Year 12 PhysicsModule 5⏱ ~20 minOptional extension · 5 MC · 3 Short AnswerLesson 9 of 18
Vertical Circular Motion: Extension
You'll be able to use conservation of mechanical energy to find speed at any point in a vertical circle.
A vertical circle combines changing velocity, radial force and energy transfer. Build a free-body diagram first, choose the inward radial direction at each location, then use energy to connect speeds at different heights.
Today's hook: A rider feels light near the top of a loop and heavy at the bottom, even though their weight $mg$ is unchanged. Which real contact force changes, and how do force and energy models explain it?
0/5TASKS
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Connect and orient
Beyond the syllabus. This whole lesson is extension: vertical circular motion tension analysis and threshold speeds are not named core relationships in Module 5. They reuse core mechanics and strong students will enjoy them — but they must not affect your core readiness, and the exam's named applications are horizontal circles, banked tracks and orbits.
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
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?
Assessment boundary
Prescribed core to retain
From L07–L08, be able to identify real forces, choose the inward radial direction and apply $\sum F_{\text{in}}=mv^2/r$ to uniform circular-motion situations. Those skills are directly aligned to the NESA Module 5 dot points.
Optional extension
This lesson applies those skills to a vertical circle where gravity changes the speed, then adds energy conservation. Vertical-circle formulae such as $\sqrt{gr}$ and $\sqrt{5gr}$ are not named formulae or a standalone dot point in the 2017 NESA syllabus. Complete this lesson for deeper transfer, not before mastering L06–L08 core.
Learning Intentions
goals
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
Scan these before reading
vocab
Vertical circular motionCircular motion in a vertical plane where gravity causes the speed to vary continuously with height.
TensionThe pulling force exerted by a string or rope along its length toward the centre of the circle.
Normal forceThe perpendicular contact force exerted by a surface on an object, replaces tension in track/loop problems.
Apparent weightThe magnitude of the normal force acting on a person, what they "feel" as their weight.
Minimum speedThe critical speed at the top of a loop at which tension (or normal force) equals zero: $v_{\text{min}} = \sqrt{rg}$.
Centripetal forceThe net force directed toward the centre of a circular path: $F_c = mv^2/r$. It is not a separate force, it is the resultant of real forces.
Cross-lesson links: L08 covered the prescribed banked-track and mass-on-a-string contexts for uniform circular motion. L09 is an optional transfer lesson: it adds varying speed and energy exchange in a vertical circle. HSC questions can assess the underlying radial-force and energy reasoning, but the specialised vertical-loop results on this page are not named syllabus formulae.
Misconceptions to fix
✗ Wrong: The normal force is always greater at the top of a loop than at the bottom.
✓ Right: The normal force (or tension) is always greater at the bottom. At the bottom, the upward normal force must overcome weight and provide centripetal force: $N = mv^2/r + mg$. At the top, gravity assists, so $N = mv^2/r - mg$.
✗ Wrong: You feel weightless at the bottom of a loop.
✓ Right: You feel heaviest at the bottom because the seat pushes up with $N = mv^2/r + mg > mg$. Weightlessness occurs at the top when $v = \sqrt{rg}$ and the normal force is zero.
✗ Wrong: Any speed is enough to complete a vertical loop.
✓ Right: For a circular, lossless, just-taut string or inside-track model, the object must have $v \geq \sqrt{rg}$ at the top. If it enters at the bottom, this corresponds to $v \geq \sqrt{5gr}$ there. These results depend on those conditions.
The normal force on a rider is greater at the top of a vertical loop than at the bottom.
Zero tension or normal force at the top is the limiting just-maintained condition; a lower speed cannot sustain the assumed string or inside-track circular path.
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Set the radial sign and build the FBD
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One radial rule, changing directions
+5 XP
Draw only real forces first. For a string, the inward contact force is tension $T$. For a car or rider in contact with a track or seat, it is a normal force $N$. “Centripetal force” is not an extra arrow; it is the inward resultant of the real-force components.
Choose inward as positive separately at every location. At the top inward is downward; at the bottom inward is upward; at a side point inward is horizontal.
The radial positive direction rotates with position, while weight always points vertically downward.
Use the free-body diagram to write the radial equation; use energy to find the speed.
At a side point of a ball-on-string vertical circle, what belongs in the inward radial equation?
