Sit by the window of the Rajdhani Express as it sweeps round a long curve at speed. The coach tilts a little towards the inside of the curve. Your cup of tea stays put on the folding table, and you do not feel pushed towards the window. Stand at a level crossing near such a curve and look along the track: the outer rail is laid higher than the inner one. Expressway curves are built the same way, tilted towards the inside.
Let the normal force do the work
On a flat bend, friction alone turns the car, and friction is unreliable: rain, oil or gravel can take it away. Engineers therefore tilt the road on a bend, raising its outer edge above its inner edge. This is called banking, and the angle of the road's surface to the horizontal is the angle of banking \(\theta\). The normal force is perpendicular to the road, so on a banked road it leans towards the centre of the turn.
Take a car on a road banked at \(\theta\), turning on a circle of radius \(R\), and suppose for now that there is no friction. Two forces act: the weight \(mg\) down, and the normal force \(N\), tilted from the vertical by \(\theta\) towards the centre. Resolve \(N\) along the vertical and the horizontal. Take the horizontal axis towards the centre, because the car accelerates towards the centre, not along the slope.
\[ N\cos\theta = mg \]
Why. the car does not rise or sink, so the vertical part of \(N\) balances the weight.
\[ N\sin\theta = \dfrac{mv_0^2}{R} \]
Why. the horizontal part of \(N\) is the only force towards the centre, so it is the centripetal force.
\[ \tan\theta = \dfrac{v_0^2}{Rg} \]
Read it. divide step 2 by step 1. Both \(N\) and \(m\) cancel.
For a given bend and a given angle, only one speed satisfies this. It is the ideal speed \(v_0 = \sqrt{Rg\tan\theta}\), also called the rated speed. At the ideal speed the tyres need no sideways friction at all, so they would take the bend even on ice. The mass has cancelled, so the ideal speed is the same for a scooter and a loaded truck.
Check the numbers of Fig 5.1. The road has \(R = 30\,\text{m}\) and \(\tan 37° = 0.75\), so \(v_0 = \sqrt{30 \times 10 \times 0.75} = \sqrt{225} = 15\,\text{m/s}\). The car needs \(mv_0^2/R = 1000 \times 225/30 = 7500\,\text{N}\) towards the centre, which is exactly \(N\sin\theta\). The normal force \(N = mg/\cos\theta = 12\,500\,\text{N}\) is larger than the weight, because it must also turn the car.
An expressway curve has a radius of \(250\,\text{m}\), and the traffic on it moves at about \(72\,\text{km/h}\). Take \(g = 10\,\text{m/s}^2\). At what angle should it be banked so that the cars need no sideways friction?
\(v_0 = 72/3.6 = 20\,\text{m/s}\). \(\tan\theta = v_0^2/(Rg) = 400/2500 = 0.16\), so \(\theta \approx 9°\).
Road curves are banked by a few degrees at most. A steep bank would make slow or stopped vehicles slide inwards.
A curve of radius \(90\,\text{m}\) is banked so that \(\tan\theta = 0.25\). Take \(g = 10\,\text{m/s}^2\). At what speed, in m/s, can a car take it with no friction?
A student resolves the weight along the slope, as for a block on an incline. For the car at its ideal speed the student writes \(N = mg\cos\theta\). What is wrong?
Other speeds: friction comes back
Real traffic does not all move at the ideal speed. A car slower than \(v_0\) needs less inward force than \(N\sin\theta\) gives. So it tends to slide down the slope, towards the inside of the bend. Static friction then points up the slope. A car faster than \(v_0\) needs more inward force; it tends to slide up the slope, outwards, and friction points down the slope. This is the band of pulls on a rough incline from the chapter on friction, with the speed in place of the pull.
Move the slider in the sim. At \(15\,\text{m/s}\) no friction is needed. At \(12\,\text{m/s}\) the turn needs \(2.2\,\text{kN}\) up the slope, within the limit of \(2.7\,\text{kN}\). At \(19\,\text{m/s}\) it needs \(3.6\,\text{kN}\) down the slope, within the limit of \(3.8\,\text{kN}\). At \(8\,\text{m/s}\) or \(22\,\text{m/s}\) the need passes the limit, and the car slides in or skids out. The limit itself changes with speed, because \(N\) grows as the car goes faster.
At the greatest speed the car is about to slide up the slope. So friction is at its limit \(\mu_s N\) and points down the slope. Resolve it too. Down the slope means partly towards the centre and partly downwards.
\[ N\cos\theta - \mu_s N\sin\theta = mg \]
Why. vertically, the normal force pushes up and the vertical part of the down-slope friction pulls down; together they balance the weight.
\[ N\sin\theta + \mu_s N\cos\theta = \dfrac{mv_{\max}^2}{R} \]
Why. horizontally, both the normal force and the down-slope friction point towards the centre; together they are the centripetal force.
\[ \dfrac{v_{\max}^2}{Rg} = \dfrac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} \]
Why. divide step 2 by step 1. \(N\) and \(m\) cancel.
\[ v_{\max}^2 = Rg\,\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} \]
Read it. divide the top and the bottom by \(\cos\theta\).
