Conservation of Energy: Roller Coaster

A car rolls along the rails with no motor. Drag it higher or lower and watch “height energy” turn into motion energy and back — and why the car can never climb above the yellow “ceiling”. Add friction and see where the energy goes.

The idea in 30 seconds

Height is a “stock” of energy

To pull the car up a hill, work is needed, and it does not vanish: a car at a height h stores the energy Ep = mgh. At the very top all the energy lies “in the height”. There is no motor on the way down — it is exactly this stock that rolls the car.

Height ⇄ speed

Going down, the height melts and the speed grows: Ep → Ek. Going up, it is the other way round. The sum Ep + Ek is always the same, so the car never rises above the yellow “ceiling” — the height it started from. And the speed at any point depends only on the difference of heights: v = √(2g·Δh), the mass does not matter.

Friction and the “extra” force

With friction, part of the energy turns into heat Q (red). The “ceiling” slowly sinks and the car stops reaching the summits. And on the bends there is also the pushing force of the rail N: in a dip it is greater than the weight (G > 1, you are pressed into the seat), on a crest it is smaller (G < 1), and at G = 0 weightlessness sets in.

Try it yourself

Press “m ×2”: the mass is twice as big, the energies doubled, but the speed at the bottom stayed the same — look at the display: v did not change.
Press “h₀ ÷2”: the height is half as big, but the speed at the bottom is smaller not 2 times, only √2 ≈ 1.41 times. The speed grows more slowly than the height.
Choose the “Hills” track and drag the car to a height of 6 m — below the summit (7 m). It will not get over the crest and will come back: the rail above the yellow “ceiling” is dimmed. There is not enough energy!
Choose “Loop” and press “loop limit”: start from h₀ = 2.5R. At the top G = 0 — that is the very border. A bit lower and a free ball would fall off. Try h₀ = 7 m with R = 3 m and look at G at the top.
Switch on “Ordinary” or “Rust”: watch the red column grow and the “ceiling” sink. In the “Bowl” the car finally stops at the bottom — all the energy has gone into heat.

Where you meet it in life

A roller coaster: the first hill is always the tallest. After that the car is driven only by the energy stored during the lift, so no later summit can be higher than the first one.
G-loads on rides and in a plane: in the dip of a loop a pilot “gets heavier” several times over, and on the crest of a hill feels almost weightless. That is why crests are made rounded, not sharp.
A skate park and a hydropower plant: in a skater’s “bowl” it is the same — fast at the bottom, you stop at the top. At a hydropower plant, the water above the dam stores potential energy, flows down and spins the turbine.

Graphs: the same in numbers

Orange is Ep, green is Ek, red is the heat Q; the yellow dashed line is the “ceiling” E/mg. On the track profile the car is the orange dot, and the green gap up to the “ceiling” is exactly its Ek.

Where is the car on the track? The profile h(s) and the “ceiling”

Where does the energy go? E(t)

How fast does it go? v(t)

What does the passenger feel? G-load G(t)

Track

R loop radius 3,0 m

h₀ starting height 8,6 m

v₀ starting push 0,0 m/s

m mass of the car 500 kg

Friction of the wheels on the rail

μ friction coefficient 0,00

Quick experiments

playback speed 1×

Show

arrows of forces and velocity

the “ceiling” E/mg and the energy column above the car

trail of the car

height ruler

“Predict it” mode

Changing the track, the height or the push starts the car again from the beginning, while the mass and the friction can be changed on the move. The motion is computed from exact equations with a 1/240 s step, so the numbers do not depend on the frame rate or on the playback speed.

KEY FORMULAS

Potential energy at a height h

$$E_{\text{p}} = mgh$$

The energy is “stored” in the height above the lowest point of the track: there h = 0 and Ep = 0.

▶ Detailed Explanation

The height h is measured from the zero level — the lowest point of the track. The higher the car, the more work was spent on lifting it and the bigger the stock Ep = mgh. Only the height matters, not the path: on a gentle and on a steep climb to the same height the stock is the same.

