ARCHETYPE: The Conical Pendulum

Every fairground has one: the chair-o-plane, a carousel of swings that fly outward as the ride spins up, chains slanting at an angle no operator chose. Nobody sets that angle — the physics negotiates it. Spin faster and the chairs swing wider; slow down and they fall back toward vertical; and no matter how furious the spin, the chains never quite reach horizontal. The ride is a conical pendulum: a mass on a string sweeping a horizontal circle while the string traces a cone, and it is the cleanest "two jobs, one force" problem in mechanics — the tension must simultaneously hold the bob's weight and pay the full centripetal bill. The negotiation has a beautiful bottom line. The tilt obeys $\tan\theta = v^2/rg$ — the same relation as a banked road, with the string's slant playing the embankment. The period obeys $T = 2\pi\sqrt{L\cos\theta/g}$ , which hides a gem: it depends only on the vertical height of the cone, not on how wide it opens. And the whole machine has a switching threshold: below a critical spin rate $\omega_{\min} = \sqrt{g/L}$ the cone refuses to open at all, and the string simply hangs straight down. JEE asks the threshold and the tension; NEET asks the period; A-Level asks the full resolve-and-divide analysis. This article drills them all — the master pattern, the impossible horizontal string, the chair-o-plane's composite radius, and the trap that catches the most students: resolving gravity along the string, which quietly solves the wrong pendulum.

Key takeaways

A conical pendulum is a bob on a string of length $L$ sweeping a horizontal circle of radius $r = L\sin\theta$ at constant speed, the string tilted at $\theta$ from the vertical.
The tension does two jobs: vertical component holds the weight ( $T\cos\theta = mg$ ), horizontal component pays the invoice ( $T\sin\theta = mv^2/r$ ). Divide: $\tan\theta = \dfrac{v^2}{rg}$ .
The tension itself: $T = \dfrac{mg}{\cos\theta} > mg$ always — and, elegantly, $T = m\omega^2 L$ when written with the angular speed.
Period: $T_{\rm p} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$ — set by the cone's height $h = L\cos\theta$ alone, not its width.
Given the spin rate: $\cos\theta = \dfrac{g}{\omega^2 L}$ , which has a solution only for $\omega \ge \omega_{\min} = \sqrt{g/L}$ — spin slower and the string hangs vertical.
A horizontal string is impossible: $\theta \to 90°$ demands $T \to \infty$ and $v \to \infty$ .
Key idea: resolve the TENSION (never gravity) into vertical and horizontal parts — one balances the weight, the other is the entire centripetal force, and every result follows from their ratio.

The core insight

In a conical pendulum the string tension splits its effort — $T\cos\theta$ balances gravity while $T\sin\theta$ supplies the whole centripetal force — so the geometry is negotiated, not chosen: $\tan\theta = v^2/rg$ , the period depends only on the cone's height through $T_{\rm p} = 2\pi\sqrt{L\cos\theta/g}$ , and the cone opens only above the threshold spin $\omega_{\min} = \sqrt{g/L}$ .

Where this lives today

The chair-o-plane still flies every fairground, and its engineering sheet is this article: chain tension rated at $m\omega^2 L$ per seat (with $\omega$ at the ride's maximum), the same for every rider mass per kilogram — which is why the ride's limit is chain strength, not passenger count symmetry. Aircraft in a coordinated banked turn obey the identical force split with wing lift as the tension: $\tan\theta = v^2/rg$ gives the bank angle for a standard-rate turn, and pilots feel $1/\cos\theta$ times their weight — the "2g turn" at $60°$ of bank is worked example 4's arithmetic in a cockpit. Centrifugal governors survive in engine overspeed cut-outs and old record players; and in the lab, the conical pendulum is the standard classroom measurement of $g$ without a stopwatch-killing swing: time laps of the cone, measure its height, and $g = 4\pi^2 h/T_{\rm p}^2$ .

The setup

Geometry first. A bob of mass $m$ hangs on a light inextensible string of length $L$ from a fixed point. Instead of swinging in a vertical plane (the simple pendulum's business), the bob is set moving in a horizontal circle at constant speed. The string sweeps a cone of half-angle $\theta$ from the vertical; the circle's radius is $r = L\sin\theta$ and its plane sits a height $h = L\cos\theta$ below the support. Unlike the vertical circle, this motion is uniform — the speed never changes, because both forces (tension, gravity) have zero component along the direction of travel.

