Vertical Circular Motion
In 1975, Knott's Berry Farm's Corkscrew became the first modern vertical loop coaster. Its loop radius was 6.1 m, requiring a minimum speed of $\sqrt{gr} = \sqrt{9.8 \times 6.1} = 7.73$ m/s at the top to maintain contact. Modern coasters use a clothoid (teardrop) loop to reduce peak g-force from 6g (circular) to 3.5g — a direct application of the vertical circle force analysis in this lesson.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
On a roller coaster loop, you travel through a complete vertical circle. Where do you feel heaviest — at the top or at the bottom of the loop? Why?
Write down your prediction before working through the lesson — you will come back to it at the end.
Warm-up — in vertical circular motion, which force is directed toward the centre of the circle at all points?
Know — Analyse Forces
- Calculate tension at the top and bottom of a vertical circle
- Apply Newton's second law with centripetal acceleration at each position
Understand — Minimum Speed
- Derive and apply $v_{\text{min(top)}} = \sqrt{rg}$ for the critical condition
- Calculate minimum entry speed $v_{\text{min(bottom)}} = \sqrt{5gr}$ using energy conservation
Can Do — Energy Method
- Use conservation of mechanical energy to find speed at any point in a vertical circle
- Calculate normal force at any point given entry speed and height
The normal force on a rider is greater at the top of a vertical loop than at the bottom.
An object falls away from a vertical circle at the top when the string tension (or normal force) reaches zero.
Core Content
Applying Newton's second law to vertical circular motion at the critical positions
Ride a roller coaster through a vertical loop: at the bottom you feel pressed hard into the seat; at the top you feel almost weightless (or, at the minimum speed, completely weightless). These sensations are measurable — at the bottom of the 1975 Knott's Berry Farm Corkscrew loop (radius 6.1 m), the normal force on a 70 kg rider travelling at 12 m/s is $N = m(v^2/r + g) = 70(23.6 + 9.8) = 2,338 \text{ N}$ — 3.4 times their weight. The forces change dramatically as height changes, because gravity speeds the object up and slows it down throughout the loop.
Vertical circular motion — forces at the top and bottom of the loop.
Detailed force analysis at multiple positions around the vertical circle.
At the Top of the Circle
Both tension (or normal force) and weight point downward, toward the centre of the circle. Taking toward the centre as positive:
$T + mg = \dfrac{mv^2}{r}$ — both forces point toward the centre (downward)
$T = \dfrac{mv^2}{r} - mg$ — tension at the top (N)
At the top, tension is minimum. If the speed is too low, tension becomes zero and the object falls away from the circular path. The minimum speed at the top occurs when $T = 0$:
$v_{\text{min(top)}} = \sqrt{rg}$ — minimum speed to maintain contact at the top
If $v < \sqrt{rg}$ at the top, the object cannot follow the circular path — it falls away (the string goes slack, or the vehicle leaves the track).
At the Bottom of the Circle
Tension points upward (toward the centre) and weight points downward (away from the centre). The net upward force provides the centripetal acceleration:
$T - mg = \dfrac{mv^2}{r}$ — tension up, weight down, net toward centre
$T = \dfrac{mv^2}{r} + mg$ — tension at the bottom (N)
Tension is always greater at the bottom than at the top for the same speed because the weight opposes the centripetal force rather than assisting it. The difference is $T_{\text{bottom}} - T_{\text{top}} = 2mg$ when the speed is the same at both points.
At the top, gravity helps provide centripetal force, so tension can be smaller (or even zero). At the bottom, tension must overcome weight and provide centripetal force, so it is always larger. This is why you feel heaviest at the bottom of a roller coaster loop.
A 0.5 kg bob on a 0.8 m string is whirled in a vertical circle. At the top, its speed is 4 m/s. Find the tension at the top and at the bottom.
- Given. $m = 0.5 \text{ kg}$, $r = 0.8 \text{ m}$, $v_{\text{top}} = 4 \text{ m/s}$
- Find. $T_{\text{top}}$ and $T_{\text{bottom}}$
- Method. Use $T = mv^2/r - mg$ at top; energy conservation then $T = mv^2/r + mg$ at bottom.