Example: m = 500 kg, h = 9 m ⇒ Ep = 500·9.8·9 = 44 100 J = 44.1 kJ. At a height of 7 m (the summit of the hill) Ep = 500·9.8·7 = 34.3 kJ.

Kinetic energy

$$E_{\text{k}} = \frac{mv^2}{2}$$

The energy of motion grows as the square of the speed.

▶ Detailed Explanation

Twice as fast means four times the energy. So at the bottom of the hill, where the speed is greatest, Ek is greatest too, and on the summit (v = 0) Ek = 0.

Example: m = 500 kg, v = 13.28 m/s ⇒ Ek = 500·13.28²/2 ≈ 44.1 kJ — all the energy the car stored at a height of 9 m.

Conservation of energy on the roller coaster

$$E_{\text{p}} + E_{\text{k}} = mgh + \frac{mv^2}{2} = E_0 = \text{const}$$

Without friction the sum of potential and kinetic energy is the same at every point of the track.

▶ Detailed Explanation

Gravity is “restoring”: what the car loses in height it gains in speed, and vice versa. The reaction force of the rail is always perpendicular to the velocity and does no work. So at every point Ep + Ek = E₀ = mgh₀ (with a push E₀ = mgh₀ + mv₀²/2). Hence the car cannot rise above h₀: at that height Ep = E₀, nothing is left for Ek — the car stops. On the screen this is the yellow “ceiling” line.

Example: E₀ = 44.1 kJ. At a height of 7 m: Ep = 34.3 kJ, so Ek = 44.1 − 34.3 = 9.8 kJ and v = √(2Ek/m) = √(2·9800/500) = 6.26 m/s.

Speed from the difference of heights

$$v = \sqrt{v_0^2 + 2g\,(h_0 - h)}$$

The mass cancels: a heavy and a light car from the same height move alike.

▶ Detailed Explanation

From mgh₀ + mv₀²/2 = mgh + mv²/2 we divide by m and get v² = v₀² + 2g(h₀ − h). With no push (v₀ = 0): v = √(2g(h₀ − h)). The speed depends only on how far the car has descended, not on the mass or the shape of the descent.

Example: a start from h₀ = 9 m. At the bottom (h = 0): v = √(2·9.8·9) = 13.28 m/s. On the 7 m summit: v = √(2·9.8·2) = 6.26 m/s. From half the height (4.5 m) the speed at the bottom is √(2·9.8·4.5) = 9.39 m/s — not half as much, only √2 times less. With a push v₀ = 6 m/s from a height of 4 m: v at the bottom = √(36 + 2·9.8·4) = 10.7 m/s.

Friction: the energy goes into heat

$$E_{\text{p}} + E_{\text{k}} + Q = E_0,\qquad Q = \mu\int N\,ds$$

The friction force μN works against the motion, and this work becomes heat Q.

▶ Detailed Explanation

The friction force of the wheels on the rail F = μN points against the velocity and on every stretch ds of the path “eats” part of the mechanical energy: Q = ∫μN ds. The energy Ep + Ek decreases, the yellow “ceiling” sinks, the red segment grows, but in total Ep + Ek + Q = E₀: the energy does not vanish, it spreads out (the wheels and the rail warm up).

Example for a straight inclined stretch: Q = μ·mg·cosα·s. With μ = 0.06, m = 500 kg, α = 30° and a path s = 10 m: Q = 0.06·500·9.8·0.866·10 ≈ 2.55 kJ. Once stopped, the car stays put if tan α ≤ μ (static friction).

Rail reaction and the G-load

$$N = m\left(g\cos\theta + \frac{v^2}{R}\right),\qquad G = \frac{N}{mg}$$

On a bend the rail pushes the car harder or lighter than its weight — this is what the passenger feels.

