Roller Coaster Energy Simulator

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

No friction: energy per kilogram is conserved	$\tfrac12v^{2}+gh=\tfrac12v_0^{2}+gh_0\ \Rightarrow\ v=\sqrt{v_0^{2}+2g(h_0-h)}$
With friction, d the horizontal distance covered	$\tfrac12v^{2}+gh=\tfrac12v_0^{2}+gh_0-\mu g\,d$

Worked example — Roller Coaster Energy Simulator

Path distance	0 m
Height	5 m
Start energy / mass	49.05 J/kg
Energy lost (friction)	0 J/kg
Speed	0 m/s
Kinetic energy	0 J
Potential energy	49.05 J
Energy check	49.05 J

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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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 takes μg joules per kilogram for every metre of horizontal distance covered, whatever the slope.

Where the formula comes from

With no friction the total energy is fixed. Kinetic plus potential is the same at every point of the ride as at the start.
$\tfrac12mv^{2}+mgh=\tfrac12mv_0^{2}+mgh_0$
Every term carries the mass, so divide it out. The mass has gone, and it has gone for good.
$\tfrac12v^{2}+gh=\tfrac12v_0^{2}+gh_0$
Solve for the speed. Only the height drop appears, and the car can only be where the square root is real: it never rises above the height its starting energy reaches.
$v=\sqrt{v_0^{2}+2g(h_0-h)},\qquad h\le h_0+\frac{v_0^{2}}{2g}$
Friction on a straight stretch at angle θ: the normal force is mg cos θ and the stretch is d/cos θ long, so the work is μmg times the horizontal distance d. Subtract it from the energy before solving for the speed.
$W_f=\mu\,(mg\cos\theta)\,\frac{d}{\cos\theta}=\mu mg\,d\ \Rightarrow\ v=\sqrt{v_0^{2}+2g(h_0-h)-2\mu g\,d}$

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. Friction is charged on the straight stretches only: the extra push a curve needs at speed, which raises the normal force in a dip, is left out. Nothing here checks whether the carriage would leave the track over a crest.
Breaks when: Step 2 is worth testing rather than trusting, and the page lets you. Press Half pipe and drag the position to the middle: the speed reads 9.9045 m/s. Change the mass from 1 kg to 500 kg and it still reads 9.9045 m/s, while the kinetic energy jumps from 49.05 J to 24,525 J. Heavier carriages carry more energy and travel no faster. Friction does not break this either: the loss per unit mass is μgd, which is also mass-free.

Problems solved in full

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.
  
  $E_0=mgh_0=1\times9.81\times5=49.05\ \mathrm{J},\qquad v_0=0$
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.
  
  $\tfrac{1}{2}mv^{2}=mg(h_0-h)\ \Longrightarrow\ v=\sqrt{2g(h_0-h)},\quad\text{real only where }h\le h_0$
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.
  
  $\text{track max}=6\ \mathrm{m}>h_0=5\ \mathrm{m}\quad\Longrightarrow\quad\text{the hill at }x=3\text{ is unreachable}$
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.
  
  $(2,3)\to(3,6):\ y=5\ \text{at}\ x=2+\tfrac{5-3}{6-3}=2.667$
5. The position control measures distance along the track, so it helps to know the whole length. Six segments, six Pythagoras calls.
  
  $s=\sum\sqrt{\Delta x^{2}+\Delta y^{2}}=2.828+3.162+5.385+3.606+2.828+3.162=20.972\ \mathrm{m}$
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.
  
  $v_{\max}=\sqrt{2g(5-3)}=6.26\ \mathrm{m/s};\qquad v_0\ \text{needed}=\sqrt{2g(6-5)}=4.43\ \mathrm{m/s}$
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.

$h_{\text{track}}=6>h_0=5\ \Longrightarrow\ \text{it stops at }x=2.667,\ \text{never reaching the drop}$
Speed at the bottom of the half-pipe 4 steps
A 250 kg car is released from rest at the 5 m rim of the half-pipe track. With no friction, how fast is it moving on the flat bottom, how much kinetic energy does it have there, and how fast is it going when it is 2 m up the far side?

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1. Energy at the start and on the bottom, with the bottom as the zero of height:
  
  $E_0=mgh_0=250(9{.}81)(5)=12\,262{.}5\ \mathrm{J},\qquad E_{\mathrm{bottom}}=\tfrac12mv^{2}$
2. The mass divides out, so the speed comes from the drop alone:
  
  $\tfrac12mv^{2}=mgh_0\ \Rightarrow\ v=\sqrt{2gh_0}=\sqrt{2(9{.}81)(5)}=9{.}9045\ \mathrm{m/s}$
3. The kinetic energy does need the mass:
  
  $E_k=\tfrac12(250)(9{.}9045)^{2}=12\,263\ \mathrm{J}$
4. At 2 m up the far side the drop from the start is only 3 m:
  
  $v=\sqrt{2g(h_0-h)}=\sqrt{2(9{.}81)(5-2)}=7{.}672\ \mathrm{m/s}$
Answer

9.9045 m/s on the bottom, with 12,263 J of kinetic energy, and 7.672 m/s at 2 m up the far side. The bottom is flat, so the car crosses all of it at the same 9.9045 m/s.