Figure

The conical pendulum — geometry of the cone and the two jobs of the tension

Two readings anchor the whole archetype:

The acceleration is horizontal, so resolve along horizontal and vertical. The bob accelerates toward the circle's centre — horizontally — at $v^2/r$ . The clean axes are therefore horizontal/vertical, and the thing to resolve is the tension: $T\cos\theta$ up, $T\sin\theta$ inward. Resolving gravity along and perpendicular to the string (the reflex imported from simple-pendulum problems) puts your axes at an angle to the acceleration and, used carelessly, produces the classic wrong answer $T = mg\cos\theta$ . The right answer is bigger than the weight, not smaller: $T = mg/\cos\theta$ .

The tilt is a negotiation, not a setting. Nobody dials in $\theta$ . Given the speed, the string swings out until the horizontal share of its tension exactly matches $mv^2/r$ — faster circles need a bigger inward share, so the cone opens wider. This self-adjustment is the same story as the chair-o-plane, the hammer throw, and an aircraft's banked turn: one tilted force splitting itself between holding weight and funding curvature, geometry set by the split.

The master solution pattern

Every conical-pendulum problem starts from the same two-line resolution of the tension, with the radius written as $r = L\sin\theta$ when the string length is given. Vertical balance (no vertical acceleration):

T\cos\theta = mg \qquad (1)

where: this line alone already yields the most-tested single fact: $T = mg/\cos\theta \ge mg$ — the string of a conical pendulum is always tenser than the hanging weight, and it grows without bound as the cone flattens. A 5 kg chair at $\theta = 60°$ hangs on $T = 2mg = 100$ N of chain tension. Horizontal (centripetal) equation:

T\sin\theta = \frac{mv^2}{r} = m\omega^2 r \qquad (2)

where: the horizontal share of the tension is the entire centripetal supply — no other horizontal force exists. Substituting $r = L\sin\theta$ into the $\omega$ form gives a gem: $T = m\omega^2 L$ , tension directly from the spin rate, no angles required. Divide (2) by (1) to eliminate $T$ :

\tan\theta = \frac{v^2}{rg} = \frac{\omega^2 r}{g} \qquad (3)

where: the banked-curve design equation reappears exactly — string tilt as bank angle. In the $\omega$ form with $r = L\sin\theta$ , the $\sin\theta$ cancels one power and (3) rearranges to $\cos\theta = \dfrac{g}{\omega^2 L}$ : the cone's opening angle as an explicit function of spin rate, with its threshold analyzed below. And the period, derived below:

T_{\rm p} = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}} \qquad (4)

where: $h = L\cos\theta$ is the cone's vertical height. Two conical pendulums of different string lengths and different angles whirl in step if their bobs' circles lie the same distance below their supports — a genuinely surprising fact and a favourite assertion–reason stem.

you want	use	notes
the tilt from the speed	$\tan\theta = v^2/rg$	banked-curve relation in costume
the tilt from the spin rate	$\cos\theta = g/\omega^2 L$	needs $\omega \ge \sqrt{g/L}$
the tension	$T = mg/\cos\theta$ or $T = m\omega^2 L$	always $> mg$ ; grows as $\omega^2$
the period	$T_{\rm p} = 2\pi\sqrt{L\cos\theta/g}$	height of cone only
the threshold spin	$\omega_{\min} = \sqrt{g/L}$	below it: hangs vertical
composite rides (arm $R$ + chain $L$ )	$r = R + L\sin\theta$ in (2)	tilt still from tension split

Derivation AThe conical pendulum solved — tilt, tension, and the height-only period

We derive the master results (1)–(4) by the two-axis routine every exam expects: resolve the tension, balance vertically, apply Newton's second law horizontally, and convert speed to period through the circle's geometry. The strategy hinges on one deliberate choice — axes aligned with the (horizontal) acceleration — and the substitution $r = L\sin\theta$ saved for last, which is what makes the period formula collapse so cleanly.