▶ Detailed Explanation

θ is the angle of the rail to the horizontal, R is the radius of the bend (with “+” for a dip and for the inner side of a loop, with “−” for a crest). On a flat stretch N = mg, G = 1. At the bottom of a dip N = m(g + v²/R): you are pressed into the seat (G > 1). On a crest N = m(g − v²/R): the faster, the lighter, and at v² = gR we get N = 0 — weightlessness.

Example: at the bottom of a loop R = 3 m, starting from h₀ = 7.5 m, v² = 2gh₀ = 147 m²/s², N = m(9.8 + 147/3) = 58.8·m, G = 6 — a six-fold G-load.

The loop: the minimum height 2.5R

$$mgh_0 = mg\cdot 2R + \frac{mv^2}{2},\quad v^2 \ge gR \;\Rightarrow\; h_0 \ge \frac{5}{2}R$$

To stay on at the top you need v ≥ √(gR), and for that — a start not lower than 2.5R.

▶ Detailed Explanation

At the top of the loop the rail presses the car down with a force N, and together with gravity they give the centripetal acceleration: N + mg = mv²/R. Without leaving the rail N ≥ 0, that is v² ≥ gR. The top is at a height 2R, so by the conservation of energy mgh₀ = mg·2R + mv²/2 ≥ mg·2R + mgR/2 = 2.5·mgR. Hence h₀ ≥ 2.5R.

Example: R = 3 m ⇒ h₀ ≥ 7.5 m; on the border at the top v = √(gR) = √(9.8·3) = 5.42 m/s. From a lower height a free ball would fall off, while a real car is held by safety wheels (the rail “pulls” it, G < 0). With a starting push the condition is h₀ + v₀²/(2g) ≥ 2.5R.

What is going on. The car moves along the rails with no motor — only because of gravity. Pulling it up to a height h above the lowest point of the track takes work, and that work does not vanish: the car “stores” potential energy Ep = mgh. If it is released with no push, at the top all the energy is potential and the speed is zero.

Conversion and the “ceiling”. The car goes down: the height decreases, the speed grows, and Ep turns into kinetic energy Ek = mv²/2. At the bottom h = 0, all the energy is kinetic — the speed is greatest. On the way up everything goes the other way round. Without friction the sum Ep + Ek at every point equals E₀ = mgh₀ — this is the law of conservation of mechanical energy. Hence the main rule of roller coasters: the car cannot rise above the height h₀ it started from (the yellow “ceiling” line on the screen), and the speed at any point depends only on the difference of heights: v = √(2g(h₀ − h)). The mass is not in this formula. Example: m = 500 kg, h₀ = 9 m ⇒ E₀ = 500·9.8·9 = 44.1 kJ; at the bottom v = √(2·9.8·9) = 13.28 m/s, on the hill summit (7 m) v = √(2·9.8·2) = 6.26 m/s.

Friction and heat. Switch on the friction of the wheels on the rail: on every metre of the way part of the mechanical energy turns into heat Q (the red segment). The “ceiling” slowly sinks and the car stops reaching the summits: first it fails to reach the summit, then it starts to rock in a dip and finally stops. The energy does not vanish: Ep + Ek + Q = E₀. If the rail is too gentle, static friction holds the car (tan α ≤ μ) — it simply will not move.

The loop and the G-load. On a bend of the rail the reaction force N acts on the car, and the passenger feels it as the G-load G = N/(mg). At the bottom of a dip G is greater than one (you are pressed into the seat), on a crest it is smaller, and at G = 0 weightlessness sets in. At the top of a vertical loop of radius R the speed must be v ≥ √(gR), otherwise a free body would fall off. From the conservation of energy comes the famous rule: with no push the loop is passed from a start not lower than h₀ = 2.5R. Example: R = 3 m ⇒ h₀ = 7.5 m, at the top v = 5.42 m/s, at the bottom the G-load is G = 6.

The simulator computes the exact motion along rails made of arcs and straight lines (not an approximation), so the numbers on the screen agree with the formulas. Compare with the simulators “Conservation of energy: the simple pendulum” and “Inclined plane with friction”.