Interactive Math & Science Lessons

Roller Coaster Energy Simulator

Drag the position slider along the track. Energy conservation determines the speed at every point.

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Lesson

The theory — Roller Coaster Energy Simulator

On a frictionless track, mechanical energy is conserved: the sum of kinetic and potential energy is the same everywhere. That single sentence fixes the speed at every point of the ride from the starting height alone — no forces, no timeline, no shape of the hill required.

What each symbol means

h
the height above the datum. Which level you call zero never changes an answer: only the drop h₀ − h reaches the formula, so moving the datum moves both heights together.
v
the speed, which the readout derives from the energy rather than from any force.
KE
kinetic energy, ½mv².
PE
potential energy, mgh — 49.05 J at the start, from 1 × 9.81 × 5.
μ
the friction coefficient. Set above zero it removes energy along the way, and the amount removed depends on distance travelled rather than height lost.

Where the formula comes from

  1. Conservation says the total is fixed: ½mv² + mgh = ½mv₀² + mgh₀.
  2. Every term carries the same factor m, so divide it out: ½v² + gh = ½v₀² + gh₀. The mass has gone, and it has gone for good.
  3. Solve for the speed: v = √(v₀² + 2g(h₀ − h)). Only the height drop appears — which is why the readout reports a start energy per unit mass, 49.05 J/kg, rather than a force.
Assumes
A point mass sliding on a track that constrains it, with no rotation, no air resistance, and enough speed to stay on the rails. Nothing here checks whether the carriage would actually leave the track over a crest.
Breaks when
Step 2 is worth testing rather than trusting, and the page lets you. At the halfway point the speed reads 7.8912 m/s; change the mass from 1 kg to 500 kg and it reads 7.8912 m/s still — identical — while KE jumps from 31.135 J to 15,568 J. Heavier carriages carry more energy and travel no faster. Friction does not break this either: the loss per unit mass is μgs, which is also mass-free.

Friction is where the shape of the track starts to matter 🖖

On a frictionless track only the drop matters, and the shape between start and finish is irrelevant. That is the assumption doing the work, and this tool has a μ control that switches it off. With friction the loss becomes μg times the distance travelled, which is a property of the path rather than the height. Its friction preset drops 5 m along 13.55 m of track and arrives at 9.209 m/s; a straight 5 m drop at the same μ = 0.05 arrives at 9.654 m/s. Same start, same finish height, 0.445 m/s given away purely for taking the longer route. The half-pipe preset is the sharp version. Frictionless, it climbs back to its starting height with exactly 0.000 m/s to spare, and its two flat middle sections add 4.00 m of path for no height at all. So the smallest friction you can dial in strands it partway up.

Speed depends only on height 🖖

On a frictionless track the ball's speed at any point is fixed entirely by how far it has dropped from its start — the shape of the track in between makes no difference. A near-vertical plunge and a long gentle slope to the same depth deliver exactly the same speed. That's why the ball can never climb higher than where it began: it would need energy it doesn't have. Add friction and each metre of track quietly skims energy away as heat.

Real loops aren't circular 🖖

A roller-coaster loop looks round but almost never is — it's a teardrop-shaped clothoid, a curve whose radius tightens toward the top. A truly circular loop fast enough to stay on the track at the crest would crush riders with punishing g-forces at the bottom. German engineer Werner Stengel introduced the clothoid loop in 1976, giving a small radius up top (little speed needed) and a gentle radius below. Same energy conservation — smarter geometry.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

  1. The mass box is set to 1 kg. Set it to 500 kg and predict what happens to the speed at the halfway point.

    Show answer
    Nothing at all — it stays 7.8912 m/s. Every term in the energy equation carries a factor of m, so it divides out and the speed depends only on the height dropped. The kinetic energy does change, from 31.135 J to 15,568 J: a heavier carriage carries far more energy and travels no faster.

  2. At the start the readout shows PE 49.05 J and KE 0 J. Where does the coaster reach its highest speed, and what is the total then?

    Show answer
    At the lowest point of the track, and the total is still 49.05 J — that is the whole content of conservation. PE has been converted into KE, not created or destroyed. With friction at zero the sum never changes, however many hills it crosses.

  3. Raise the friction coefficient above zero. Does the energy lost depend on how far the coaster has travelled, or on how much height it has lost?

    Show answer
    On distance travelled. The table reports the loss as μgs, where s is path distance — which is why the readout tracks distance separately from height. A long flat stretch costs energy while a vertical drop of the same length costs almost none. Note it is also mass-free, so friction does not break the previous answer either.

Problem solved in full

  1. A 1 kg car released from rest at 5 m on the track 6 steps

    A 1 kg car released from rest at 5 m on the track 0,5; 2,3; 3,6; 5,1; 7,4; 9,2; 10,5. Find where it ends up — and it is not the far end.

    1. With no friction the total energy is fixed by the release height and never changes. One kilogram and 5 metres is all the input this problem has.

    2. Rearranged for speed, the same statement becomes a restriction: the car can only be where the square root is real. That is the whole solution, before any track is drawn.

    3. Read the track for its maximum. The peak sits above the release point, which settles the question — and note the drop to 1 m beyond it never enters the argument.

    4. Find the turning point exactly. It is on the climb from (2,3) to (3,6), where the straight segment crosses back through the release height.

    5. The path length matters because friction would act along it, not along the horizontal. Six segments, six Pythagoras calls.

    6. Two speeds close it out: the fastest the car actually goes on its short excursion, and the launch speed that would unlock the rest of the ride.

    Answer

    It turns around at x = 2.667 m and never sees the rest of the track. The hill at x = 3 stands at 6 m and the car started at 5, so the energy simply is not there; the readout says stopped for every position beyond that point, which is the tool agreeing rather than failing. What makes this worth working is that the answer needs no simulation at all. Energy conservation turns a differential equation into an inequality — h ≤ h₀ — and the shape of the track between the start and the obstacle is irrelevant. Give the car 4.43 m/s at the top and the whole 20.972 m opens up, including a 9.90 m/s run through the dip at x = 5. Below that speed, none of it exists.

Learning path

Energy instead of forces

Leads to momentum-collision

References (1)

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Example problems

  • Classic - Released at 5 m, the car turns back at x = 2.667 m: the hill at x = 3 stands at 6 m and the energy is simply not there. The worked problem below settles it without simulating anything.
  • Big drop - An 8 m drop over 16.45 m of frictionless track. The car clears every hill and arrives at 12.528 m/s, the fastest anything moves on this page.
  • Half pipe - Three track points sit at zero height, so the car crosses 4.00 m of flat bottom at an unchanging 9.905 m/s, then climbs back to exactly 5 m with 0.000 m/s left.
  • With friction - At μ = 0.05 the 13.55 m of track burns 6.65 J/kg, which is why the car arrives at 9.209 m/s instead of the 9.905 it would reach with no friction at all.