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Analyse the top condition
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Top of the vertical circle
+5 XP
At the top, inward is downward. Weight and the inward contact force both point toward the centre:
String or inside-track case
$T+mg=mv_{\text{top}}^2/r$ or $N+mg=mv_{\text{top}}^2/r$
$T=mv_{\text{top}}^2/r-mg$; replace $T$ with $N$ for a normal-force problem.
The just-maintained limiting condition is $T=0$ for a string, or $N=0$ for an inside track. This gives $v_{\text{top}}=\sqrt{gr}$. Zero contact force is the boundary: below this required speed, a string would need to push or a track would need to pull, so the assumed circular path cannot continue.
Applicability
$v_{\text{top}}=\sqrt{gr}$ is not a universal speed for every vertical-circle device. It applies to the just-taut string or just-maintained inside-contact case at the top. A rigid rod can push as well as pull, and a constrained track can have different contact geometry.
At the top, choose downward/inward positive: $T+mg=mv^2/r$. For just-maintained string tension, $T=0$, so $v_{\text{top,min}}=\sqrt{gr}$.
At $T=0$ the ball has already left the circular path.
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Analyse the bottom condition
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Bottom of the vertical circle
+5 XP
At the bottom, inward is upward. Tension or normal force points inward; weight points outward:
Bottom radial equation
$T-mg=mv_{\text{bottom}}^2/r$, so $T=mv_{\text{bottom}}^2/r+mg$
For a rider, apparent weight is the normal force $N$, not gravity. The seat must support the rider and provide the upward radial resultant, so $N>mg$ whenever $v>0$ at the bottom.
Worked Example 1, Tension at Top and Bottom
A 0.50 kg bob on a 0.80 m string has speed 4.0 m s$^{-1}$ at the top. Find the tension there and, assuming no losses, at the bottom.
Top FBD. $T_{\text{top}}+mg=mv_{\text{top}}^2/r$.
Top tension. $T_{\text{top}}=0.50(4.0)^2/0.80-0.50(9.8)=5.1\ \text{N}$.
Energy, top to bottom. With $h=0$ at the bottom, $v_{\text{bottom}}^2=v_{\text{top}}^2+4gr=47.36\ \text{m}^2\text{s}^{-2}$.
Check. Units are newtons and $T_{\text{bottom}}>T_{\text{top}}$. Here energy also makes the bottom speed larger.
At the bottom, choose upward/inward positive: $T-mg=mv^2/r$. For the same speed, $T_{\text{bottom}}-T_{\text{top}}=2mg$; for lossless motion between top and bottom, the speed also changes.
Why does a rider feel heavier at the bottom of a loop?
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Transfer energy between points
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Energy connects locations
+5 XP
Force equations use the speed at one location. To find how speed changes with height, use conservation of mechanical energy when resistive losses are negligible.
Choose a zero level and state it
$\tfrac12mv_A^2+mgh_A=\tfrac12mv_B^2+mgh_B$
Here the bottom is $h=0$, so the top is $h=2r$.
Bottom
$h=0$ GPE minimum speed maximum
Rising
GPE increases KE decreases speed decreases
Top
$h=2r$ GPE maximum speed minimum
For a just-complete string/inside-track loop, combine $v_{\text{top}}^2=gr$ with $\tfrac12mv_{\text{bottom}}^2=\tfrac12mv_{\text{top}}^2+mg(2r)$ to obtain $v_{\text{bottom}}=\sqrt{5gr}$.
Assumptions are part of the answer
$v_{\text{bottom}}=\sqrt{5gr}$ assumes release or entry at the bottom, a circular path of radius $r$, negligible losses, and the just-maintained condition at the top. It is not a general bottom-speed formula.
Worked Example 2, Just-complete loop
A toy car must just complete a circular loop of radius 2.5 m. Find its minimum bottom speed under the stated ideal assumptions.
Condition at top. $N=0$, so $v_{\text{top}}^2=gr$.
Check contact. $N>0$, equivalently $v_{\text{top}}=11.7\ \text{m s}^{-1}>\sqrt{gr}=6.26\ \text{m s}^{-1}$, so the car maintains contact in this ideal model.
Interactive, vertical-circle calculator
model
Change mass, speed and radius. Compare top and bottom contact forces, then test where the top force reaches zero.
