Centripetal Acceleration — Roller Coaster Hill (Worked Example)

Medium

Problem

A roller coaster crosses the top of a hill (radius of curvature 12 m) at 18 m/s. What is its centripetal acceleration there?

Given

v = 18 m/s
r = 12 m

Find

Centripetal acceleration a.

Principle

Over the hilltop the track curves downward, so the centre of the circular path is below — acceleration points down, reducing the normal force (the 'airtime' feeling).

Formula

a = \frac{v^2}{r}

a = v²/r

Solution

Write the formula: $a = v^2/r$
Substitute: $a = 18^2 / 12$
Calculate: $a = 324 / 12 = 27\ \mathrm{m/s^2}$ downward

27 m/s² downward

Unit check

(m/s)² ÷ m = m/s² — correct.

What this means physically

27 m/s² downward exceeds g (9.8 m/s²) — the track must pull down on the car (via the wheels' grip) or it would fly off. Designers tune hill shapes so this stays within safe limits.

Common mistake

Pointing the acceleration upward. Centripetal always aims at the circle's centre — over a hilltop, that is below the car.