?
Your turn. at the least speed the car is about to slide down the slope. Which way does friction point, and what is the vertical balance?
\[ N\sin\theta - \mu_s N\cos\theta = \dfrac{mv_{\min}^2}{R} \]
Why. horizontally, the up-slope friction has a part pointing away from the centre, which subtracts from \(N\sin\theta\).
\[ v_{\min}^2 = Rg\,\dfrac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} \]
Read it. divide step 2 by step 1 and divide the top and the bottom by \(\cos\theta\).
Put \(\mu_s = 0\) into both limits, and each becomes \(Rg\tan\theta\): the band closes to the single ideal speed. Put \(\theta = 0\) into the upper limit, and it becomes \(\mu_s Rg\), the flat road of Lesson 4. The general formula contains both earlier results.
For the curve of the sim, \(R = 30\,\text{m}\), \(\tan\theta = 0.75\) and \(\mu_s = 0.25\). Take \(g = 10\,\text{m/s}^2\). Find the least and the greatest safe speeds.
\(Rg = 300\,\text{m}^2/\text{s}^2\). The two products are \(\mu_s\tan\theta = 0.1875\), so \(1 - \mu_s\tan\theta = 0.8125\) and \(1 + \mu_s\tan\theta = 1.1875\).
\(v_{\min}^2 = 300 \times \dfrac{0.75 - 0.25}{1.1875}\) and \(v_{\max}^2 = 300 \times \dfrac{0.75 + 0.25}{0.8125}\).
\(v_{\min}^2 \approx 126\), so \(v_{\min} \approx 11.2\,\text{m/s}\). \(v_{\max}^2 \approx 369\), so \(v_{\max} \approx 19.2\,\text{m/s}\).
The car holds its line from about \(40\,\text{km/h}\) to about \(69\,\text{km/h}\). The ideal speed, \(54\,\text{km/h}\), lies inside the band.
A curve of radius \(55\,\text{m}\) is banked at \(37°\), so \(\tan\theta = 0.75\), and \(\mu_s = 0.5\). Take \(g = 10\,\text{m/s}^2\). What is the least speed, in m/s, at which a car can go round without sliding down?
A truck takes a banked curve slower than its ideal speed and does not slide. Which way does friction from the road act on its tyres?
Railway curves
A train on a level curve has no sideways friction to rely on. The flange, the rim on the inner side of each wheel, would press against the outer rail. The outer rail's push on the flanges would then be the centripetal force. That push is large for a heavy train. It wears both the rails and the wheels, and it can even lift a wheel over the rail.
So the outer rail is laid higher than the inner rail by a height \(h\) called the cant. With the rails a distance \(d\) apart, the track is banked at an angle whose sine is \(h/d\). The angle is small, so \(\sin\theta \approx \tan\theta\), and the design speed of the curve satisfies \(\tan\theta = v^2/(Rg)\). This gives the cant:
\[ h \approx d\,\frac{v^2}{Rg} \]
The rails of a broad-gauge track are about \(1.68\,\text{m}\) apart. A curve has a radius of \(1000\,\text{m}\), and trains take it at \(90\,\text{km/h}\). Take \(g = 10\,\text{m/s}^2\). By how much should the outer rail be raised?
\(v = 90/3.6 = 25\,\text{m/s}\). \(\tan\theta = v^2/(Rg) = 625/10\,000 = 0.0625\).
\(h \approx d\tan\theta = 1.68 \times 0.0625 \approx 0.105\,\text{m}\), about \(10\,\text{cm}\). Faster trains or sharper curves need more.
On a metre-gauge line the rails are \(1\,\text{m}\) apart. A curve of radius \(225\,\text{m}\) is designed for trains at \(15\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). By how many centimetres should the outer rail be raised?
Now return to the tea on the Rajdhani. The curve is canted for the train's speed. At that speed, the rails' push on the coach is tilted just enough to supply \(mv^2/R\) by itself. The same is true for the table's push on the cup. So the cup needs no sideways friction, and it does not slide. On the expressway, the tilt does the same job for cars at the design speed.
A railway curve carries fast express trains and slow goods trains. A cant chosen for the express would make a slow goods train press on the inner rail, and a stopped train would lean inwards. So Indian Railways choose a cant for an average speed and limit the "cant deficiency" for faster trains. Road engineers face the same choice between fast cars and slow, heavy trucks, which is why road banking stays small.
- Writing \(N = mg\cos\theta\) for a car turning on a banked road.The car accelerates horizontally. Balance the vertical forces: \(N\cos\theta = mg\) at the ideal speed.
- Drawing friction up the slope at every speed.Compare with the ideal speed. Slower: friction up the slope. Faster: down the slope. At \(v_0\): none.
- Swapping the signs in the band of speeds.At the greatest speed friction helps turn the car: \(\tan\theta + \mu_s\) on top. At the least speed it opposes: \(\tan\theta - \mu_s\).