$v_{\mathrm{bottom}}=9{.}9045\ \mathrm{m/s},\quad E_k=12\,263\ \mathrm{J},\quad v(2\ \mathrm{m})=7{.}672\ \mathrm{m/s}$
The least launch speed that clears the hill 3 steps
On the Classic track the car starts at 5 m and the hill at x = 3 is 6 m high. With no friction, what is the least launch speed that carries it over the hill, and how fast is it then moving in the 1 m dip at x = 5?

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1. At the crest the car may arrive with no speed to spare. Energy from the start to the crest:
  
  $\tfrac12v_0^{2}+gh_0=\tfrac12v_{\mathrm{top}}^{2}+gh_{\mathrm{top}},\qquad v_{\mathrm{top}}\ge0$
2. Solve for the launch speed:
  
  $v_0=\sqrt{2g(h_{\mathrm{top}}-h_0)}=\sqrt{2(9{.}81)(6-5)}=\sqrt{19{.}62}=4{.}4294\ \mathrm{m/s}$
3. Energy from the start to the dip, 4 m below the start:
  
  $v=\sqrt{v_0^{2}+2g(h_0-h)}=\sqrt{19{.}62+2(9{.}81)(5-1)}=\sqrt{98{.}1}=9{.}9045\ \mathrm{m/s}$
Answer

4.4294 m/s, and then 9.9045 m/s in the dip. That is the speed of a 5 m drop from rest, because the launch supplied exactly the 1 m of height the hill asked for.

$v_0=4{.}4294\ \mathrm{m/s},\qquad v_{\mathrm{dip}}=9{.}9045\ \mathrm{m/s}$
A track with friction 3 steps
A 1 kg car is released from rest at 5 m on the track with friction, which ends at ground level 10 m further on. The friction coefficient is 0.05. How fast does it arrive, and how much energy has friction taken?

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1. On each straight stretch friction takes μmg times the horizontal distance, so over the whole track it takes:
  
  $W_f=\mu mg\,d=0{.}05(1)(9{.}81)(10)=4{.}905\ \mathrm{J}$
2. Energy at the start equals energy at the end plus the work done against friction:
  
  $mgh_0=\tfrac12mv^{2}+W_f\ \Rightarrow\ v=\sqrt{2gh_0-2\mu g\,d}=\sqrt{98{.}1-9{.}81}=9{.}3963\ \mathrm{m/s}$
3. Without friction the same drop would give:
  
  $\mu=0:\quad v=\sqrt{2(9{.}81)(5)}=9{.}9045\ \mathrm{m/s}$
Answer

9.3963 m/s, with 4.905 J taken by friction. The two dips and the hill in between do not change the loss: one straight 10 m ramp from 5 m would arrive at the same speed.

$v=9{.}3963\ \mathrm{m/s},\qquad W_f=4{.}905\ \mathrm{J}$
How high up the far side? 3 steps
The half-pipe has a 5 m rim, a 2 m slope down, 4 m of flat bottom and a 2 m slope up to the far rim, which is 5 m high. Released from rest with a friction coefficient of 0.05, how high up the far side does the car climb?

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1. Let the car stop at height h on the far slope. That slope rises 5 m over 2 m of horizontal distance, so reaching h takes 2h/5 of it, and the horizontal distance from the start is:
  
  $d=2+4+\tfrac25h=6+0{.}4h$
2. Energy at the start equals the potential energy at the stop plus the work against friction:
  
  $gh_0=gh+\mu g\,(6+0{.}4h)$
3. Solve for h:
  
  $h=\frac{h_0-6\mu}{1+0{.}4\mu}=\frac{5-0{.}3}{1{.}02}=4{.}6078\ \mathrm{m}$
Answer

4.6078 m, about 0.39 m short of the rim. It then rolls back down, and every crossing of the bottom costs it more.

$h=4{.}6078\ \mathrm{m}$
The loop: how high must it start? 3 steps
A car rolls without friction from rest down a slope into a circular vertical loop of radius 4.0 m. From what height must it start to stay on the track at the top of the loop, and how hard does the track push on a 70 kg rider at the bottom?
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1. At the top the track can only push down, so gravity alone must at least supply the centripetal force:
  
  $mg\le\frac{mv_{\mathrm{top}}^{2}}{R}\ \Rightarrow\ v_{\mathrm{top}}^{2}\ge gR$
2. Energy from the start to the top of the loop, which is 2R up:
  
  $gh=\tfrac12v_{\mathrm{top}}^{2}+g(2R)\ \ge\ \tfrac12gR+2gR\ \Rightarrow\ h\ge\tfrac52R=10\ \mathrm{m}$
3. At the bottom the speed comes from the full drop, and the track must supply the weight plus the centripetal force:
  
  $v_{\mathrm{bottom}}^{2}=2gh=5gR,\qquad N=mg+\frac{mv_{\mathrm{bottom}}^{2}}{R}=6mg=6(70)(9{.}81)=4120\ \mathrm{N}$
Answer

At least 10 m, two and a half times the radius; and the track pushes on the rider with 4,120 N, six times their weight. That six is why real loops are not circles: a clothoid keeps the radius small at the top, where the speed is low, and large at the bottom, where it is high.

