Physics · Mechanics · Grade 7

Roller coaster

Why doesn't the car fall off at the top of the loop?

The situation

At an amusement park roller coaster, a car is pulled back up a 40-metre tower and released. The car has no engine — only gravity moves it. It plunges down, goes round a vertical loop upside down, then climbs another hill. Why don't the passengers fall out at the very top of the loop, and from what minimum height must the car be released?

Topic

Mechanical work and energy. Fairground hill: potential and kinetic energy

Key ideas

Conservation of mechanical energy
Centripetal force
Work done against friction

Formulas

\frac{m v^{2}}{2} + m g y + Q = m g h, Q = \int (μ N + \frac{1}{2} ρ C_{x} S v^{2}) d s

Conservation of energy: kinetic + potential + heat = initial energy. The work done against friction and air resistance turns into heat.

v = \sqrt{2 g (h - y)}

Speed at any height without friction — the mass does not appear in the formula.

N = m (g \cos θ + v^{2} κ)

Normal reaction force: $θ$ is the slope of the track, $κ = 1 / r$ is the curvature of the track.

N_{t o p} = m (\frac{v^{2}}{R} - g) \geq 0 \Rightarrow h \geq 2.5 R

Condition for passing the top of the loop without falling.

A = F s \cos α, \sum A = Δ E_{k}, P = \frac{A}{t} = F v \cos α, F_{s} = - \frac{d U}{d s}

Work, the work–kinetic energy theorem, power, and the link between a conservative force and potential energy.

Theory

Energy conversion

At the top of the tower the car has only potential energy $m g h$ . As it descends this turns into kinetic energy, and it returns again as the car goes round the loop and climbs the hill. Friction and air resistance turn a little energy into heat every metre — in the columns on the right the total always stays equal to $m g h$ .

What holds the car in the loop?

At the top of the loop both the gravitational force and the rail's push $N$ point downwards — towards the centre. Their sum gives the centripetal force: $m g + N = m v^{2} / R$ . If the speed is $v^{2} < g R$ , $N$ comes out negative — the rail would have to pull the car. An ordinary car would fall off in that case. Real rides have wheels that grip the underside of the rail (up-stop wheels); our model also holds the car on the rail, but it gives a warning when $N < 0$ . From $v^{2} = 2 g (h - 2 R) \geq g R$ it follows that $h \geq 2.5 R$ .

Why doesn't mass matter?

Both the gravitational force and the friction force $μ N$ are proportional to the mass, so the acceleration $a = F / m$ does not depend on the mass. Only air resistance $\frac{1}{2} ρ C_{x} S v^{2}$ does not depend on the mass — that is why, with air resistance switched on, a heavy car is a little faster.

Model simplifications

The car is a point mass on the rail: its position is determined by the distance $s$ travelled along the track. The track consists of straight lines and circular arcs (the curvature is known exactly), and the loop is an exact circle. Friction is $F = μ | N |$ , air resistance $C_{x} S = 0.9$ m², $ρ = 1.2$ kg/m³. The loop is shown shifted sideways by 2.4 m (so that the track does not cross itself) — this does not affect the physics. The brake is a constant deceleration.

Work and power

If a force moves a body through a distance $s$ , it does work: $A = F s \cos α$ (J). On the way down the gravitational force acts along the motion — its work is positive; on the way up the hill it acts against the motion — negative. Friction and air resistance always act against the motion, so their work is always negative. In the “Work and power” readouts, the sum of the work done by all the forces equals the change in kinetic energy: $\sum A = Δ E_{k}$ — work is a measure of the change in energy. Power is the rate at which work is done: $P = A / t = F v \cos α$ (W). It describes the performance of a mechanism: if a chain lift raises a 600 kg car by 40 m in 30 s, it needs $P = m g h / t \approx 7.8$ kW (without friction). There are two kinds of potential energy: that of a body raised above the ground, $m g h$ , and that of an elastically deformed body, $k x^{2} / 2$ — a drawn bow fires an arrow using this energy (the “Elasticity and deformation” experiment).

Potential energy curve

The car moves like a bead on a wire, in one dimension: its position depends only on $s$ , and $E = \frac{1}{2} m v^{2} + U (s)$ . On the “U(s)” graph the track profile is drawn on an energy scale, the dashed line is the total energy $E$ , and the segment between them is $E_{k}$ . The car can only move where $U (s) \leq E$ ; at the point where $U = E$ it stops and turns back (the turning point). The force along the track is $F_{s} = - d U / d s$ : the steeper the curve, the greater the force. $d U / d s = 0$ means equilibrium: stable at the bottom of a dip (U minimum), unstable at the top of a hill (U maximum). Gravity is a conservative force: its work depends only on the initial and final heights, and is zero over a closed path. Friction is non-conservative (dissipative): the $E$ line slowly falls as energy goes into heat. If the forces do not depend explicitly on time (homogeneity of time), $E$ is conserved — see the “Hamiltonian mechanics” experiment for details. These laws are the basis for designing rides, hydroelectric power stations and car brakes.

Did you know?

In modern rides the loop is not a circle but is built in the shape of a teardrop (a clothoid) that narrows towards the top: at the bottom of a circular loop the passengers feel a very large load — even at $h = 2.5 R$ it is $N = 6 m g$ .
A steel roller coaster car has three kinds of wheels: running wheels on top of the rail, guide wheels at the side, and wheels that grip the underside of the rail (up-stop). The last kind stops the car from lifting off the rail.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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