Step 1. Free-body diagram: exactly two forces — tension $T$ along the string toward the support, weight $mg$ down. Resolve $T$ : vertical $T\cos\theta$ , horizontal $T\sin\theta$ pointing at the circle's centre. Why we resolve T and not g: the acceleration is horizontal ( $v^2/r$ toward the centre). Choosing axes along/perpendicular to the acceleration keeps Newton's law one-dimensional per axis. Resolving gravity along the string aligns axes with neither the acceleration nor the constraint — legal but treacherous, and the source of the $T = mg\cos\theta$ blunder.

Step 2. Vertical: no vertical acceleration, so $T\cos\theta = mg$ , giving $T = mg/\cos\theta$ .

Step 3. Horizontal: $T\sin\theta = \dfrac{mv^2}{r}$ . Divide by Step 2: $\tan\theta = \dfrac{v^2}{rg}$ — the negotiation equation.

Step 4. Period. With $v = \dfrac{2\pi r}{T_{\rm p}}$ , equation (3) reads $\tan\theta = \dfrac{4\pi^2 r}{g T_{\rm p}^2}$ . Substitute $r = L\sin\theta$ and use $\tan\theta = \sin\theta/\cos\theta$ : the $\sin\theta$ cancels, leaving

T_{\rm p} = 2\pi\sqrt{\frac{L\cos\theta}{g}}

Why the width of the cone dropped out: widening the cone at fixed $L$ enlarges the circle (more distance per lap) but exactly in proportion boosts the required speed through the tension split. The two effects cancel, leaving only the height $h = L\cos\theta$ — so the period is that of a simple pendulum whose length equals the cone's height.

Step 5. Tension via the spin rate: from Step 3 in $\omega$ form, $T\sin\theta = m\omega^2 L\sin\theta$ , so $T = m\omega^2 L$ — a one-line result worth memorizing: double the spin rate, quadruple the chain tension. Verified: $L = 1.5$ m, $\theta = 30°$ , $g = 9.8$ : $T_{\rm p} = 2\pi\sqrt{1.5 \times 0.866/9.8} = 2.29$ s; $v = \sqrt{rg\tan\theta} = \sqrt{0.75 \times 9.8 \times 0.577} = 2.06$ m/s; self-consistent with $v = 2\pi r/T_{\rm p} = 4.71/2.29 = 2.06$ m/s ✓. Small-angle limit: $\cos\theta \to 1$ gives $2\pi\sqrt{L/g}$ , the simple pendulum ✓ (a small-angle conical pendulum is two perpendicular swings in step). Treatment: Kleppner Ch. 2, Halliday Ch. 6.

Go deeper · Derivation B · 2 minThe threshold spin and the impossible horizontal string

Derivation BThe threshold spin and the impossible horizontal string

We derive the two boundary behaviours that JEE-Advanced mines: the minimum spin rate for the cone to open, and the divergence that forbids a horizontal string. The strategy: write the tilt as a function of $\omega$ and read the function's domain — the physics lives where solutions exist.

Step 1. From the master pattern with $r = L\sin\theta$ : $\ T = m\omega^2 L$ and $T\cos\theta = mg$ combine to

\cos\theta = \frac{g}{\omega^2 L}

Step 2. A cosine cannot exceed 1. The equation has a tilted solution only when $\dfrac{g}{\omega^2 L} \le 1$ , i.e.

\omega \ge \omega_{\min} = \sqrt{\frac{g}{L}}

Below the threshold, the only equilibrium is $\theta = 0$ : the string hangs straight down, rotating about its own line, cone closed. Why the cone refuses to open at low spin: at small tilt the available inward force is $\approx mg\,\theta$ while the required is $m\omega^2 L\,\theta$ — both proportional to $\theta$ . If $\omega^2 L < g$ , demand falls short of supply at every angle, and the tilt collapses back to zero. At exactly $\omega = \omega_{\min}$ the balance is neutral; beyond it, hanging vertical becomes unstable and the cone snaps open. Note $\omega_{\min} = \sqrt{g/L}$ is precisely the frequency of the same string as a simple pendulum — no coincidence: the conical pendulum "opens" when you drive it at its natural swing rate.