- Solve (top). $T_{\text{top}} = \dfrac{0.5 \times 16}{0.8} - (0.5 \times 9.8) = 10 - 4.9 = 5.1 \text{ N}$
- Solve (speed at bottom). $v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr = 16 + 4 \times 9.8 \times 0.8 = 47.36$, so $v_{\text{bottom}} = 6.88 \text{ m/s}$
- Solve (bottom). $T_{\text{bottom}} = \dfrac{0.5 \times 47.36}{0.8} + (0.5 \times 9.8) = 29.6 + 4.9 = 34.5 \text{ N}$
Top: $T = mv^2/r - mg$ (gravity assists centripetal force; minimum speed when $T=0$: $v_\text{min} = \sqrt{rg}$). Bottom: $T = mv^2/r + mg$ (gravity opposes). Tension is always greater at the bottom; $T_\text{bottom} - T_\text{top} = 2mg$ at the same speed.
Pause — copy the highlighted top/bottom tension equations into your book before moving on.
At the top of a vertical loop, which expression gives the tension $T$ in the string?
Finding the critical entry speed needed to maintain circular motion throughout
We just saw that the top is the critical point where tension is smallest. That raises a question: what minimum speed at the bottom guarantees the object gets over the top without losing contact? This card answers it → set $T=0$ at the top → $v_\text{top} = \sqrt{rg}$, then energy conservation → $v_\text{bottom} = \sqrt{5gr}$.
For an object to complete a vertical circle, it must maintain contact with the track (or keep the string taut) at all points. The critical position is the top of the loop, where the normal force or tension is smallest.
Critical Condition at the Top
At the top, the object is just about to lose contact when the normal force (or tension) becomes zero. At this point, gravity alone provides the centripetal force:
$mg = \dfrac{mv_{\text{min}}^2}{r}$
$v_{\text{min(top)}} = \sqrt{rg}$ — minimum speed at the top of the loop
Relating Bottom Speed to Top Speed
Using conservation of mechanical energy, with the bottom of the loop as the reference level ($h = 0$):
$\dfrac{1}{2}mv_{\text{bottom}}^2 = \dfrac{1}{2}mv_{\text{top}}^2 + mg(2r)$
Substituting $v_{\text{top}}^2 = rg$ for the minimum case:
$v_{\text{bottom}}^2 = rg + 4gr = 5gr$
$v_{\text{min(bottom)}} = \sqrt{5gr}$ — minimum entry speed to complete the loop
This is a key result: to just complete a vertical loop of radius $r$, the speed at the bottom must be $\sqrt{5}$ times the minimum speed at the top. The factor of 5 arises because the object must gain enough kinetic energy to reach the top (converting $2mgr$ of kinetic energy to gravitational potential energy) and still have enough speed to maintain contact ($\tfrac{1}{2}mrg$ of kinetic energy).
At minimum speed, total energy at the bottom is $\tfrac{1}{2}m(5gr) = 2.5mgr$. At the top: kinetic energy $= \tfrac{1}{2}m(rg) = 0.5mgr$ and potential energy $= mg(2r) = 2mgr$. Sum $= 2.5mgr$ — energy is conserved.
A toy car must complete a vertical loop-the-loop of radius 2.5 m. What is the minimum speed at the bottom of the loop?
- Given. $r = 2.5 \text{ m}$, $g = 9.8 \text{ m/s}^2$
- Find. $v_{\text{min(bottom)}}$
- Method. Use $v_{\text{min(bottom)}} = \sqrt{5gr}$
- Solve. $v_{\text{min(bottom)}} = \sqrt{5 \times 9.8 \times 2.5} = \sqrt{122.5} = 11.1 \text{ m/s}$
Minimum speed at top: $v_\text{min(top)} = \sqrt{rg}$ (from $T = 0$). Minimum entry speed at bottom (energy conservation): $v_\text{min(bottom)} = \sqrt{5gr}$. Factor of 5 = PE rise of $2mgr$ plus KE of $\tfrac{1}{2}mrg$ needed at the top.
Add the highlighted loop-completion speeds to your notes before the check below.
To just complete a vertical loop of radius $r$, the minimum speed at the bottom must be:
Using conservation of mechanical energy to find speed at any point in a vertical circle
We just saw the minimum-speed condition using energy conservation. That raises a question: how do we find the speed — and the normal force — at any arbitrary point in the loop, not just the minimum case? This card answers it → apply $\tfrac{1}{2}mv_A^2 + mgh_A = \tfrac{1}{2}mv_B^2 + mgh_B$, then Newton's 2nd Law at that point.
Unlike uniform circular motion, vertical circular motion has a speed that changes continuously as the object rises and falls. Speed is maximum at the bottom and minimum at the top. We use conservation of mechanical energy to relate speeds at different heights.
