Centripetal Force and Vertical-Circle Dynamics

A motorway exit ramp is quietly reading your speedometer. Its curve is banked at an angle computed from $v^2/rg$ so that, at the posted speed, the road pushes you around the bend without asking your tyres for any sideways grip at all. Drive it in ice-storm conditions at exactly that speed and you are safe; misjudge by enough and no bank can save you. A roller coaster performs a more theatrical version of the same arithmetic: at the top of the loop, upside down, the track must push you downward — and the whole trick of not falling out of the sky is arriving fast enough that gravity alone cannot supply the required pull toward the centre. You already own the kinematics: circular motion at speed $v$ demands an acceleration $v^2/r$ pointed at the centre, every instant. This article is about the dynamics — the question of who pays. Centripetal force is not a new force of nature; it is a requirement, an invoice of magnitude $mv^2/r$ that must be settled by the ordinary forces on the scene: friction, gravity, tension, a normal force, or some alliance of them. Identify the payers, resolve them toward the centre, set the sum equal to $mv^2/r$ — that three-step routine solves the banked curve, the vertical loop, the conical pendulum, and every spin-cycle, centrifuge, and orbit you will ever meet. The routine comes with famous traps — phantom "centrifugal forces," the urge to add a centripetal arrow to a free-body diagram — and this article disarms them. Three classic geometries, fully worked, with the numbers your intuition should carry.

Key takeaways

Centripetal force is a requirement, not a force: circular motion at speed $v$ demands net inward force $F_{\rm net} = mv^2/r$ , and ordinary forces (friction, gravity, normal, tension) must supply it.
Flat curve: friction pays the whole bill — maximum cornering speed $v_{\max} = \sqrt{\mu_s g r}$ , mass-independent.
Banked curve: at the design speed $v^2 = gr\tan\theta$ the normal force alone pays; friction extends the safe range to $v_{\max}^2 = gr\,\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}$ .
Vertical circle: the invoice is hardest to pay at the top, where gravity helps — minimum top speed $v_{\rm top}^2 = gR$ (feels weightless there); tension/normal force at the bottom exceeds the top value by $6mg$ .
To just complete a frictionless loop of radius $R$ , release from height $h = \tfrac52 R$ above the bottom — where riders briefly pull $6g$ .
Conical pendulum: the string's tilt satisfies $\tan\theta = v^2/rg$ with period $T = 2\pi\sqrt{L\cos\theta/g}$ — a horizontal string would demand infinite speed.
Key idea: never draw a "centripetal force" arrow — draw the real forces, resolve toward the centre, and set the inward net equal to $mv^2/r$ ; the classic geometries differ only in who pays.

The core insight

An object in circular motion is not in equilibrium — the real forces acting on it must sum to a net inward (centripetal) force of magnitude $mv^2/r$ ; solving circle dynamics means identifying which physical forces supply that requirement, with the banked curve (normal + friction), the vertical loop (gravity + normal/tension), and the conical pendulum (tension + gravity) as the canonical castings.

Where this lives today

Every road you drive was checked against this article. Highway design codes specify superelevation (banking) rates from equation (3) with friction margins from (5) — curve radii, posted speeds, and bank angles are solved together so that the friction reserve stays comfortable in rain. Racing pushes the same equations to their edges: NASCAR's Talladega banks at 33° so that 300 km/h stock cars corner mostly on geometry; velodrome bends at 42–45° let track cyclists sweep by at 60–75 km/h with their wheels perpendicular to the boards; and a Formula 1 car inverts the logic — downforce adds an aerodynamic term to $N$ , letting flat corners be taken at speeds equation (2) alone would forbid (the paddock's party line that an F1 car "could drive on the ceiling" above ~150 km/h is precisely a statement about that extra normal force). The vertical-circle physics runs the amusement industry and aerospace medicine alike. Coaster loops are certified against apparent-weight envelopes — the clothoid geometry holds riders between the weightless crest and roughly 4g at the trough, with equation (6)'s bottom-heavy asymmetry as the fundamental constraint. Fighter pilots live in the same equations at higher stakes: a 9g turn is $v^2/r$ chosen by the flight computer against the pilot's blood-pressure physiology, and centrifuge training rigs — human-rated conical pendulums driven in reverse — rehearse exactly the loading this article computes. At the smallest scale, laboratory centrifuges cast the sample tube as the circling body and the rotor as the payer: at 100,000 rpm an ultracentrifuge demands about $10^6 g$ of centripetal acceleration from its titanium, applying equation (1) a million times more intensely than any road.

The physical picture

Uniform circular motion feels steady, but it is perpetual acceleration: velocity's direction changes every instant, and from the kinematics article you know the acceleration points at the centre with magnitude $v^2/r$ . Newton's second law then makes a blunt demand: some real force, or combination of real forces, must point inward and total exactly $mv^2/r$ . Nothing about the circle generates this force; the circle happens only if the force is there. Swing a ball on a string and the tension pays; let go, and with the payer gone the ball does not fly outward — it flies straight, along the tangent, exactly as Newton's first law says an unforced body must. The "outward fling" you feel in a turning car is your body trying to go straight while the car's door comes curving into you. That reframe — centripetal force as an invoice to be paid by ordinary forces — turns every circular-dynamics problem into the same short procedure. Draw the free-body diagram with only the real forces: gravity, normal force, tension, friction. Resolve them along two directions: toward the centre of the circle, and along the local vertical (or whatever direction has no acceleration). The no-acceleration direction gives a balance equation; the centre direction gives $\sum F_{\rm inward} = mv^2/r$ . Two equations, and the geometry problems of this article all fall to them. The interesting physics is who pays, case by case. On a flat road the only inward force available is tyre friction — so cornering ability is capped at $\mu_s mg$ , and the invoice $mv^2/r$ outruns it at $v_{\max} = \sqrt{\mu_s gr}$ , mass cancelling (a loaded truck and a bicycle skid at the same speed on the same curve). Bank the road and the normal force — enormous but useless when vertical — tilts and acquires an inward component: the road itself starts paying, friction is relieved, and at one magic speed relieved entirely. In a vertical loop the payer roster changes with position: at the bottom the track pushes up while gravity pulls down, so the normal force must cover the invoice plus your weight (the heavy-in-the-seat feeling); at the top both gravity and track push toward the centre, gravity covers part or all of the bill, and the normal force can drop to zero — the weightless crest, and the origin of the minimum-speed condition. The conical pendulum makes tension the payer, its tilt self-adjusting so the horizontal component of tension exactly finances the circle.