Step 3. The opposite extreme, $\theta \to 90°$ : $\cos\theta \to 0$ forces $\omega^2 = g/(L\cos\theta) \to \infty$ , and with it $T = m\omega^2 L \to \infty$ . A horizontal cone needs infinite spin and infinite tension. Why no finite effort suffices: a horizontal string has zero vertical component of tension — nothing to hold the weight. Every real whirl must keep enough tilt for $T\cos\theta$ to equal $mg$ ; the faster you spin, the smaller that reserved vertical slice can be as a fraction, but it can never be zero. A hammer thrower's 7.26 kg ball on ~2 kN of pull sags only about $2°$ below horizontal — close enough to fool the eye, never exact.

Step 4. Useful summary of the regimes: $\omega < \sqrt{g/L}$ : hangs vertical. $\omega = \sqrt{g/L}$ : threshold. $\omega > \sqrt{g/L}$ : $\cos\theta = g/\omega^2 L$ , tension $m\omega^2 L$ , period $2\pi/\omega$ — and as $\omega \to \infty$ , $\theta \to 90°$ asymptotically. Verified: $L = 0.4$ m, $g = 10$ : $\omega_{\min} = 5$ rad/s ✓ (worked example 4); at $\omega = 10$ rad/s: $\cos\theta = 10/(100 \times 0.4) = 0.25$ , $\theta = 75.5°$ , $T = 40m$ — all self-consistent with $T\cos\theta = 10m = mg$ ✓.

Figure

The cone versus the spin rate — closed below the threshold, asymptotically flat above it

Limiting cases

case	result	meaning
$\theta \to 0$	$T_{\rm p} \to 2\pi\sqrt{L/g}$ , $T \to mg$	simple-pendulum limit: two tiny swings in quadrature
$\omega < \sqrt{g/L}$	$\theta = 0$ only	cone closed: the string hangs and spins in place
$\omega = \sqrt{g/L}$	threshold	vertical hang goes unstable; the cone is born
$\theta \to 90°$	$T, \omega, v \to \infty$	horizontal string impossible — always a sag
$\omega$ doubled (open cone)	$T \to 4T$	tension rides $\omega^2$ via $T = m\omega^2 L$
same $h = L\cos\theta$ , different $L$	same period	the height-only period: cones of equal depth tick together
in a lift accelerating up at $a$	$g \to g + a$ everywhere	threshold rises, cone narrows — $g_{\rm eff}$ toolkit

Common trapResolving gravity along the string, and the phantom T = mg cos θ

This trap catches more students than any other in circular dynamics, because the reflex is trained by a different pendulum. In a simple pendulum at the extreme of its swing, the bob's acceleration is tangential, and resolving gravity along/perpendicular to the string gives the familiar $T = mg\cos\theta$ . Import that habit into the conical pendulum and you have quietly assumed the wrong acceleration. Here the bob accelerates horizontally — toward the circle's centre — and the string direction has a component along that acceleration. The correct vertical balance is $T\cos\theta = mg$ , giving $T = mg/\cos\theta$ : the conical string is always tenser than the weight, not slacker. The two formulas differ by a factor $\cos^2\theta$ — at $60°$ that is a factor of four, an error no examiner misses. The immunization: before resolving anything, ask where does the acceleration point? — then resolve forces along and perpendicular to that. Horizontal acceleration → resolve the tension. Tangential acceleration → resolve gravity. The pendulum's name tells you which world you are in.

Worked examples

1 · Board level — the standard cone (A-Level/ISC pattern)

A $0.40$ kg bob on a $1.5$ m string moves as a conical pendulum with the string at $30°$ to the vertical. Find (a) the tension, (b) the bob's speed, (c) the period. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. (a) Vertical balance: $T\cos 30° = mg = 3.92$ N, so $T = \dfrac{3.92}{0.866} = \boxed{4.5\ \mathrm{N}}$ — above the weight, as it must be.

Step 2. Radius: $r = L\sin 30° = 0.75$ m. (b) From $\tan\theta = v^2/rg$ : $v^2 = rg\tan 30° = 0.75 \times 9.8 \times 0.577 = 4.24$ , so $v = \boxed{2.1\ \mathrm{m/s}}$ .

Step 3. (c) Period: $T_{\rm p} = 2\pi\sqrt{\dfrac{L\cos 30°}{g}} = 2\pi\sqrt{\dfrac{1.30}{9.8}} = \boxed{2.3\ \mathrm{s}}$ .

