Acceleration on a bend

Problem

A car enters a bend at constant speed $v$ in magnitude. Explain why it has acceleration even without changing $\abs{v}$, and identify the real force that causes it.

The velocity $\vec v$ is tangent to the path; the acceleration $\vec a_c$ points towards the centre $O$.

Solution

Velocity is a vector. Acceleration is defined as $\vec a = \dd\vec v/\dd t$, i.e. the rate at which the velocity vector changes — not just its magnitude. On a bend the magnitude $\abs{v}$ stays constant, but the direction of $\vec v$ (always tangent to the path) keeps rotating. A vector that changes direction has a non-zero derivative: hence: $\vec a = \frac{\dd\vec v}{\dd t} \neq 0$

Direction and magnitude of the acceleration. This acceleration is purely centripetal: it is perpendicular to $\vec v$ and points towards the centre $O$ of the bend. Its magnitude is: $a_c = \frac{v^2}{r}$

The real force. By Newton’s second law, every acceleration requires a real force in the same direction. Here the centripetal force $F_c = mv^2/r$ is supplied by the static friction between the tyres and the asphalt (the tyres do not slip: they roll, so it is static, not kinetic, friction). It is the lateral grip of the tyres that “pulls” the car towards the inside of the bend.

$\ev{\vec a = \frac{\dd\vec v}{\dd t} \neq 0,\quad a_c = \frac{v^2}{r},\ \text{supplied by tyre–asphalt static friction}}$

If the friction is not enough (bend too tight or too fast, wet surface) the car fails to turn and continues nearly in a straight line: it “skids outward”.

Links

Topics: Dinamica Concepts: Accelerazione centripeta · Moto circolare uniforme · Attrito statico · Forza centripeta