Skier at terminal speed

Problem

Explain why a skier reaches a terminal speed on a slope of constant gradient. Identify the forces at play and indicate how to estimate the order of magnitude of $v_\text{lim}$.

Along the slope the component of the weight pushes forward; friction and air resistance oppose it, the latter growing with speed.

Solution

Forces at play. The skier is subject to:

• the component of the weight along the slope, $mg\sin\theta$, which accelerates it downhill (driving, constant);
• the dynamic friction between snow and skis, $F_a = \mu_d N = \mu_d\, mg\cos\theta$, opposing the motion (roughly constant);
• air resistance $F_\text{aria} = \tfrac{1}{2}\rho\, C_d\, A\, v^2$, opposing the motion and increasing with the square of the speed.

Why a terminal speed exists. At low speed air resistance is negligible and the skier accelerates with $a = g(\sin\theta - \mu_d\cos\theta)$ (positive acceleration as long as $\mu_d < \tan\theta$). As $v$ grows, the term $\propto v^2$ becomes increasingly important until the sum of the braking forces equals the driving force: the resultant vanishes, the acceleration goes to zero, and the speed stays constant. This value is the terminal speed (steady-state motion).

Estimating $v_\text{lim}$. At equilibrium the parallel component of the weight balances friction plus air resistance: $mg\sin\theta = \mu_d\, mg\cos\theta + \tfrac{1}{2}\rho\, C_d\, A\, v_\text{lim}^2$ Isolating: $v_\text{lim} = \sqrt{\frac{2mg(\sin\theta - \mu_d\cos\theta)}{\rho\, C_d\, A}}$ With typical values ($m \sim 70\;\mathrm{kg}$, $\theta \sim 25^\circ$, $\mu_d \sim 0{,}05$, $\rho \approx 1{,}2\;\mathrm{kg/m^3}$, $C_d A \sim 0{,}3$–$0{,}5\;\mathrm{m^2}$) one gets an order of magnitude of a few tens of metres per second, that is: $\ev{v_\text{lim} \sim 50\text{--}100\;\mathrm{km/h}}$

Linked items

Topics: Dinamica · Attrito Concepts: Forza peso · Attrito dinamico Skills: Scomposizione sul piano inclinato · Stime di Fermi Objects: Piano inclinato