ARCHETYPE: Motion in a Vertical Circle

Whirl a bucket of water over your head. At the top of the swing the bucket is upside down, the water is directly above your face — and nothing spills. Every board in this archetype's list has asked why, and the answer is more subtle than "centrifugal force holds it in": at the top of a fast-enough vertical circle, gravity is not fighting the motion at all. Gravity is the centripetal force, fully employed in curving the water's path. The water does not fall out because falling is exactly what it is doing — falling along a circle. The vertical circle is the first genuinely non-uniform circular motion exams pose. Unlike the flat curve or the conical pendulum, the speed here changes constantly — fastest at the bottom, slowest at the top — so two chapters must cooperate: Newton's second law along the radius at each point, and energy bookkeeping to connect the speeds at different heights. Out of that cooperation fall three celebrated results: the minimum top speed $\sqrt{gL}$ below which a string goes slack, the minimum bottom speed $\sqrt{5gL}$ to complete the loop, and the marvellous 6mg theorem — the tension at the bottom exceeds the tension at the top by six times the weight, no matter how fast the circle is whirled. This article drills the full archetype: the tension map around the circle, string versus rod (they have different minimums, and examiners know you'll mix them), the three regimes of an underpowered swing, and the bucket, the pilot and the coaster that carry the same physics.

Key takeaways

In a vertical circle the speed varies with height: energy conservation gives $v^2 = u^2 - 2gL(1-\cos\theta)$ , with $u$ the bottom speed and $\theta$ measured from the bottom.
Radial Newton's law at the extremes: top $T_{\rm top} + mg = \dfrac{mv_{\rm top}^2}{L}$ ; bottom $T_{\rm bot} - mg = \dfrac{mv_{\rm bot}^2}{L}$ .
String minimums: $v_{\rm top} \ge \sqrt{gL}$ (tension just reaches zero; gravity alone curves the path) and hence $v_{\rm bot} \ge \sqrt{5gL}$ .
The 6mg theorem: $T_{\rm bot} - T_{\rm top} = 6mg$ always — independent of speed, mass, or radius.
Rod or bead: a rod can push, so $v_{\rm top}$ may be zero; the rod's minimum bottom speed is $\sqrt{4gL}$ , not $\sqrt{5gL}$ .
Underpowered string ( $\sqrt{2gL} < u < \sqrt{5gL}$ ): the string goes slack above the horizontal at $\cos\theta = \dfrac{2gL - u^2}{3gL}$ and the bob flies off as a projectile.
Key idea: solve every vertical-circle problem with radial Newton's law at the point in question plus energy conservation between points — and remember the critical condition is $T = 0$ , never $v = 0$ .

The core insight

A body on a vertical circle obeys radial Newton's law at each point while energy conservation links the speeds at different heights; for a string the survival condition at the top is tension reaching zero with gravity alone paying $mv^2/L$ — which forces $v_{\rm top}^2 \ge gL$ , $v_{\rm bot}^2 \ge 5gL$ , and the universal $6mg$ bottom-to-top tension gap.

Where this lives today

Roller-coaster loops are the tension map with passengers inside. A circular loop taken at minimum speed loads riders at $6g$ at the bottom — beyond the blackout threshold for most humans — which is why every modern loop is a clothoid: the radius is large at the bottom (shrinking $v^2/R$ exactly where speed peaks) and small at the top (so the crest clears the $v^2 \ge gR$ bar with margin at modest speed). Fighter pilots fly the same physics with their own bodies: pulling out of a dive is the bottom of a vertical circle, and the " $g$ -suit" squeezes the legs precisely because $N = mg + mv^2/R$ drains blood downward; aerobatic routines quote manoeuvres in multiples of $mg$ straight off equation (2). And astronaut training's "vomit comet" flies the top of the map continuously — a parabolic arc holding the $T = 0$ , pure-free-fall condition for 25 seconds at a time, manufacturing weightlessness without leaving the atmosphere.

The setup

A bob of mass $m$ on a light inextensible string of length $L$ (or a bead on a ring, a ball inside a track, a coaster on a loop of radius $R$ ) moves in a vertical circle. Two forces act throughout: gravity, constant and down; and the string's tension (or track's normal force), always along the radius. The radial force equation changes character around the circle — at the bottom tension and gravity oppose, at the top they cooperate — while the tangential component of gravity speeds the bob down the descending half and brakes it up the ascending half.

