# 6.1.4 Derivation of v2/r (beyond syllabus)

The H2 syllabus does not require students to derive $\displaystyle \frac{{{{v}^{2}}}}{r}$. But the derivation was an eye-opener to me and I still remember the sense of wonder when I first read it when I was 17 years old.

Let’s consider an object travelling at constant speed v along a circular arc of radius r. The object travels from s1 to s2 in time duration $\Delta t$, undergoing a displacement of $\Delta s$ and an angular displacement of $\Delta \theta$.

Since we are geniuses, we detect two isosceles triangles both subtended by angle Δθ. The first one, which I call the displacement triangle, is formed by the displacement vectors s1 and s2. It has two equal sides r, and the third side represents Δs. The second triangle, which I call the velocity triangle, is formed by the velocity vectors v1 and v2. It has two equal side v, and the third side represents the change in velocity Δv. (Think vector subtraction)

Since these two triangles are similar triangles, we can write down $\displaystyle \frac{{\Delta s}}{r}=\frac{{\Delta v}}{v}$

Plot twist. Since we are looking for the instantaneous acceleration, we should be looking at an infinitesimally small Δt. Meaning an infinitesimally small Δθ. Meaning our isosceles triangles should actually be needle-shaped wedges, with Δθ approaching 0°, and the other two angles approaching 90°.

Let’s examine the velocity triangle first. Do you realize that as Δθ approaches 0°, Δv becomes perpendicular to velocity? Since acceleration is in the direction of velocity change, this confirms that the acceleration (during uniform circular motion) is in the centripetal direction.

We now examine the displacement triangle. Do you realize that as Δθ approaches zero, the circular arc vΔt becomes indistinguishable from the chord Δs?

Going back to earlier equation, and substituting Δs for vΔt, we have \displaystyle \begin{aligned}\frac{{\Delta s}}{r}&=\frac{{\Delta v}}{v}\\\frac{{v\Delta t}}{r}&=\frac{{\Delta v}}{v}\\\frac{{\Delta v}}{{\Delta t}}&=\frac{{{{v}^{2}}}}{r}\\{{a}_{c}}&=\frac{{{{v}^{2}}}}{r}\end{aligned}

Isn’t it awesome? To look for the instantaneous $\displaystyle \frac{{dv}}{{dt}}$, we look for $\displaystyle \displaystyle \frac{{\Delta v}}{{\Delta t}}$ when $\displaystyle \displaystyle \Delta t\to 0$, which inevitably means that $\Delta v\to 0$. So $\displaystyle \frac{{\Delta v}}{{\Delta t}}$ is a near zero number, divided by another near zero number. The outcome is neither zero nor infinity. It is $\displaystyle \frac{{{{v}^{2}}}}{r}$. Welcome to circular motion.

Video Explanation

The Similar Triangle Proof for v2/r