Magnitude versus vector

Problem

In uniform circular motion the magnitude $\abs{v}(t)$ is constant while the vector $\vv v(t)$ traces a circle in the plane. Sketch the two graphs and discuss why “constant magnitude” does not mean “zero acceleration”.

Solution

Graph of the magnitude $\abs{v}(t)$. In uniform circular motion the magnitude of the velocity never changes: in the $\abs{v}$–$t$ plane this gives a horizontal line at height $\abs{v} = \omega R$, constant in time.

Graph of the vector $\vv v(t)$. If we plot the velocity vector in the “velocity plane” ($v_x$, $v_y$), its tip traces a circle of radius $\abs{v}$: the components oscillate as $\vv v(t) = \big(-\omega R \sin(\omega t),\; \omega R \cos(\omega t)\big)$ The magnitude stays constant, but the direction keeps rotating.

Why the acceleration is not zero. Acceleration is the derivative of the vector velocity, not of its magnitude: $\vv a = \frac{\dd \vv v}{\dd t}$ Even though $\abs{v}$ is constant, $\vv v$ changes direction, so $\dd\vv v/\dd t \ne 0$. Differentiating shows that $\vv a$ is perpendicular to $\vv v$ (points towards the centre) and has magnitude $\abs{a} = \omega\,\abs{v} = \frac{\abs{v}^2}{R} = \omega^2 R$ This is precisely the centripetal acceleration. Moral: “constant speed” in the scalar sense (magnitude) does not imply a constant velocity vector; it is the change in direction that generates the acceleration.

$\ev{\abs{v}\text{ constant (horizontal line)}; \quad \vv v(t)\text{ traces a circle}; \quad \frac{\dd\vv v}{\dd t} \perp \vv v,\; \abs{a} = \omega\abs{v}}$

Linked items

Topics: Dinamica Concepts: Moto circolare uniforme · Accelerazione centripeta · Velocità angolare Skills: Scomposizione vettoriale