Figure

The vertical circle — the invoice is mv squared over L everywhere, but the payers change roles

Two structural facts organize everything:

The speed is not yours to choose point by point. Once the bob passes the bottom at speed $u$ , energy conservation dictates the speed everywhere else: rising a height $h$ costs $v^2 = u^2 - 2gh$ (the string does no work — it pulls along the radius while the bob moves along the tangent). For a circle, $h = L(1 - \cos\theta)$ with $\theta$ from the bottom, so the top ( $h = 2L$ ) always runs $4gL$ of $v^2$ poorer than the bottom. This borrowed result from the energy chapter is the same formula as free fall, $v^2 = u^2 - 2gh$ , because gravity is the only force doing work.

A string can only pull. Tension is non-negative; the string cannot push the bob outward. So at the top, where the required centripetal force is $mv_{\rm top}^2/L$ and gravity already supplies $mg$ inward, the string supplies only the shortfall: $T_{\rm top} = mv_{\rm top}^2/L - mg$ . If the bob is too slow, that expression would go negative — impossible — and instead the string goes slack and the bob departs the circle. A rod (or the inside of a track approached from outside, or a bead on a wire) can push, and that single mechanical difference splits every minimum-speed answer in two.

The master solution pattern

Every vertical-circle question is two tools used at the right points. Tool one, radial Newton's law at the point of interest, with inward positive; tool two, energy conservation between that point and wherever the speed is known. At the top, tension and gravity both point toward the centre:

T_{\rm top} + mg = \frac{m v_{\rm top}^2}{L} \qquad (1)

where: the critical condition is $T_{\rm top} = 0$ (string on the verge of slack), giving $v_{\rm top,\min}^2 = gL$ . At this minimum the bob is momentarily in free fall — gravity alone curves it, and a rider at the crest of a loop taken at minimum speed feels weightless. Below this speed the string cannot "un-pull"; the circle is lost before the top. At the bottom, tension fights gravity:

T_{\rm bot} - mg = \frac{m v_{\rm bot}^2}{L} \qquad (2)

where: the bottom is the tension's worst moment — it must carry the weight and the full invoice at the circle's highest speed. Feeding the minimum-top condition through the energy link $v_{\rm bot}^2 = v_{\rm top}^2 + 4gL$ gives $v_{\rm bot,\min} = \sqrt{5gL}$ and $T_{\rm bot} = 6mg$ at that minimum. At a general angle $\theta$ from the bottom, the radial component of gravity is $mg\cos\theta$ outward (below the horizontal) and the energy link fixes $v^2$ :

T = \frac{mu^2}{L} - 2mg + 3mg\cos\theta \qquad (3)

where: $u$ is the bottom speed. This tension map is the archetype's master formula: it decreases smoothly from the bottom ( $\theta = 0$ ) to the top ( $\theta = 180°$ ), and the $6mg$ theorem sits inside it in plain sight — the $3mg\cos\theta$ term swings from $+3mg$ to $-3mg$ across the map, so $T(0°) - T(180°) = 6mg$ whatever the value of $u$ . Setting $T = 0$ locates where an underpowered string goes slack.

you want	use	notes
min top speed (string/track)	$v_{\rm top}^2 = gL$ , from $T = 0$ in (1)	free-fall crest; rider weightless
min bottom speed (string)	$v_{\rm bot} = \sqrt{5gL}$	energy link adds $4gL$
min bottom speed (rod/bead)	$v_{\rm bot} = \sqrt{4gL}$	rod can push: $v_{\rm top} = 0$ allowed
tension anywhere	map (3)	slack where $T = 0$
bottom-vs-top tension	$T_{\rm bot} - T_{\rm top} = 6mg$	universal, speed-independent
bucket of water at top	$v^2 \ge gr$	N of bucket floor plays the string's role

Derivation AThe minimum speeds and the 6mg theorem

We derive the three string classics: the minimum top speed, the minimum bottom speed, and the universal $6mg$ tension gap. The strategy: radial Newton's law at the two special points — where all forces are conveniently radial — with energy conservation as the bridge between their speeds. Watch the signs: gravity is a payer of centripetal force at the top and a burden on the tension at the bottom; that asymmetry is the entire story.

