Banking of roads — derivation: Worked examples

This child page puts the parent derivation to work. We won't re-derive the formulas — we will use them on every kind of situation the topic can throw at you. Before touching numbers, let's map the whole battlefield so no scenario surprises you.

Recall The three formulas we will reuse (from the parent)

Design (frictionless): $tan θ = \frac{v ^{2}}{r g}$ , so $v_{0} = r g tan θ$ . Maximum with friction: $v_{ma x} = r g \frac{μ + tan θ}{1 - μ tan θ}$ . Minimum with friction: $v_{min} = r g \frac{tan θ - μ}{1 + μ tan θ}$ . Here $g = 9.8 m/s^{2}$ throughout unless a problem says otherwise.

The scenario matrix

Every banking problem lives in one of these cells. Each cell asks a different physical question, so each needs its own worked example.

Cell	What is special about it	Physical question
A. Pure design	$μ = 0$ , ask for the one perfect speed	What is $v_{0}$ ?
B. Max with friction	$μ > 0$ , going too fast	How fast before flying out?
C. Min with friction	$μ > 0$ , going too slow	How slow before slipping in?
D. Degenerate: flat road	$θ = 0$	Does the formula collapse to the friction limit?
E. Degenerate: $tan θ \leq μ$	very steep OR very rough	Can the car stand still ( $v_{min} = 0$ )?
F. Limiting: $μ tan θ \to 1$	denominator $\to 0$	Why does $v_{ma x} \to \infty$ ?
G. Inverse problem	given $v$ , find $θ$	Design a road for a target speed
H. Real-world word problem	km/h, real curve	Is the posted limit safe?
I. Exam twist	friction direction ambiguous	Which way does $f$ point, and why?

We now hit all nine cells with 9 examples.

Cell A — Pure design speed

Cell B — Maximum speed with friction

Cell C — Minimum speed with friction

Cell D — Degenerate: flat road ( $θ = 0$ )

Cell E — Degenerate: $tan θ \leq μ$ (car can stand still)

Cell F — Limiting: $μ tan θ \to 1$ (denominator vanishes)

Cell G — Inverse problem (design a road)

Cell H — Real-world word problem

Cell I — Exam twist: which way does friction point?

This is the classic trap. The direction of friction is not fixed — it depends on whether the given speed is above or below the design speed. Look at the figure.

Worked example I: decide friction's direction, then solve

A curve has $r = 100 m$ , $θ = 2 0^{\circ}$ , $μ = 0.30$ . A car moves at $v = 30 m/s$ . Which way does friction point, and is the car safe?

Forecast: First find the design speed. If $30 m/s$ is above it, the car tends outward → friction down-slope; use the max comparison.

Design speed: $v_{0} = r g tan 2 0^{\circ} = 100 \times 9.8 \times 0.36397 = 356.7 = 18.89 m/s$ . Why first? It's the pivot: below it friction points up, above it friction points down.
Since $30 > 18.89$ , the car tends to slide up/outward, so friction acts down the slope (red arrow in the figure). Why? Friction opposes the tendency of relative sliding, not gravity.
Check the ceiling: $v_{ma x} = r g \frac{μ + tan θ}{1 - μ tan θ} = 980 \frac{0.30 + 0.36397}{1 - 0.30 \times 0.36397}$ .
Ratio $= \frac{0.66397}{0.89081} = 0.74535$ ; $v_{ma x} = 980 \times 0.74535 = 730.4 = 27.03 m/s$ .

Verify: $30 m/s > 27.03 m/s = v_{ma x}$ → the car exceeds the maximum and would skid outward. Unsafe. Direction call (down-slope) confirmed because $v > v_{0}$ . ✓

Recall check

Friction points which way when

v > v_{0}

Down the slope (car tends to slide outward/up).

Friction points which way when

v < v_{0}

Up the slope (car tends to slip inward/down).

When is

v_{min} = 0

When

tan θ \leq μ

— friction alone can hold the car even at rest.

Why does

v_{ma x} \to \infty

μ tan θ \to 1

The denominator

1 - μ tan θ \to 0

; friction's inward grip grows with

N

faster than any speed can overpower.

How do you find the banking angle for a target design speed?

θ = arctan \frac{v ^{2}}{r g}

Connections

Centripetal force — every example balances forces to supply $m v^{2} / r$ .
Friction — sets the $v_{min}$ – $v_{ma x}$ range in cells B, C, E, F, I.
Uniform Circular Motion — origin of the $v^{2} / r$ requirement.
Inclined plane — same resolve-into-axes method used here.
Conical pendulum — shares the design formula $tan θ = v^{2} / r g$ .
Newton's Second Law — applied per axis in each solution.
Back to the parent derivation.