Q.A cyclist rides in a circle of radius $20\ \text{m}$ at a speed of $5\ \text{m/s}$ on a level road. At what angle to the vertical must the cyclist lean to avoid sliding, and what is the minimum coefficient of friction between the tyres and the road that would allow this turn to be taken safely? (Take $g = 9.8\ \text{m/s}^2$.)

Given: $r = 20\ \text{m}$, $v = 5\ \text{m/s}$, $g = 9.8\ \text{m/s}^2$.\n\nLean angle. For a cyclist relying on leaning (and friction) to turn on a level road, the required lean angle from the vertical satisfies $\tan\theta = \frac{v^2}{rg} = \frac{5^2}{20 \times 9.8} = \frac{25}{196} \approx 0.1276$ $\theta = \tan^{-1}(0.1276) \approx 7.3^\circ$\n\nMinimum coefficient of friction. On a level road, since the road provides no help from banking, friction alone must supply the entire centripetal force needed for this turn, giving the same numerical requirement: $\mu_{\min} = \frac{v^2}{rg} \approx 0.1276 \approx 0.128$ This equality is not a coincidence -- both th …

Apply $\tan\theta = v^2/(rg)$ for the lean angle, and separately note that the minimum required coefficient of friction on a level road, …

• Treating the lean-angle formula and the friction-coefficient formula as needing entirely different derivations, missing that they reduce to the same expression here. …