Step 1. Top of the circle, inward positive. Both the string ( $T_{\rm top}$ ) and gravity ( $mg$ ) point toward the centre:

T_{\rm top} + mg = \frac{mv_{\rm top}^2}{L}

Why the crisis is here: a string can only pull. If $v_{\rm top}^2 < gL$ , the invoice $mv_{\rm top}^2/L$ is smaller than what gravity alone supplies, so the equation demands $T_{\rm top} < 0$ — a push the string cannot deliver. The string goes slack and the bob leaves the circle. The survival threshold is $T_{\rm top} = 0$ : $v_{\rm top,\min} = \sqrt{gL}$ , a crest curved by gravity alone.

Step 2. Bottom of the circle, inward (upward) positive. Tension pulls up, gravity drags down:

T_{\rm bot} - mg = \frac{mv_{\rm bot}^2}{L}

Step 3. Energy bridge across the height $2L$ (string does no work; gravity is the only worker):

\tfrac12 mv_{\rm bot}^2 = \tfrac12 mv_{\rm top}^2 + mg(2L) \;\;\Rightarrow\;\; v_{\rm bot}^2 = v_{\rm top}^2 + 4gL

Feed the threshold through: $v_{\rm bot,\min}^2 = gL + 4gL = 5gL$ , so $v_{\rm bot,\min} = \sqrt{5gL}$ .

Step 4. Subtract Step 1 from Step 2, using the bridge:

T_{\rm bot} - T_{\rm top} = \frac{m(v_{\rm bot}^2 - v_{\rm top}^2)}{L} + 2mg = \frac{m(4gL)}{L} + 2mg = 6mg

Why the speed cancelled: the gap has two fixed contributions — $2mg$ from gravity switching sides (burden at the bottom, payer at the top), and $4mg$ from the energy-mandated speed difference across the $2L$ climb. Neither depends on how fast the circle is whirled. Any string rated below "top requirement $+\ 6mg$ " fails at the bottom, however gentle the swing looks up top. *Verified: at the minimum, $T_{\rm top} = 0$ gives $T_{\rm bot} = 6mg$ directly from (2): $m(5gL)/L + mg = 6mg$ ✓. Numbers for $m = 1$ kg, $L = 1$ m, $g = 10$ : $v_{\rm top} = 3.16$ m/s, $v_{\rm bot} = 7.07$ m/s, $T_{\rm bot} = 60$ N — worked example 1. Standard treatment: NCERT Ch. 5–6, HC Verma Ch. 8, Halliday Ch. 6.*

Go deeper · Derivation B · 2 minThe tension map and where the string goes slack

Derivation BThe tension map and where the string goes slack

We derive the general-angle formula (3) and the slack-point condition for an underpowered swing — the JEE-Advanced half of the archetype. The strategy: write the radial equation at angle $\theta$ from the bottom, then eliminate the local speed with the energy bridge to the bottom. Watch which component of gravity enters: only the radial part $mg\cos\theta$ appears in the force equation; the tangential part $mg\sin\theta$ changes the speed, and the energy method has already accounted for it.

Step 1. Geometry: at angle $\theta$ from the bottom, the bob has risen $h = L(1 - \cos\theta)$ . For $\theta < 90°$ gravity's radial component $mg\cos\theta$ points outward (away from the centre); the string's $T$ points inward.

Step 2. Radial Newton's law, inward positive:

T - mg\cos\theta = \frac{mv^2}{L}

Step 3. Energy bridge from the bottom (speed $u$ ):

v^2 = u^2 - 2gL(1 - \cos\theta)

Substitute:

T = \frac{mu^2}{L} - 2mg + 3mg\cos\theta

Why $3mg\cos\theta$ : one $mg\cos\theta$ from gravity's direct radial pull, plus $2mg\cos\theta$ smuggled in through the speed the energy bridge assigns at that height. Students who quote " $T$ falls by $mg\cos\theta$ " have counted only the first and missed two-thirds of the effect.