Step 4. Cross-check (the habit that catches slips): $v = 2\pi r/T_{\rm p} = 4.71/2.29 = 2.06$ m/s ✓ consistent with (b).

2 · Board level — from spin rate to geometry (NEET/JEE-Main bridge)

A $0.2$ kg bob on a $0.5$ m string is whirled as a conical pendulum at $\omega = 7.0$ rad/s. Find (a) the cone's angle, (b) the tension, two ways. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) $\cos\theta = \dfrac{g}{\omega^2 L} = \dfrac{10}{49 \times 0.5} = 0.408$ , so $\theta = \boxed{65.9°}$ — a wide cone; check the threshold first: $\omega_{\min} = \sqrt{10/0.5} = 4.5$ rad/s $< 7$ ✓, the cone is open.

Step 2. (b) Route one: $T = \dfrac{mg}{\cos\theta} = \dfrac{2.0}{0.408} = \boxed{4.9\ \mathrm{N}}$ .

Step 3. Route two, the one-liner: $T = m\omega^2 L = 0.2 \times 49 \times 0.5 = \boxed{4.9\ \mathrm{N}}$ ✓ — same number, no trigonometry.

Step 4. Note what the agreement teaches: $T = m\omega^2 L$ is exact for any open cone, so on MCQs that give $\omega$ and $L$ , the tension needs no angle at all — a 15-second answer.

3 · Competitive level — the chair-o-plane's composite radius (JEE-Main pattern, ~3 minutes)

A fairground swing hangs on a $2.0$ m chain from the rim of a rotating disc of radius $3.0$ m. In steady rotation the chain makes $37°$ with the vertical. For a $50$ kg rider, find (a) the speed, (b) the chain tension. (Take $g = 10\ \mathrm{m/s^2}$ , $\sin 37° = 0.6$ , $\cos 37° = 0.8$ .)

Step 1. The circle's radius is the disc's rim plus the chain's horizontal reach: $r = R + L\sin 37° = 3.0 + 2.0(0.6) = 4.2$ m. (Using $r = L\sin\theta$ alone is the designed error.)

Step 2. The force analysis is unchanged — the tilt is still set by the tension split: $\tan 37° = \dfrac{v^2}{rg}$ , so $v^2 = 0.75 \times 4.2 \times 10 = 31.5$ , giving $v = \boxed{5.6\ \mathrm{m/s}}$ .

Step 3. (b) Vertical balance: $T = \dfrac{mg}{\cos 37°} = \dfrac{500}{0.8} = \boxed{625\ \mathrm{N}}$ .

Step 4. Sanity: the inward supply $T\sin 37° = 375$ N should equal $\dfrac{mv^2}{r} = \dfrac{50 \times 31.5}{4.2} = 375$ N ✓. Note the tension formula never saw the disc: $T = mg/\cos\theta$ holds whatever the geometry of the radius, because it is pure vertical bookkeeping. Timing note: ~3 min. The single decision that scores the problem is $r = R + L\sin\theta$ ; everything after is the standard pattern. Rides with tilted arms simply add their own horizontal reach the same way.

4 · Competitive level — the threshold and beyond (JEE-Advanced flavour, ~3.5 minutes)

A bob hangs on a $0.40$ m string from the axis of a turntable. (a) Below what spin rate does the string simply hang vertical? (b) At $\omega = 10$ rad/s, find the cone's angle and the string tension in units of the bob's weight. (c) Can any $\omega$ make the string horizontal? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) Threshold: $\omega_{\min} = \sqrt{g/L} = \sqrt{25} = \boxed{5.0\ \mathrm{rad/s}}$ . Below this, $\cos\theta = g/\omega^2L$ would exceed 1 — no tilted solution exists, and the vertical hang is stable.

Step 2. (b) $\cos\theta = \dfrac{g}{\omega^2 L} = \dfrac{10}{100 \times 0.4} = 0.25$ , so $\theta = \boxed{75.5°}$ .

Step 3. Tension: $T = m\omega^2 L = m \times 100 \times 0.4 = 40m$ newtons $= \dfrac{40m}{10m} = \boxed{4\ \text{weights}}$ . (Check: $mg/\cos\theta = mg/0.25 = 4mg$ ✓.)

