Circular Motion — Uniform

Parametrise the position of a body moving anticlockwise on a circle of radius $r$ centred at the origin:

Differentiating once gives the velocity, and again gives the acceleration:

The speed is $|\dot{\mathbf{r}}| = r\omega = v$, so $\omega = v/r$. The acceleration has magnitude $|\ddot{\mathbf{r}}| = r\omega^2 = v^2/r$ and direction $-\mathbf{r}/|\mathbf{r}|$ — pointing back toward the centre. Hence:

Worked through the hook. Car on a flat bend, $r = 50$ m, $v = 15$ m s$^{-1}$. Centripetal acceleration $a = v^2/r = 225/50 = 4.5$ m s$^{-2}$, directed horizontally toward the centre. On a flat (unbanked) road only one force has a horizontal component: friction. So static friction between tyres and road supplies the centripetal force $f = ma = 4.5m$ N.

$\mathbf{r}(t) = (r\cos\omega t, r\sin\omega t) \Rightarrow \ddot{\mathbf{r}} = -\omega^2 \mathbf{r}$ (toward centre) · $\omega = v/r$ (rad s$^{-1}$); $a = v^2/r = \omega^2 r$; $T = 2\pi/\omega$ · Radial Newton's law: $\sum F_{\text{radial}} = mv^2/r$ · Direction of $a$ is toward the centre, always

Pause — copy $\omega = v/r$, $a = v^2/r = \omega^2 r$, $T = 2\pi/\omega$, $\ddot{\mathbf{r}} = -\omega^2\mathbf{r}$ (centripetal toward centre), and the radial Newton's law $\sum F_{\text{radial}} = mv^2/r$ into your book.

We just saw that $\mathbf{r}(t) = (r\cos\omega t, r\sin\omega t)$ gives $\ddot{\mathbf{r}} = -\omega^2 \mathbf{r}$ (centripetal, toward centre), with $a = v^2/r = \omega^2 r$ and the radial Newton's law $\sum F_{\text{radial}} = mv^2/r$. That raises a question: when a road is banked at angle $\theta$, how do we find the design speed where no friction is needed? This card answers it → resolve $N$: vertical $N\cos\theta = mg$, radial $N\sin\theta = mv^2/r$; divide to get $\tan\theta = v^2/(rg)$, so $v = \sqrt{rg\tan\theta}$.

On a flat road, friction supplies the centripetal force — risky in wet conditions. A banked road tilts inward at an angle $\theta$. The normal reaction $N$ now has a horizontal component pointing toward the centre, and at the right speed this alone supplies the centripetal force (no friction needed).

Free-body diagram: forces on the car are weight $mg$ down and normal reaction $N$ perpendicular to the road surface. Resolve $N$ into horizontal ($N\sin\theta$, toward centre) and vertical ($N\cos\theta$, up) components.

Divide the radial equation by the vertical one:

This is the design speed equation: at speed $v = \sqrt{rg\tan\theta}$, the car rounds the bend with no sideways friction at all. Slower and friction must act outward (up the slope); faster and friction must act inward (down the slope).

Banked road: $N\cos\theta = mg$, $N\sin\theta = mv^2/r$ · Divide: $\tan\theta = v^2/(rg)$ · Design speed: $v = \sqrt{rg\tan\theta}$ — no friction needed · Only horizontal component of $N$ is centripetal — resolve carefully

Pause — copy the banked-road equilibrium equations ($N\cos\theta = mg$, $N\sin\theta = mv^2/r$), the design-speed formula $v = \sqrt{rg\tan\theta}$, and the note that only the horizontal component of $N$ is centripetal into your book.