Step 4. The slack point. Set $T = 0$ :

\cos\theta_{\rm slack} = \frac{2gL - u^2}{3gL}

Read its regimes. If $u^2 \ge 5gL$ : the right side is $\le -1$ , no solution — the string never slackens; full circles. If $u^2 \le 2gL$ : the right side is $\ge 0$ … but check the speed first: $v = 0$ occurs at $\cos\theta = 1 - u^2/2gL$ , before $T$ can vanish below the horizontal — the bob stops and swings back, an oscillating pendulum. In between, $2gL < u^2 < 5gL$ : the string goes slack at an angle above the horizontal ( $\cos\theta_{\rm slack} < 0$ ), with speed still nonzero, and the bob departs on a projectile arc from that point. Why the three regimes matter: "will it complete the circle / oscillate / leave the circle?" is a standard multiple-choice stem, answered instantly by placing $u^2$ against the landmarks $2gL$ and $5gL$ . *Verified: limits — $u^2 = 5gL$ gives $\cos\theta_{\rm slack} = -1$ (slack exactly at the top, the critical circle) ✓; $u^2 = 2gL$ gives $\cos\theta_{\rm slack} = 0$ with $v = 0$ there too (barely reaches the horizontal) ✓; map (3) at $\theta = 0,\pi$ reproduces (2),(1) and the $6mg$ gap ✓. Worked example 4 runs $u^2 = 4gL$ : slack at $\cos\theta = -2/3$ , i.e. $41.8°$ above the horizontal, with $v^2 = \tfrac{2}{3}gL$ .*

Figure

The tension map — T around the circle for three bottom speeds

String versus rod — the halved minimum

Replace the string with a light rigid rod and the top constraint dissolves: a rod can push, so $T < 0$ is allowed and the bob merely needs to arrive at the top, $v_{\rm top} \ge 0$ . The energy bridge then gives the rod's minimum bottom speed as $v_{\rm bot} = \sqrt{4gL} = 2\sqrt{gL}$ — it must only climb the height $2L$ , not also fund a free-fall crest. A bead threaded on a vertical ring and a ball inside a grooved track share the rod's rule; a coaster on the inside of a loop and water in a bucket share the string's. Exams flip between the two in consecutive parts, and the entire difference is one physical question: can the constraint push?

Limiting cases

case	result	meaning
$v_{\rm top}^2 = gL$ exactly	$T_{\rm top} = 0$	free-fall crest; rider momentarily weightless
$u^2 \ge 5gL$ (string)	full circles, $T > 0$ throughout	the map (3) never touches zero
$2gL < u^2 < 5gL$ (string)	slack above the horizontal, then projectile	the three-regime middle band
$u^2 \le 2gL$ (string)	oscillates below the horizontal	$v$ hits zero before $T$ does
rod, $u^2 = 4gL$	just crawls over the top with $v_{\rm top} = 0$	pushing constraint halves the energy demand
any speed, any radius	$T_{\rm bot} - T_{\rm top} = 6mg$	the gap is structural, not dynamical
$L \to 0$	all thresholds $\to 0$	a tiny circle is easy — thresholds scale as $\sqrt{L}$

Common trapAt the top the bob can afford to be at rest, since the string will hold it

For a string, it cannot — and this error inverts the entire archetype. A string holds by pulling inward, which only adds to the centripetal demand: at the top the bob needs $mv^2/L$ of inward force just to stay on the circle, and gravity supplies $mg$ of it whether you like it or not. If $v^2 < gL$ the bob needs less inward force than gravity provides — the surplus cannot be cancelled (the string cannot push outward), so the bob departs the circular path early and the string folds. The minimum condition is therefore about tension reaching zero, not speed reaching zero: $T_{\rm top} = 0$ , $v_{\rm top} = \sqrt{gL}$ , a crest taken in pure free fall. Reserve " $v_{\rm top} = 0$ " for the rod, the bead on a wire, and the groove — constraints that can push. And its cousin trap: "the water in the inverted bucket is held by the reaction of centrifugal force". No — the water stays because the bucket floor pushes it down, helping gravity curve the water into the circle. Everything at the top points inward; nothing needs to point out.

Worked examples

1 · Board level — the stone on a string (CBSE/NEET pattern, 3 marks)

A $1.0$ kg stone is whirled in a vertical circle on a $1.0$ m string. Find (a) the minimum speed at the top for the string to stay taut, (b) the corresponding speed at the bottom, (c) the tension at the bottom at that minimum. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) Critical condition at the top is $T = 0$ : gravity alone pays $mv^2/L$ , so $v_{\rm top} = \sqrt{gL} = \sqrt{10} = \boxed{3.2\ \mathrm{m/s}}$ .

Step 2. (b) Energy bridge over height $2L$ : $v_{\rm bot}^2 = v_{\rm top}^2 + 4gL = 10 + 40 = 50$ , so $v_{\rm bot} = \sqrt{5gL} = \boxed{7.1\ \mathrm{m/s}}$ .

