = mg$ and sideways, toward the centre of the curve, with friction $f = mv^2/r$. For the cyclist not to topple, their resultant must pass through the centre of mass, i.e. point along the bicycle. With $\beta$ the lean from the vertical:

\[ \tan\beta = \frac{f}{N}. \]

Step 2: Limit from friction

\[ f \le \mu N \;\Longrightarrow\; \tan\beta \le \mu. \]

\[\beta_{\max} = \arctan\mu \ \text{from the vertical}\]

Equivalently, the bicycle must make at least $90^\circ - \arctan\mu$ with the road. For example, $\mu = 0.5$ allows a lean of up to 26.6° from the vertical.

$N$ $f \le \mu N$ $mg$ $\beta$ $N + f$ $\tan\beta = \frac{f}{N} \le \mu$ $\beta_{\max} = \arctan\mu$

A cyclist leaning in a curve (seen from behind, centre of the curve to the left). The road's total force $N + f$ must point along the bicycle through the centre of mass; its tilt is limited by $f \le \mu N$.

Result

\[\boxed{\,\beta_{\max} = \arctan\mu \ \text{from the vertical}\,}\]

Remark

About the original solution. The result $\arctan\mu$ is correct as the lean from the vertical, but the derivation ($N = mg\cos\theta$, $f = mg\sin\theta$) is that of a block on an inclined plane, not of a cyclist on a level road. The statement's "moving in a straight line" must mean riding through a curve at constant speed: on a straight road there is no sideways force, and a leaning cyclist would simply fall over.