Centripetal Acceleration covered circles at steady speed, where $a = v^2/r$ points to the center. Position, Velocity and Acceleration Vectors showed that acceleration can point anywhere relative to the velocity. Here we split it into its two meaningful parts.
Key idea
On a curved path, acceleration has two perpendicular parts: $$a_t = \dfrac{dv}{dt} \text{ (along the path)}, \qquad a_c = \dfrac{v^2}{r} \text{ (toward the center of curvature)},$$ and $|\vec{a}| = \sqrt{a_t^2 + a_c^2}$. The tangential part changes the speed; the centripetal part changes the direction.
Why it works
Write the velocity as speed times a unit vector along the path, $\vec{v} = v\,\hat{T}$. The product rule gives $$\dfrac{d\vec{v}}{dt} = \dfrac{dv}{dt}\hat{T} + v\,\dfrac{d\hat{T}}{dt}.$$ The first term lies along the path. For the second, on a circle of radius $r$ the unit vector $\hat{T}$ turns at the angular rate $v/r$, and a turning unit vector changes at that rate in the perpendicular direction, toward the center. So the second term has size $v \cdot v/r = v^2/r$.
Worked example: A car speeding up on a curve
A car on a curve of radius $50\ \text{m}$ has speed $v(t) = 10 + 2t$ (m/s, s). Find the size of its acceleration at $t = 5\ \text{s}$.
- $a_t = dv/dt = 2\ \text{m/s}^2$.
- At $t = 5$: $v = 20\ \text{m/s}$, so $a_c = 400/50 = 8\ \text{m/s}^2$.
- $|\vec{a}| = \sqrt{2^2 + 8^2} = \sqrt{68} = 8.25\ \text{m/s}^2$.
Answer: $8.25\ \text{m/s}^2$.
Worked example: Distance along the path
A bead moves on a circle of radius $8\ \text{m}$ with arc length $s(t) = t^3$ (m, s). Find $a_t$, $a_c$ and $|\vec{a}|$ at $t = 2\ \text{s}$.
- Speed: $v = ds/dt = 3t^2 = 12\ \text{m/s}$ at $t = 2$.
- $a_t = dv/dt = 6t = 12\ \text{m/s}^2$.
- $a_c = v^2/r = 144/8 = 18\ \text{m/s}^2$.
- $|\vec{a}| = \sqrt{144 + 324} = 21.6\ \text{m/s}^2$.
Answer: $a_t = 12$, $a_c = 18$, $|\vec{a}| \approx 21.6\ \text{m/s}^2$.
Common mistake
Taking the whole acceleration to be $v^2/r$ while the speed is changing. For the car above that gives $8\ \text{m/s}^2$ instead of $8.25\ \text{m/s}^2$, and the direction is wrong too: the true acceleration tilts forward of the center. Check: if the speedometer reading is changing, $a_t \neq 0$ and the acceleration can't point straight at the center.
The angle between $\vec{a}$ and $\vec{v}$ tells the story at a glance: under $90^\circ$ the object is speeding up, exactly $90^\circ$ means steady speed, and over $90^\circ$ means slowing down. A car braking into a bend has an acceleration that points backward and inward at once.
The same split works on any curve, with $r$ the radius of the circle that best fits the path at that point (the radius of curvature). At the top of a projectile's arc, for example, $a_t = 0$ and $a_c = g$, so $r = v^2/g$ there.