Step 3. (c) Bottom equation: $T_{\rm bot} = \dfrac{mv_{\rm bot}^2}{L} + mg = \dfrac{1(50)}{1} + 10 = \boxed{60\ \mathrm{N}} = 6mg$ .

Step 4. Sanity: the $6mg$ theorem predicts $T_{\rm bot} - T_{\rm top} = 60 - 0 = 6mg$ ✓. State the physical read for the last mark: at minimum, the crest is free fall ( $T=0$ ) and the bottom loads the string with six weights.

2 · Board level — the bucket of water (ICSE/CBSE pattern)

A bucket of water is whirled in a vertical circle of radius $0.9$ m. (a) What minimum speed at the top keeps the water in? (b) How many revolutions per minute does the corresponding uniform whirl correspond to, roughly? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) The water presses on the bucket floor; the floor's normal force $N$ plays the tension's role: $N + mg = mv^2/r$ with survival condition $N \ge 0$ . Minimum: $v_{\rm top} = \sqrt{gr} = \sqrt{9} = \boxed{3.0\ \mathrm{m/s}}$ .

Step 2. (b) Treat the whirl as roughly uniform at $3.0$ m/s: period $T_{\rm rev} = \dfrac{2\pi r}{v} = \dfrac{2\pi(0.9)}{3.0} = 1.9$ s.

Step 3. Rate: $60/1.9 \approx \boxed{32\ \mathrm{rev/min}}$ — a lazy, entirely human arm speed. Slower than this and the water leaves the bucket at the crest (both then fall, briefly together).

Step 4. The concept mark: at the minimum the water is in free fall at the top — the bucket does not "hold the water up"; bucket and water fall together along the curve, like the ISS and its astronauts.

3 · Competitive level — the tension pair and the 6mg audit (JEE-Main pattern, ~2.5 minutes)

A $0.5$ kg bob on a $2.0$ m string passes the bottom of its vertical circle at $12$ m/s. Find the tension at the bottom and at the top, and verify the tension gap. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Bottom: $T_{\rm bot} = \dfrac{mu^2}{L} + mg = \dfrac{0.5 \times 144}{2} + 5 = 36 + 5 = \boxed{41\ \mathrm{N}}$ .

Step 2. Energy bridge to the top: $v_{\rm top}^2 = 144 - 4gL = 144 - 80 = 64$ — comfortably above $gL = 20$ : the circle completes.

Step 3. Top: $T_{\rm top} = \dfrac{mv_{\rm top}^2}{L} - mg = \dfrac{0.5 \times 64}{2} - 5 = 16 - 5 = \boxed{11\ \mathrm{N}}$ .

Step 4. Audit: $T_{\rm bot} - T_{\rm top} = 41 - 11 = 30$ N $= 6mg = 6(0.5)(10)$ ✓ — the theorem holds at any speed, which is also the fastest way to get one tension from the other on an MCQ. Timing note: ~2.5 min, or ~60 s using the theorem: compute $T_{\rm bot}$ , subtract $6mg$ . Options are built around forgetting the $-mg$ at the top (16 N) or the $+mg$ at the bottom (36 N).

4 · Competitive level — the underpowered swing (JEE-Advanced flavour, ~4 minutes)

A bob on a string of length $L$ passes the bottom at $u = 2\sqrt{gL}$ ( $u^2 = 4gL$ ). (a) Show it cannot complete the circle. (b) Find where the string goes slack. (c) Find its speed there. What happens next?

Step 1. (a) Full circles need $u^2 \ge 5gL$ ; here $u^2 = 4gL < 5gL$ . And oscillation needs $u^2 \le 2gL$ ; $4gL$ exceeds that too. The bob is in the middle regime: it will leave the circle mid-flight.

Step 2. (b) Slack condition from the tension map: $\cos\theta_{\rm slack} = \dfrac{2gL - u^2}{3gL} = \dfrac{2gL - 4gL}{3gL} = -\dfrac{2}{3}$ , so $\theta_{\rm slack} = 131.8°$ from the bottom — that is $\boxed{41.8°\ \text{above the horizontal}}$ .

Step 3. (c) Speed there from the energy bridge: $v^2 = u^2 - 2gL(1 - \cos\theta) = 4gL - 2gL\left(1 + \tfrac23\right) = 4gL - \tfrac{10}{3}gL = \boxed{\tfrac{2}{3}gL}$ — nonzero: the bob is still moving when the string folds.