Step 4. (c) No. Horizontal means $\cos\theta = 0$ , requiring $\omega \to \infty$ and $T = m\omega^2 L \to \infty$ . Every increase in $\omega$ flattens the cone further ( $\cos\theta \propto 1/\omega^2$ ) but the string always reserves enough tilt for $T\cos\theta$ to hold the weight — an asymptote, never an arrival. Timing note: ~3.5 min. Assertion–reason setters love (a) and (c): "the string of a conical pendulum can be made horizontal at sufficiently high speed" is FALSE, and "below $\sqrt{g/L}$ the pendulum does not tilt" is TRUE — both one-line justifications from $\cos\theta = g/\omega^2L$ .

Exam traps

$T = mg\cos\theta$ . The simple-pendulum reflex applied to the wrong pendulum — see the pitfall box. Conical strings obey $T = mg/\cos\theta$ : always more than the weight.

Radius versus string length. $r = L\sin\theta$ , not $L$ — and in composite rides, $r = R + L\sin\theta$ . The period formula, by contrast, wants $L\cos\theta$ , the height. Three different lengths live in this problem; label them.

Angle with the vertical versus the horizontal. $\theta$ here is from the vertical (so $\cos\theta$ holds the weight). A paper measuring from the horizontal swaps every sine and cosine. Redraw and relabel before computing.

Forgetting the threshold. $\cos\theta = g/\omega^2L > 1$ is not an error in your algebra — it is physics telling you the cone is closed. Answer "the string stays vertical", not a complex angle.

Claiming the horizontal whirl. No finite speed achieves $\theta = 90°$ . Any option offering a finite $\omega$ for a horizontal cone is wrong by structure, not by arithmetic.

Uniform-motion amnesia. The conical pendulum runs at constant speed (no tangential force), so no energy bridge is needed — do not import vertical-circle machinery. Conversely, do not export this uniformity back to vertical circles.

Period misread. $T_{\rm p}$ depends only on $L\cos\theta$ : a wider cone at the same string length whirls faster (shorter period). "Period is independent of the angle" is false; "period depends only on the cone's height" is the true version.

Which $g$ . A-Level (CIE) uses $9.81$ ; JEE/NEET declare $10$ almost always; ISC $9.8$ . The threshold and period scale as $\sqrt{g}$ — use the stated value.

Origins · 1 minJames Watt and how this was discovered

James Watt

1736–1819 · Glasgow & Birmingham

Watt did not invent the steam engine — he made it governable, and the governor he popularized in 1788 is two conical pendulums in brass. A pair of heavy balls hangs from a spinning spindle geared to the engine; as the engine races, $\omega$ rises, the cones open wider (our $\cos\theta = g/\omega^2L$ exactly), and the rising collar throttles the steam valve; as the engine lugs, the balls droop and the valve reopens. The machine computes its own speed correction from the tilt of a conical pendulum, decades before anyone wrote "feedback control". Maxwell's 1868 paper On Governors, analyzing when Watt's device hunts or settles, is a founding document of control theory — a straight line runs from the fairground chair-o-plane's slant to the cruise control in every car.

History

1659

— Huygens analyzes the conical pendulum while perfecting clocks, deriving the cone's period and its relation to the simple pendulum.

1673

— Huygens' Horologium Oscillatorium publishes the theory of circular and pendular motion that the conical pendulum obeys.

1788

— Watt fits the centrifugal governor — twin conical pendulums — to a steam engine; industry acquires automatic speed control.

1868

— Maxwell's On Governors analyzes the device's stability, founding feedback-control theory.

1900

s — Chair-o-plane rides industrialize the physics: every rider hangs at cos θ = g/ω²L, tension m ω² L in the chains.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 7 · Kleppner & Kolenkow Ch. 2 (the conical pendulum example) · CIE 9702 past papers, Motion in a Circle, for the resolve-and-divide pattern.

Provenance. All results are standard. A verifier should recompute: example 1 ( $T = 4.5$ N, $v = 2.06$ m/s, $T_{\rm p} = 2.29$ s, cross-checked via $2\pi r/T_{\rm p}$ ); example 2 ( $\theta = 65.9°$ , $T = 4.9$ N both routes); example 3 ( $r = 4.2$ m, $v = 5.6$ m/s, $T = 625$ N, audit $375 = 375$ N); example 4 ( $\omega_{\min} = 5$ rad/s, $\theta = 75.5°$ , $T = 4mg$ ); the threshold identity $\omega_{\min} = \sqrt{g/L}$ and the height-only period.