$h_{\min}=2{.}5R=10\ \mathrm{m},\qquad N_{\mathrm{bottom}}=6mg=4120\ \mathrm{N}$

Supplementary problems, answers only

A 600 kg car moving at 2.0 m/s at the top of a 30 m drop runs down a frictionless track. How fast is it moving when it is 10 m above the ground?

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19.91 m/s.

$v=\sqrt{2{.}0^{2}+2(9{.}81)(30-10)}=\sqrt{396{.}4}=19{.}91\ \mathrm{m/s}$
A sled moving at 8.0 m/s at the foot of a frictionless slope slides up it. How high does it rise?

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3.262 m.

$h=\frac{v^{2}}{2g}=\frac{8{.}0^{2}}{2(9{.}81)}=3{.}262\ \mathrm{m}$
A block is released from rest 3.0 m up a straight ramp and slides on along a level floor. The friction coefficient is 0.15 everywhere. What horizontal distance from the starting point does it cover before it stops?

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20 m, for any ramp steep enough to start sliding at all (tan θ > 0.15, steeper than 8.53°): friction takes μg per metre of horizontal distance, and 3.0/0.15 = 20. On a gentler ramp the block never moves.

$mgh=\mu mg\,d\ \Rightarrow\ d=\frac{h}{\mu}=\frac{3{.}0}{0{.}15}=20\ \mathrm{m}$
A car starts from rest at the top of a 45 m drop and covers 60 m of horizontal distance on the way down, with a friction coefficient of 0.02. How fast is it moving at the bottom?

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29.315 m/s.

$v=\sqrt{2g(h-\mu d)}=\sqrt{2(9{.}81)(45-0{.}02\times60)}=29{.}315\ \mathrm{m/s}$
A car starts from rest 12 m above the bottom of a frictionless circular loop. What is the largest loop radius it can get round without leaving the track?

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4.8 m, because the start must be at least 2.5 times the radius above the bottom.

$h\ge\tfrac52R\ \Rightarrow\ R\le\frac{12}{2{.}5}=4{.}8\ \mathrm{m}$

Friction charges for the horizontal distance, not the track

On a straight slope at angle θ the normal force is mg cos θ, and the slope is d/cos θ long for a horizontal distance d. The two cosines cancel, so friction takes μmgd whatever the angle: a steep plunge and a gentle ramp across the same horizontal distance lose the same energy, and a vertical drop loses none. With friction drops 5 m across 10 m of horizontal track through two dips and a hill, loses 4.905 J/kg, and arrives at 9.3963 m/s, exactly what one straight ramp from 5 m to the ground over the same 10 m would give. The shape comes back only through the curves, which this track leaves out: a real dip presses the car harder into the rails and takes more than the formula charges. The Half pipe is the sharp case. Frictionless, it climbs back to its starting height with nothing to spare, so the smallest friction strands it partway up the far side: at μ = 0.05 it stops at 4.6078 m.

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 horizontal distance 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.

Press Half pipe, drag the position to the middle of the flat bottom and read the speed. Now change the mass from 1 kg to 500 kg. Predict the speed first.

Show answer

It stays 9.9045 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 49.05 J to 24,525 J: a heavier car carries far more energy and travels no faster.
At the start the readout shows a potential energy of 49.05 J and a kinetic energy of 0 J. Where does the car on this track reach its highest speed, and what is the total then?

Show answer

At the lowest point it actually reaches: 3 m up, at x = 2, where it moves at 6.2642 m/s. The 1 m dip at x = 5 is lower, but it lies past the 6 m hill the car cannot climb. The total there is still 49.05 J, 19.62 J of it kinetic.
Raise the friction coefficient above zero. Does the energy lost depend on how far the car has travelled along the track, or on something else?

Show answer

On the horizontal distance it has covered. On a slope the normal force shrinks by cos θ while the length grows by 1/cos θ, so the table charges μgd with d measured across, not along. A vertical drop costs nothing; a level run costs the most per metre of track. The loss is also mass-free, so friction does not break the previous answer either.

Learning path

Energy instead of forces

References (1)

Why real loops are not circles, and what that does to the forces at the top: A.-M. Pendrill, "Roller coaster loop shapes revisited." Physics Education 51(3), 030106, 2016.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.