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Banked tracks

Forces and motion · level 7, lesson 4 of 6

A question to start

Racing tracks and fast railway curves tilt inwards on bends. Why? And if engineers on Mars built a banked curve with the same tilt and radius as one on Earth, would it be built for the same speed?

The idea

On a banked curve, the road tilts towards the inside by an angle \(\theta\). The normal force \(N\) from the road is then tilted too. Its horizontal part, \(N\sin\theta\), points to the centre and can supply the turn, with no friction needed.

With no friction, vertically: \(N\cos\theta = mg\). Horizontally: \(N\sin\theta = mv^2/r\). Divide one by the other:

\[\tan\theta = \frac{v^2}{rg} \qquad\Rightarrow\qquad v_\text{design} = \sqrt{rg\tan\theta}.\]

At this design speed, a vehicle goes round with no sideways friction at all, even on ice. Faster, friction must pull it in; slower, friction must hold it from sliding down.

NmgN sin θθcentre of turn to the left

For the same radius and bank angle, the design speed goes as \(\sqrt g\). A banked bend on Mars is designed for 0.62 times the Earth speed. To design for the same speed on Mars, the bank must be steeper: \(\tan\theta\) must be 2.6 times bigger.

A bend of radius 50 m Mars Earth
Design speed with a 20° bank 8.2 m/s, 30 km/h 13.4 m/s, 48 km/h
Bank needed for 13.4 m/s 44° 20°

Worked examples

1. Mars, 20° bank, 50 m. \(v = \sqrt{50 \times 3.71 \times 0.364} = \sqrt{67.5} = 8.2\) m/s.

2. The bank for 13.4 m/s on Mars. \(\tan\theta = 13.4^2 / (50 \times 3.71) = 179.6/185.5 = 0.968\), so \(\theta = 44°\). That is very steep for a road; on Mars, it is easier to slow down.

3. A cyclist leaning. A cyclist leans into a turn for the same reason: the road's push through the tyres then points partly inwards. The lean angle obeys the same rule, \(\tan\theta = v^2/(rg)\).

Try it

  1. Derive \(\tan\theta = v^2/(rg)\).
  2. A track on Earth has a 30° bank and radius 100 m. What is its design speed?
  3. What would the design speed be for the same track on Mars?
  4. A cyclist on Mars goes round a bend of radius 10 m at 5 m/s. At what angle does she lean?
Answers
  1. \(N\cos\theta = mg\) and \(N\sin\theta = mv^2/r\). Divide: \(\tan\theta = v^2/(rg)\).
  2. \(\sqrt{100 \times 9.81 \times 0.577} = 23.8\) m/s.
  3. \(\sqrt{100 \times 3.71 \times 0.577} = 14.6\) m/s.
  4. \(\tan\theta = 25 / 37.1 = 0.674\), \(\theta = 34°\) from upright. (On Earth she would lean only 14°.)

Common mistake

Drawing friction on a banked curve at the design speed. At exactly the design speed, no friction is needed. Draw only the normal force and the weight.

Where it leads