Setting Up the Energy Equation
Taking the bottom of the circle as the reference level ($h = 0$), mechanical energy is conserved (no friction):
$\dfrac{1}{2}mv_A^2 + mgh_A = \dfrac{1}{2}mv_B^2 + mgh_B$
total mechanical energy at A equals total mechanical energy at B
At the bottom of the loop: $h = 0$, so $PE = 0$ and $KE = \tfrac{1}{2}mv_{\text{bottom}}^2$.
At the top of the loop: $h = 2r$, so $PE = mg(2r)$ and $KE = \tfrac{1}{2}mv_{\text{top}}^2$.
For a roller coaster entering a loop at known speed, we find the speed at any height using energy conservation, then calculate the normal force using Newton's second law at that point.
A roller coaster car of mass 300 kg starts from rest at a height of 15 m above the ground and enters a vertical loop of radius 4 m. Find (a) the speed at the top of the loop, and (b) the normal force on the car at the top.
- Given. $m = 300 \text{ kg}$, $h_{\text{start}} = 15 \text{ m}$, $r = 4 \text{ m}$, $v_{\text{start}} = 0$. Height at top of loop: $h_{\text{top}} = 2r = 8 \text{ m}$ above the bottom of the loop; height drop from start to top = $15 - 8 = 7 \text{ m}$.
- Find. (a) $v_{\text{top}}$ (b) $N_{\text{top}}$
- Method. Energy conservation from start to top, then $N + mg = mv^2/r$ at the top.
- Solve (a). $mgh_{\text{start}} = \tfrac{1}{2}mv_{\text{top}}^2 + mgh_{\text{top}}$
$300 \times 9.8 \times 15 = \tfrac{1}{2} \times 300 \times v_{\text{top}}^2 + 300 \times 9.8 \times 7$
$44100 = 150v_{\text{top}}^2 + 20580$
$v_{\text{top}}^2 = 156.8$, $v_{\text{top}} = 12.5 \text{ m/s}$ - Solve (b). $N = \dfrac{mv^2}{r} - mg = \dfrac{300 \times 156.8}{4} - (300 \times 9.8) = 11760 - 2940 = 8820 \text{ N}$
Energy conservation (no friction): $\tfrac{1}{2}mv_A^2 + mgh_A = \tfrac{1}{2}mv_B^2 + mgh_B$. Height from bottom to top $= 2r$. Once speed is found, apply Newton's 2nd Law at that point to find $N$ or $T$.
Pause — write the highlighted energy equation and the two-step method into your book before the check below.
Fill the gap. For a vertical loop, the height from the bottom to the top is _____ × $r$.
Top of circle: $T + mg = mv^2/r$ → $T = mv^2/r - mg$
Bottom of circle: $T - mg = mv^2/r$ → $T = mv^2/r + mg$
Minimum speed at top: $v_{\text{min(top)}} = \sqrt{rg}$ (when $T = 0$)
Minimum speed at bottom: $v_{\text{min(bottom)}} = \sqrt{5gr}$
Energy conservation: $\tfrac{1}{2}mv_A^2 + mgh_A = \tfrac{1}{2}mv_B^2 + mgh_B$
Three of these statements are correct. Pick the odd one out.
The loop on a modern roller coaster is not a perfect circle — it uses a clothoid shape (Euler spiral) with continuously varying radius. This reduces dangerous g-force variation. In a circular loop with $v = 15 \text{ m/s}$ and $r = 5 \text{ m}$ at the bottom:
$a_c = v^2/r = 225/5 = 45 \text{ m/s}^2 \approx 4.6g$
The rider feels $N = m(a_c + g) \approx 5.6g$. Most people experience greyout above 5g, so clothoid loops use a larger radius at the bottom (reducing peak g-force) and a smaller radius at the top (ensuring sufficient speed). This keeps g-forces in a safe range of approximately 2–3.5g throughout.
A bucket of water is whirled in a vertical circle. The water stays in the bucket at the top because:
Key Definitions
- Vertical circular motion: speed varies because gravity is not always perpendicular to motion
- Minimum speed at top: $v_{\text{min}} = \sqrt{rg}$ — tension/normal force is zero
- Clothoid loop (enrichment): non-circular loop with varying radius to control g-forces
Force Equations
- Top: $T = mv^2/r - mg$
- Bottom: $T = mv^2/r + mg$
- Difference: $T_{\text{bottom}} - T_{\text{top}} = 2mg$ (same speed)
Energy Method
- $\tfrac{1}{2}mv_A^2 + mgh_A = \tfrac{1}{2}mv_B^2 + mgh_B$
- $v_{\text{min(bottom)}} = \sqrt{5gr}$
- $v_{\text{min(top)}} = \sqrt{rg}$
Key Principles
- Tension always greater at the bottom than the top
- You feel heaviest at the bottom of the loop
- Minimum speed at top: $T = 0$, gravity alone provides centripetal force
Match each formula to its correct description.