Modern roller-coaster loops are often clothoid-like rather than circular, so the radius changes around the loop to manage peak acceleration. Within this optional extension lesson, calculations use the stated circular-loop model. The clothoid design detail is contextual only.
At the top of a bucket-of-water circle, what keeps the water following the circular path?
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Respond and consolidate
Wrap-up, Summary
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
Minimum bottom speed for a circular, lossless, just-complete loop
Tension at the top of a vertical circle
Conservation of mechanical energy
Activities
Activity 1, Tension Calculations
ApplyBand 4
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 _____.
Activity 2, Minimum Entry Speed
ApplyBand 5
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 $v_{\text{min(bottom)}} = \sqrt{5gr}$.
Analyse6 marks
A mass on the end of a string is swung so that it travels in a horizontal circle. Analyse the forces acting on the mass, distinguishing clearly between the individual forces and the net force. (6 marks)
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Unpack the verb
Analyse is this dot point's own verb, and the dot point names a mass on a string as one of the situations to analyse. The distinction the question asks for is the one that makes every circular-motion problem work.
Plan: list the real forces, state what the net force must be, then show how the real forces combine to produce it, and say what centripetal force is and is not.
Getting the ontology right first — which entries on the page are forces and which is a requirement — prevents the double-counting that spoils these answers.
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List the real forces
Only two forces act on the mass: its weight $mg$ vertically downwards, and the tension $T$ in the string, directed along the string towards the point of support.
Two, and only two. If a third appears in a free-body diagram for this system, something has gone wrong.
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State what the net force must be
The mass travels in a horizontal circle at constant speed, so its acceleration is horizontal, directed towards the centre of that circle, with magnitude $v^2/r$. The net force must therefore be horizontal and inward, of magnitude $mv^2/r$.
The motion determines the net force, not the other way round. This is a requirement read off the geometry before any force is resolved.
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Show how the real forces produce it
Because the circle is horizontal and below the support, the string must be inclined to the vertical. Resolving the tension, its vertical component balances the weight, $T\cos\theta = mg$, while its horizontal component is unbalanced and supplies the centripetal requirement, $T\sin\theta = mv^2/r$.
The string cannot be horizontal: if it were, there would be nothing to balance the weight. That is a useful consistency check on any diagram.
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Say what centripetal force is
The centripetal force is not a third force acting on the mass. It is the name for the net inward force, which here is the horizontal component of the tension. Adding a separate centripetal force to the free-body diagram would count the same physical effect twice.
This is the distinction the question asks for, stated directly. Naming the double-counting error makes the point concrete.
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Give the consequence
It follows that the tension is always greater than the weight in this arrangement, since $T\cos\theta = mg$ with $\cos\theta < 1$, and that swinging faster requires a larger angle and a larger tension.
A consequence drawn from the analysis is stronger than a restatement of it, and this one is testable: the string rises towards the horizontal as the speed increases.
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The full-mark answer
Only two forces act on the mass: its weight $mg$ vertically downwards and the tension $T$ along the string towards the support. The mass moves in a horizontal circle at constant speed, so its acceleration is horizontal and directed towards the centre of that circle with magnitude $v^2/r$, which means the net force on it must be horizontal and inward with magnitude $mv^2/r$. Since the circle lies below the support the string is inclined to the vertical, and resolving the tension gives a vertical component that balances the weight, $T\cos\theta = mg$, and a horizontal component that is unbalanced and supplies the centripetal requirement, $T\sin\theta = mv^2/r$. The string cannot be horizontal, because then nothing would balance the weight. The centripetal force is not an additional force acting on the mass: it is the name given to the net inward force, which here is simply the horizontal component of the tension, and adding it separately to a free-body diagram would count the same effect twice. It follows that the tension always exceeds the weight in this arrangement, and that swinging the mass faster requires both a larger angle from the vertical and a greater tension.
Six marks: the two real forces, the net force read from the motion, the resolution into components, the impossibility of a horizontal string, the centripetal-force distinction, and a testable consequence.
Evaluate5 marks
A student applies the uniform circular motion relationships to a bucket of water being swung in a vertical circle, treating the speed as constant all the way round. Evaluate this approach. (5 marks)
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Unpack the verb
This dot point names evaluate and restricts its relationships to objects "executing uniform circular motion". The task is to judge whether this situation qualifies.
