A bus runs quickly along a rolling road in the Nilgiri hills. At the top of each rise, the children on the back seat lift a little off their seats and laugh; their stomachs seem to float. At the bottom of each dip they are pressed hard into the seat, and the bus groans on its springs. The speed of the bus hardly changes. Nobody pushed the children up or down. Yet the seat pushes on them with very different forces at the top and at the bottom.
A road that curves up and down
Take a car moving at a steady speed \(v\) along a road that rises into a hump and then falls into a dip. The speed does not change, but the direction of the velocity does, because the path is curved. So the car accelerates towards the centre of the curve it is on at that moment. Over a hump that centre lies below the road. In a dip it lies above the road.
Look first at the very top of a hump of radius \(R\). Two forces act on the car along the vertical: its weight \(mg\) down, and the normal force \(N\) of the road up. Both lie along the radius. The engine and the brakes act along the road, so they do not enter the equation along the radius. The centre is below, so take downwards as positive.
\[ mg - N = F_{\text{in}} \]
Why. the net force towards the centre, which is below the car, is the weight minus the upward normal force.
\[ F_{\text{in}} = \dfrac{mv^2}{R} \]
Why. the centripetal requirement for a body at speed \(v\) on a curve of radius \(R\).
\[ N = mg - \dfrac{mv^2}{R} \]
Read it. put step 2 into step 1 and solve for \(N\).
At the bottom of a dip, the centre is above the car. Now the net force must point upwards, so the normal force must be larger than the weight: \(N - mg = mv^2/R\). The road pushes on the car with more than its weight. Over a hump the road pushes with less.
The scale inside
Put a weighing scale on the seat and let a passenger sit on it. The scale reads the push it gives the passenger, divided by \(g\). This is the passenger's apparent weight, exactly as in the lift of the chapter on accelerated objects. The same two formulas hold for the passenger alone, with the passenger's mass. So the reading falls over a hump and rises in a dip.
Check the numbers of the figure at \(10\,\text{m/s}\). The centripetal acceleration is \(v^2/R = 100/20 = 5\,\text{m/s}^2\), half of \(g\). At the top, \(N = m(g - 5) = mg/2\), so the scale shows half of \(60\,\text{kg}\). At the bottom, \(N = m(g + 5) = 1.5\,mg\), so it shows \(90\,\text{kg}\). At \(5\,\text{m/s}\) the change is only a quarter as large, and at \(12\,\text{m/s}\) it is larger still.
Watch the arrow on the sides of the hump, too. There it is even shorter than at the top. On a slope the weight is tilted from the radius, so only part of it, \(mg\cos\phi\), points towards the centre, where \(\phi\) is the tilt of the road. The road need push only \(mg\cos\phi - mv^2/R\). This is why a car driven fast at a steady speed first loses its grip where the hump begins, not only at its top.
A real road is not made of circles. It does not matter. At every point, a curved road has a radius of curvature, the radius of the circle that fits it best there. The chapter on the kinematics of circular motion showed how to find it. Use that radius in the same formulas. A sharp crest has a small radius and a large effect. A gentle one has a large radius and a small effect.
A car of mass \(1000\,\text{kg}\) moves at a steady \(15\,\text{m/s}\). It passes the top of a hump and then the bottom of a dip, each of radius of curvature \(50\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). Find the normal force on the car at each point.
The centripetal acceleration is \(v^2/R = 225/50 = 4.5\,\text{m/s}^2\), so \(mv^2/R = 4500\,\text{N}\). The weight is \(10\,000\,\text{N}\).
At the top of the hump, \(N = 10\,000 - 4500 = 5500\,\text{N}\), a little over half the weight.
At the bottom of the dip, \(N = 10\,000 + 4500 = 14\,500\,\text{N}\). The tyres and springs carry almost one and a half times the weight.
A \(50\,\text{kg}\) student sits on a weighing scale in a car. The car passes the bottom of a dip of radius \(25\,\text{m}\) at \(10\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). What does the scale read, in kilograms?
A car crosses two crests at the same speed. One has a radius of curvature of \(20\,\text{m}\), the other \(80\,\text{m}\). At which crest is the normal force on the car smaller?
How fast is too fast?
The formula \(N = mg - mv^2/R\) gives a smaller normal force at every higher speed. But a road can only push on a car. It cannot pull it down. So \(N\) can fall to zero and no further. When \(N = 0\), the weight alone does the whole centripetal job: \(mg = mv^2/R\). This gives the speed at which the car just loses its contact with the top of the hump.
\[ v = \sqrt{gR} \]
The mass has cancelled. A loaded lorry and a light scooter leave a hump at the same speed. At that speed a passenger on a scale reads zero: he is weightless for an instant, like a person in a freely falling lift. The children in the bus felt a small part of this feeling at the top of each rise.
The figure shows why the limit is \(\sqrt{gR}\). A body with no push from below falls like a projectile. The chapter on the kinematics of circular motion found the radius of curvature of a projectile at the top of its path: \(v^2/g\). If that radius is smaller than \(R\), the free path bends more sharply than the road, and the car presses on the road. If it is larger, the road falls away faster than the car falls. The car leaves the road. The two radii are equal when \(v^2/g = R\).
An old humpback bridge has a radius of curvature of \(40\,\text{m}\) at its top. Take \(g = 10\,\text{m/s}^2\). Above what speed, in km/h, does a car lose contact with the top of the bridge?
A loaded truck three times as heavy as a car crosses the same hump. Compared with the car, the truck leaves the top of the hump at
In a dip there is no such speed limit, because the road can push as hard as needed. The danger in a dip is the opposite one: a very large \(N\). A fast car in a sharp dip can bottom out its springs or burst a tyre. So road builders limit the sharpness of dips as well as crests.
The top of a hump on an amusement-park ride has a radius of curvature of about \(10\,\text{m}\). Estimate the speed, in m/s, at which the riders feel weightless at the top.
The children on the back seat lift off at the top of a rise and are pressed down at the bottom of a dip. Explain why, with the forces on one child.
Only the weight \(mg\) and the seat's push \(N\) act on the child along the vertical. At the top of a rise the centre of the curve is below the child. So the net force points down and equals \(mv^2/R\), and \(N = mg - mv^2/R\), less than the weight. If the bus is fast enough, \(N\) would have to be negative, and since the seat cannot pull, the child leaves it. At the bottom of a dip the centre is above, so \(N = mg + mv^2/R\), more than the weight, and the child is pressed into the seat.
- Writing \(N = mg\) at the top of a hump because the speed is constant.A constant speed on a curve still means an acceleration \(v^2/R\) towards the centre. Write the net force along the radius.
- Getting the sign wrong in the dip.Ask where the centre is. Below the car: \(mg - N = mv^2/R\). Above the car: \(N - mg = mv^2/R\).
- Letting \(N\) become negative.A road only pushes. When the formula gives \(N < 0\), the car has already left the road; the limit is \(v = \sqrt{gR}\).