Activities
Practise calculating tension at the top of a vertical circle
- A 0.3 kg ball on a 0.6 m string moves in a vertical circle. At the top, its speed is 3.5 m/s. Find the tension in the string at the top.
- For the same ball, find the minimum speed at the top to keep the string taut. Show all working.
Activity check — for a 0.3 kg ball on a 0.6 m string at the top with speed 3.5 m/s, the tension (to 2 decimal places, in N) is _____.
Determine the minimum speed needed to complete a vertical loop
- Find the minimum speed required at the bottom of a vertical loop of radius 3.0 m to complete the loop.
- A roller coaster designer wants to reduce the minimum entry speed. Should they increase or decrease the loop radius? Justify using the formula $v_{\text{min(bottom)}} = \sqrt{5gr}$.
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(2 marks) 1. A 0.4 kg stone on a 0.5 m string is whirled in a vertical circle. At the top, the tension is 3.2 N. Calculate the speed of the stone at the top of the circle.
1 mark: correct method using $T + mg = mv^2/r$ · 1 mark: correct answer with units
ApplyBand 5(3 marks) 2. A roller coaster car of mass 250 kg approaches a vertical loop of radius 6.0 m. Calculate (a) the minimum speed required at the bottom to complete the loop, and (b) the normal force on the car at the bottom at this minimum speed.
1 mark: correct $v_{\text{min(bottom)}} = \sqrt{5gr}$ · 1 mark: correct method for $N$ · 1 mark: correct $N$ with units
EvaluateBand 6(4 marks) 3. A roller coaster designer is choosing between Loop A (radius 8.0 m) and Loop B (radius 12.0 m). Cars enter each loop at 18 m/s. Evaluate which loop design is safer for riders by comparing the normal force at the bottom of each loop. Use physics principles to justify your answer.
1 mark: correct formula $N = mv^2/r + mg$ · 1 mark each: correct $N_A$ and $N_B$ · 1 mark: physics-based comparison and conclusion
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (2 marks): At the top: $T + mg = mv^2/r$. $3.2 + (0.4 \times 9.8) = 0.4 v^2 / 0.5$. $3.2 + 3.92 = 0.8v^2$. $v^2 = 7.12 / 0.8 = 8.9$. $v = 2.98 \text{ m/s}$ (1 mark method, 1 mark answer).
Q2 (3 marks): (a) $v_{\text{min(bottom)}} = \sqrt{5gr} = \sqrt{5 \times 9.8 \times 6} = \sqrt{294} = 17.1 \text{ m/s}$ (1 mark). (b) $N - mg = mv^2/r$, so $N = \dfrac{250 \times 294}{6} + (250 \times 9.8) = 12250 + 2450 = 14700 \text{ N}$ (1 mark method, 1 mark answer). Note: this equals $6mg$.
Q3 (4 marks): At the bottom: $N = mv^2/r + mg$ (1 mark). Loop A ($r = 8.0$ m): $N_A = m(18)^2/8 + mg = 40.5m + mg = 41.5m$ N. Loop B ($r = 12.0$ m): $N_B = m(18)^2/12 + mg = 27m + mg = 28m$ N (1 mark each). Loop B is safer because the larger radius produces a smaller centripetal acceleration ($a_c = v^2/r$) and therefore a smaller normal force on the rider at the bottom (lower g-force). However, the designer must also verify that the speed at the top of Loop B remains above $\sqrt{rg}$ to maintain contact throughout (1 mark).
Five timed questions on vertical circular motion. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaCircular motion orbits — maintain the right speed to keep your craft on track. Lighter than the boss — pure practice that hammers home the force-speed relationship.
At the start you were asked about the 1975 Knott's Berry Farm Corkscrew vertical loop (radius 6.1 m) and where riders feel heavier — top or bottom.
The answer is the bottom. At the bottom: $N = mv^2/r + mg$ — the seat pushes up to both support weight and provide centripetal force. At the top: $N = mv^2/r - mg$ — gravity now assists centripetal force, so the seat pushes far less. The minimum speed at the top where $N = 0$ gives $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 6.1} = 7.73 \text{ m/s}$. Modern coasters use larger-radius loops at the top (clothoid shape) to limit the bottom-of-loop g-force to around 3.5g rather than 6g. Has your prediction held up?