Prerequisites. Centripetal Force: Dynamics of Circular Motion · Components & Resolution of Vectors · Banked Curves (the same $\tan\theta$ law). Leads to. Rotor & Well of Death · The Simple Pendulum · Pendulum in Lift, Cart & on Incline · Centripetal Force Dynamics (UG deep dive).

Why does the string tilt at all when I spin it? Couldn't it stay vertical and just rotate?

Below $\omega_{\min} = \sqrt{g/L}$ it does exactly that. Above it, the vertical hang becomes unstable: give the bob the tiniest sideways displacement and the inward force available at that tilt ( $\approx mg\theta$ ) is less than the force needed to hold that circle ( $m\omega^2 L\theta$ ), so the displacement grows — the string swings out until the growing tension's horizontal share catches up with the demand, settling at $\cos\theta = g/\omega^2L$ . The cone is the stable compromise; verticality survives only at low spin.

How is this different from the simple pendulum? They look like the same object.

Same hardware, different motion, different mathematics. The simple pendulum swings in a vertical plane: speed varies, acceleration is partly tangential, and energy methods matter. The conical pendulum sweeps a horizontal circle at constant speed: acceleration is purely centripetal, no energy bridge needed, and the whole solution is the two-line tension split. The deep connection appears in the small-angle limit: a tiny conical motion is two perpendicular simple-pendulum swings a quarter-cycle apart, which is why both share the period $2\pi\sqrt{L/g}$ as $\theta \to 0$ .

Where does the centripetal force come from here — there's nothing pulling toward the centre except empty space?

The string does it, by geometry. Tension acts along the string, and a tilted string has a horizontal component pointing from the bob toward the axis — that component is the centripetal force, all of it. Nothing else is needed: the bob is not "balanced" horizontally, it is deliberately unbalanced inward, which is exactly what moving on a circle requires. The vertical component, meanwhile, holds the weight. One force, two jobs, split by the tilt.

Why does the period depend only on the height of the cone and not its width?

Open the cone wider at fixed $L$ : the circle grows ( $r = L\sin\theta$ increases), so each lap is longer — but the tension also tilts further inward, demanding a faster speed in exact proportion. The two effects cancel in the lap time, leaving $T_{\rm p} = 2\pi\sqrt{L\cos\theta/g}$ : only the depth of the bob below the support survives. Equivalently: the "restoring machinery" per unit horizontal displacement is $g/h$ , just as for a simple pendulum of length $h$ .

What happens to the cone inside an accelerating lift?

Replace $g$ with $g_{\rm eff} = g + a$ (lift accelerating up). Every result inherits the swap: the threshold spin rises to $\sqrt{(g+a)/L}$ , the cone narrows at a given $\omega$ ( $\cos\theta = (g+a)/\omega^2L$ ), the period shortens. In free fall $g_{\rm eff} = 0$ : any spin at all opens the cone to horizontal — the one place the horizontal string is legal, because there is no weight left to hold.

Is the chair-o-plane really safe if the tension grows as ω²?

That $\omega^2$ scaling is precisely what the designers rate. The operating spin is fixed by motor gearing, so the maximum chain tension per rider is $m\omega_{\max}^2 (L + \ldots)$ -scale, known exactly, with chains rated several-fold above it. The self-limiting feature is elegant: tension per unit mass, $\omega^2 L$ , is the same for a child and an adult — heavier riders load their chains proportionally, not disproportionately. What the ride cannot do, ever, is fly the chains horizontal; the sag you see at full speed is equation (1) holding everyone's weight.

If I give the bob more speed but keep the same string, does it rise? Doesn't that violate energy conservation?

It rises, and energy conservation is perfectly happy — you paid. Spinning the pendulum up does work on the bob: some becomes kinetic energy (larger $v$ ), some becomes potential energy (the bob's circle sits higher, $h = L\cos\theta$ shrinks as $\theta$ opens). In steady state afterwards, no energy flows at all: tension is perpendicular to the velocity, gravity does zero net work per lap, and the cone whirls forever in the ideal case. Energy questions arise only during the spin-up and spin-down transients.