Plan: state the condition uniform circular motion requires, test the vertical circle against it, then judge what may and may not be carried over.
An evaluation of a model starts from the model's own stated conditions, not from whether the answer looks reasonable.
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State the condition
Uniform circular motion requires the speed to be constant, which means the net force must be perpendicular to the velocity at all times so that no work is done on the object.
The condition is about work, not about the shape of the path. That is what makes it testable in the next step.
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Test the vertical circle against it
In a vertical circle the weight $mg$ always acts downwards while the velocity is tangential, so except at the very top and bottom the weight has a component along the direction of motion. That component does work, so the object speeds up as it descends and slows as it rises, and the speed is not constant.
Identify exactly where the condition fails and where it momentarily holds. At the top and bottom the weight is perpendicular to the velocity, which is why those two points are the ones usually analysed.
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Say what breaks and what survives
So the motion is not uniform circular motion and any result assuming constant speed throughout — a single value of $v$, a constant period, or a constant net force — is invalid. What does survive is the centripetal requirement itself: at any instant, $F_{\text{net, inward}} = mv^2/r$ still holds with the instantaneous speed at that point, because that follows from the geometry of circular motion and not from the speed being constant.
This is the discriminating judgement. The relationship is still true instant by instant; it is the assumption of a single $v$ everywhere that fails.
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Deliver the verdict
The approach is therefore invalid as applied. A correct treatment analyses specified points separately, using the instantaneous speed at each and linking different points by conservation of energy rather than by assuming the speed is the same at both.
Saying what the correct treatment does is what makes this an evaluation rather than a rejection, and it points at the energy method without needing any threshold result.
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The full-mark answer
Uniform circular motion requires the speed to be constant, which in turn requires the net force to be perpendicular to the velocity at every instant so that no work is done on the object. A bucket swung in a vertical circle does not meet that condition: the weight always acts downwards while the velocity is tangential, so except at the very top and the very bottom the weight has a component along the direction of motion, and that component does work — the bucket speeds up as it descends and slows as it rises. The motion is therefore not uniform, and any result that assumes a single constant speed throughout, such as a constant period or a constant net force, is invalid here. What does carry over is the centripetal requirement itself: at any instant the net inward force still equals $mv^2/r$ using the instantaneous speed at that point, because that follows from the geometry of circular motion rather than from the speed being constant. The student's approach is therefore invalid as applied; the correct treatment analyses specified points individually with the speed each has at that point, and connects different points using conservation of energy rather than assuming equal speeds.
Five marks: the condition stated in terms of work, the test against the vertical circle, the two points where it momentarily holds, what fails and what survives, and a verdict that names the correct method.
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Apply vertical circular motion independently
Quick recall, vertical circular motion
+5 XP
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.
Optional extension short answer, 9 marks
+5 XP
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). A 300 kg car enters each loop at 18 m/s. Evaluate which loop design produces the smaller normal force at the bottom. Use physics principles to justify your answer and identify one further safety check.
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.
Q3 (4 marks): At the bottom, $N=m(v^2/r+g)$ (1 mark). Loop A: $N_A=300(18^2/8.0+9.8)=\mathbf{1.51\times10^4\ N}$ (1 mark). Loop B: $N_B=300(18^2/12.0+9.8)=\mathbf{1.10\times10^4\ N}$ (1 mark). Loop B produces the smaller bottom normal force because its larger radius gives a smaller $v^2/r$. This comparison alone does not establish overall safety: the designer must also use energy to find the top speed and verify $N_{\text{top}}\geq0$ throughout the loop (1 mark).
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Retrieve and reflect
Check what actually stuck
Take the full module quiz
quiz
The full module quiz prioritises prescribed Module 5 content. Treat this vertical-circle lesson as optional transfer practice rather than a prerequisite for core mastery.
Circular 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.
How did your thinking change?
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 the radial resultant. At the top: $N = mv^2/r - mg$, gravity assists the inward resultant, so the seat pushes far less. In the circular, just-contact model, $N = 0$ gives $v_{\min} = \sqrt{gr}$. Real coaster loops are not perfect circles: a larger radius near the bottom reduces peak acceleration, while a smaller radius near the top helps maintain contact. Has your prediction held up?