Step 4. Next: with the string slack, the only force is gravity — the bob becomes a projectile, launched at $\sqrt{2gL/3}$ along the tangent at that point, on a parabola until (in the classic follow-up) the string snaps taut again. The examiner's favourite check: at the slack point, gravity's radial component $mg\cos(180° - \theta)\ldots$ equals exactly $mv^2/L = \tfrac23 mg$ ✓ — the circle hands over to free flight with no jerk. Timing note: ~4 min. The regime placement (Step 1) is instant marks; the slack formula is worth memorizing in the form $\cos\theta = (2gL - u^2)/3gL$ . Advanced variants ask for the projectile's peak or where it crosses the circle — set up axes at the slack point and it is kinematics from there.

Exam traps

Using $v = 0$ at the top for a string. The minimum condition is $T = 0$ with $v_{\rm top} = \sqrt{gL}$ . Zero top speed is legal only for rods, beads and grooves — constraints that can push.

Mixing the two minimums. String: $\sqrt{5gL}$ at the bottom. Rod: $\sqrt{4gL} = 2\sqrt{gL}$ . Consecutive exam parts love the switch; the physical question to ask is always "can it push?"

Treating the motion as uniform. $v$ changes with height — every equation must carry the energy bridge. "Find the tension at the side point given the bottom speed" is unsolvable without $v^2 = u^2 - 2gL$ .

Only counting gravity's direct pull in the map. $T$ falls by $3mg\cos\theta$ -worth per the map, not $mg\cos\theta$ : two-thirds of the variation rides in through the speed. Quote equation (3), not an improvised one.

Forgetting the theorem. $T_{\rm bot} - T_{\rm top} = 6mg$ regardless of speed — the fastest route between the two tensions, and a free consistency check on any pair you compute separately.

Where it breaks. A string that will snap does so at the bottom (maximum tension, maximum speed); an underpowered swing goes slack above the horizontal ( $\cos\theta_{\rm slack} < 0$ ), never below.

Radius versus length. For a coaster loop the "string length" is the loop radius $R$ ; for a bob it is the string length $L$ . The $\tfrac52 R$ release-height classic belongs to the energy chapter's loop-the-loop module — here, know that it follows from $v_{\rm top}^2 = gR$ plus the energy bridge.

Which $g$ . CBSE/ISC: $9.8$ unless stated; JEE/NEET: usually a declared $10$ ; AP allows $10$ for estimates. The landmark speeds scale as $\sqrt{g}$ — do not mix.

Origins · 1 minÉmilie du Châtelet and how this was discovered

ÉD

Émilie du Châtelet

1706–1749 · Cirey & Paris

The Marquise du Châtelet translated Newton's Principia into French — still the standard translation — and fought Europe's philosophers over the quantity that makes vertical circles solvable: vis viva, the "living force" $mv^2$ . Against the Cartesian view that motion is measured by $mv$ , she championed and publicized Willem 's Gravesande's experiments (brass balls dropped into clay: double the speed, four times the dent) showing that motion's "cost" scales with the square of speed. That squared bookkeeping — our kinetic energy — is exactly what links the top and bottom of a vertical circle: the crest runs $4gL$ of $v^2$ poorer than the base, no matter the mass, which is where the 6mg theorem's speed-independence is born. She completed her Newton commentary while pregnant at 42, working eighteen-hour days against a deadline she sensed correctly; she died days after the birth, the manuscript finished.

History

1673

— Huygens publishes the laws of centrifugal tendency: circular motion demands a centre-seeking force growing as v²/r.

1687

— Newton's Principia supplies the dynamics; du Châtelet's French translation (1749) carries it to the Continent with vis-viva commentary.

1740

— Du Châtelet's Institutions de physique argues the mv² bookkeeping that becomes energy conservation — the vertical circle's second tool.

1901

— The Flip-Flap Railway at Coney Island runs a perfectly circular loop; its ~12g bottom loads injure necks and teach engineers the 6mg lesson the hard way.

1976

— The Revolution at Six Flags debuts the modern clothoid (teardrop) loop: wide radius at the bottom, tight at the top, taming both ends of the tension map.

Going further

Texts. NCERT Physics Class 11, Ch. 5–6 · HC Verma, Concepts of Physics Vol. 1, Ch. 8 (the vertical-circle problem set) · Halliday–Resnick–Walker Ch. 6 and Morin Ch. 3 for the loop family.

