On a highway near Nashik, a flat bend carries a sign: "Sharp curve, 50 km/h". Most days, cars take it at 60 without trouble. After the first monsoon shower, a car at the same speed slides wide, off the edge of the road and into the soft shoulder. The driver says the car "was thrown outwards". The road and the curve were the same as the day before. Only the water was new.
Only friction can turn the car
Look at a car taking a flat, level bend of radius \(R\) at a steady speed \(v\), seen from the front. Three real forces act on it. Its weight \(mg\) pulls down. The road pushes up with the normal force \(N\). And the road pushes sideways on the tyres with friction \(f\). The car does not rise or sink, so \(N = mg\). The weight and the normal force are vertical, so friction is the only horizontal force. It must do the centripetal job.
The tyres roll; they do not slide over the road sideways. So this friction is static. It is the same static friction that pushes a walking foot forward in the chapter on friction. Here the tyres tend to slide outwards, away from the centre, and friction on them points inwards, against that tendency. Static friction supplies whatever is needed, \(f = mv^2/R\), up to its limit \(\mu_s N = \mu_s mg\).
\[ N = mg \]
Why. the car neither rises nor sinks, so the vertical forces balance.
\[ f = \dfrac{mv^2}{R} \]
Why. friction is the only horizontal force, so it is the centripetal force.
\[ \dfrac{mv^2}{R} \le \mu_s mg \]
Why. static friction cannot exceed its limit \(\mu_s N\).
\[ v_{\max} = \sqrt{\mu_s g R} \]
Read it. the mass cancels. The largest speed makes the two sides equal.
The formula carries three lessons for a driver. The mass has cancelled, so a loaded lorry skids at the same speed as an empty one. A bend of four times the radius allows twice the speed, because \(R\) sits under a square root. And a wet road, with a smaller \(\mu_s\), lowers the safe speed.
A flat bend has a radius of \(50\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). Find the greatest safe speed, in m/s and in km/h, (a) on a dry road with \(\mu_s = 0.8\), (b) on a wet road with \(\mu_s = 0.45\).
(a) \(v_{\max} = \sqrt{0.8 \times 10 \times 50} = \sqrt{400} = 20\,\text{m/s}\), which is \(72\,\text{km/h}\).
(b) \(v_{\max} = \sqrt{0.45 \times 10 \times 50} = \sqrt{225} = 15\,\text{m/s}\), which is \(54\,\text{km/h}\). A little water cuts \(18\,\text{km/h}\) from the safe speed.
What "thrown outwards" really is
Now let the car enter the bend too fast. The turn needs more friction than the road can give. Friction reaches its limit and stays there, pointing towards the centre of the car's own path. A smaller inward force can only turn the car on a wider circle, of radius \(v^2/(\mu g)\) instead of \(R\). So the car runs wide. Its path starts along the lane and curves more gently, until it crosses the outer edge.
In every chip of the figure, friction points inwards and the car keeps turning. Nothing pushes it outwards. It leaves the road because it turns too little, not because a force throws it out. This is the stone of Lesson 1 again. The stone left along the tangent when its inward force vanished. The car curves away from the lane when its inward force falls short.
A scooter takes a flat bend of radius \(20\,\text{m}\). The coefficient of static friction between its tyres and the road is \(0.45\). Take \(g = 10\,\text{m/s}^2\). What is the greatest speed, in m/s, at which it can take the bend without skidding?
On a flat bend the greatest safe speed is \(v\). A second bend on the same road, with the same tyres, has twice the radius. Its greatest safe speed is
Why a cyclist leans
A cyclist taking a bend does not stay upright. He leans in, towards the centre of the turn. Try to take a bend sitting bolt upright, and the cycle tips you outwards. The reason lies in where the road pushes. The road pushes on the tyre at one point, the contact patch. Its push has two parts: \(N\) upwards and friction \(f\) towards the centre. Their sum is the road's total force on the tyre.
Suppose that total force did not point at the rider's centre of mass. It would then turn the rider and the cycle over, as a push at the edge of a book spins it. So the rider leans until the line from the tyre through the centre of mass lies exactly along the road's total force. The angle of the lean from the vertical is then the angle between that total force and \(N\):
\[ \tan\theta = \frac{f}{N} = \frac{mv^2/R}{mg} = \frac{v^2}{Rg} \]
Move the slider from \(7\) to \(12\,\text{m/s}\). The lean grows from about \(14°\) to about \(36°\). Twice the speed would give four times the tangent, because \(\tan\theta\) grows as \(v^2\). At \(10\,\text{m/s}\) on this bend, \(\tan\theta = 100/200 = 0.5\) and the rider leans \(27°\). Notice that the mass has cancelled once again. A heavy rider and a light one lean at the same angle at the same speed.
The lean has a limit. The ratio \(f/N\) is \(\tan\theta\), and friction can be at most \(\mu_s N\). So \(\tan\theta\) can be at most \(\mu_s\). The greatest lean is the angle of friction of the chapter on friction. A rider who leans further, to take the bend faster, asks for more friction than the road can give. The tyres then slide out from under him. On the slider's bend with \(\mu_s = 0.8\), that happens above \(\sqrt{0.8 \times 10 \times 20} \approx 12.6\,\text{m/s}\).
A motorcyclist takes a flat bend of radius \(30\,\text{m}\) at \(54\,\text{km/h}\). Take \(g = 10\,\text{m/s}^2\). At what angle to the vertical, in degrees, must the rider lean?
A cyclist leans into a flat bend at a steady speed. Which statements are true?
A real cycle also steers slightly into the turn and uses the gyroscopic effect of its spinning wheels to fall into the lean smoothly. Those are torques, the subject of the unit on rotational mechanics. The balance of forces here does not depend on them. Once the lean is steady, \(\tan\theta = v^2/(Rg)\) holds for a cycle, a motorcycle or a running athlete.
Motorcycle racers lean so far that a knee brushes the track, at angles above \(55°\). By \(\tan\theta \le \mu_s\), that needs \(\mu_s\) above \(1.4\). Racing tyres, warmed up to become soft and sticky, give it. On a dusty city road, with \(\mu_s\) near \(0.6\), the tyres would slide out at a lean of about \(31°\).
- Using kinetic friction for a car turning without skidding.The tyres roll and do not slide sideways, so the friction is static, with limit \(\mu_s N\).
- Saying a skidding car is thrown outwards along the radius.Friction still points inwards; it is only too small. The car runs wide on a gentler curve.
- Thinking a heavier vehicle can take a flat bend faster.The mass cancels: \(v_{\max} = \sqrt{\mu_s g R}\) for every vehicle.