Provenance. All results are standard. A verifier should recompute: the landmark speeds ( $\sqrt{gL}$ , $\sqrt{5gL}$ , rod $\sqrt{4gL}$ ); the map $T = mu^2/L - 2mg + 3mg\cos\theta$ and its $6mg$ span; example 1 ( $3.2$ , $7.1$ m/s, $60$ N); example 3 ( $41$ N, $11$ N, gap $30$ N $= 6mg$ ); example 4 ( $\cos\theta = -2/3$ , $v^2 = \tfrac23 gL$ , radial handover check).

Prerequisites. Centripetal Force: Dynamics of Circular Motion · Tension & Ideal Strings · Conservation of Mechanical Energy (used as the speed bridge). Leads to. Loop-the-Loop & Curved Tracks · Rotor & Well of Death · Limiting & Extreme Cases (technique) · Centripetal Force Dynamics (UG deep dive).

Why doesn't the water fall out of the bucket at the top? It's upside down!

The water is falling — along the circle. At the crest the water needs an inward (downward) force of $mv^2/r$ to keep curving, and gravity supplies $mg$ of it. If $v^2 \ge gr$ , gravity alone is not even enough, and the bucket floor must push down to top it up. "Falling out" would mean the water accelerating downward faster than the bucket — but the bucket's crest acceleration is at least $g$ , and the free water can manage only exactly $g$ . The bucket outruns any escape attempt.

Where does the energy equation come from if we haven't finished the energy chapter?

From work done by gravity — and you already own the special case. For any motion where gravity is the only force doing work (the string pulls perpendicular to the path, so it does none), speeds and heights trade by $v^2 = u^2 - 2gh$ : the same formula as vertical free fall, now applied along a curved path. The energy chapter (next unit) derives why path shape does not matter; here you may use the result, and every board expects you to.

Why is the critical condition T = 0 and not v = 0?

Because a string fails by slackening, not by stopping. As the bob slows near the top, the tension shrinks first — the string is the adjustable payer, gravity is fixed. Tension hits zero while the bob still has speed $\sqrt{gL}$ ; any slower and the string would need to push outward to keep the bob on the circle, which strings cannot do. Speed reaching zero is the failure mode of the climb (oscillating regime), and it happens only when the bob is still below the horizontal.

Is the 6mg theorem really independent of everything? Even mass and radius?

Of speed, yes — that is the miracle: $2mg$ comes from gravity switching from burden (bottom) to payer (top), and $4mg$ from the energy-mandated $v^2$ difference across the $2L$ climb; both are fixed. Of mass, no — the gap is $6mg$ , so it scales with $m$ . Of radius, yes — $L$ cancels between the invoice and the energy bridge. Conditions: the speeds must be linked by energy conservation (no motor, negligible friction) and the path must be a circle. A powered bob (constant speed) breaks the link and the gap becomes $2mg$ .

What actually happens after the string goes slack?

Projectile motion. At the slack point the bob still has speed (equation: $v^2 = \tfrac{2}{3}gL$ for the $u^2 = 4gL$ case), directed along the tangent. With the string folded, gravity is the only force, so the bob flies a parabola — rising, cresting, falling — until the string straightens again (a jerk, and typically an energy loss) or the bob returns to the circle's interior. Advanced problems chase that parabola; the handover itself is smooth because $T = 0$ means the string was already contributing nothing.

Does the bob move fastest at the bottom even when whirled hard?

Always — energy conservation is indifferent to how hard you whirl. Whatever happens at the top, the bottom runs $v_{\rm bot}^2 = v_{\rm top}^2 + 4gL$ : the descent converts $2L$ of height into speed, every lap. That is also why the string's danger point is always the bottom: maximum speed and gravity-as-burden coincide there, stacking the tension to its map maximum $mu^2/L - 2mg + 3mg$ .

How is the coaster loop different from the bob on a string?

Structurally identical, with the track's normal force playing the tension — the track (from inside) can only push inward toward the loop's centre... which is the same one-signed constraint as a string's pull-only. Same minimums, same map, same 6mg. The differences are practical: loops use the energy of a preceding drop rather than a whirling arm (release height $h \ge \tfrac52 R$ for a frictionless circular loop — the loop-the-loop module), and real loops are clothoid-shaped to keep the bottom g-load humane.