Figure

One invoice, three payers — who supplies mv²/r in the classic geometries

The math: the requirement and its three classic castings

The master equation is Newton's second law aimed at the centre. For a body of mass $m$ moving on a circle of radius $r$ at speed $v$ ,

\sum F_{\rm inward} = \frac{mv^2}{r} = m\omega^2 r \qquad (1)

where: the left side is the sum of the inward components of real forces — count forces toward the centre as positive, away as negative — and the right side is the kinematic requirement, written either with speed $v$ or angular velocity $\omega = v/r$ . Equation (1) is not a new law; it is $F = ma$ with the known circular acceleration inserted. Two disciplines keep it honest: no "centripetal force" arrow ever appears on the free-body diagram (the right side is the total, not a participant), and no "centrifugal force" appears either — that phantom belongs to rotating reference frames, a later article, and has no place in an inertial-frame analysis. On a flat curve, friction is the only candidate payer, and its ceiling caps the speed:

\frac{mv^2}{r} \le \mu_s N = \mu_s mg \qquad\Longrightarrow\qquad v_{\max} = \sqrt{\mu_s\, g\, r} \qquad (2)

where: the mass has cancelled — cornering limits depend on tyre grip and geometry, not on how heavy the vehicle is. For $\mu_s = 0.9$ (dry tyres) and $r = 50$ m, $v_{\max} = \sqrt{0.9 \times 9.8 \times 50} = 21$ m/s (76 km/h); rain drops $\mu_s$ to ~0.5 and the same bend's limit to 57 km/h. Note it is static friction — the contact patch is not sliding sideways — the same subtlety as rolling. Bank the road at angle $\theta$ and the normal force tilts, gaining the inward component $N\sin\theta$ . Requiring zero friction defines the design speed:

\tan\theta = \frac{v_{\rm design}^2}{g\,r} \qquad (3)

where: the same relation, read backwards, gives the bank angle a highway engineer chooses for a posted speed. The full friction-assisted speed range is derived below. The identical equation governs the conical pendulum — a bob circling on a string tilted at $\theta$ from the vertical — and an aircraft in a coordinated banked turn, where the wings' lift plays the role of the normal force. One geometry, three costumes. In a vertical circle the payers change with position, and the crisis is at the top:

v_{\rm top}^2 \ge g\,R \qquad \text{(minimum speed to maintain contact/taut string)} \qquad (4)

where: at the minimum, gravity alone pays the entire invoice $mv^2/R$ and the normal force (or string tension) drops to zero — the rider is momentarily in free fall while curving, which is why loop crests feel weightless. Below this speed the track cannot pull (normal forces only push) and the body leaves the circle early. The derivation below adds the two companion classics: the $6mg$ tension gap between bottom and top, and the $\tfrac52 R$ release height.

Derivation AThe banked curve with friction — the full safe-speed range

We are deriving the complete banked-curve result: a car on a curve of radius $r$ banked at $\theta$ , with static friction coefficient $\mu_s$ , can round it without sliding for speeds $v_{\min} \le v \le v_{\max}$ , with the design speed (3) — where friction is unneeded — in the interior. The strategy is the standard two-axis resolution, with one non-obvious choice made deliberately: we resolve into horizontal and vertical components (not along/perpendicular to the incline), because the acceleration is horizontal — toward the circle's centre — and aligning axes with the acceleration keeps the right side of Newton's law on one axis only. Watch friction change direction between the two speed limits: at $v_{\max}$ it grips down-slope (fighting an outward slide), at $v_{\min}$ up-slope (fighting an inward slide).

Step 1. Free-body diagram at the fast limit: weight $mg$ down; normal force $N$ perpendicular to the road surface (components $N\cos\theta$ up, $N\sin\theta$ inward); friction $f = \mu_s N$ at its ceiling, pointing down the slope (components $\mu_s N\sin\theta$ down... resolved: $\mu_s N\cos\theta$ inward, $\mu_s N\sin\theta$ downward). Why friction points down-slope here: at excessive speed the car tends to slide up and outward across the banking; static friction opposes the impending slide.

Step 2. Write the two axis equations. Vertical (no acceleration): $N\cos\theta - \mu_s N\sin\theta - mg = 0$ , so $N = \dfrac{mg}{\cos\theta - \mu_s\sin\theta}$ . Horizontal (centripetal): $N\sin\theta + \mu_s N\cos\theta = \dfrac{mv_{\max}^2}{r}$ . Why the vertical equation matters as much as the horizontal: banking + friction change $N$ itself — it is no longer $mg\cos\theta$ or $mg$ ; forgetting this recalibration is the classic error in banked-curve problems.

Step 3. Divide the horizontal equation by the vertical one to eliminate $N$ :

v_{\max}^2 = g\,r\;\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} \qquad (5)

Why the denominator can vanish: if $\mu_s\tan\theta \ge 1$ (steep bank, grippy tyres), no speed can break the car loose outward — the faster you go, the harder the geometry presses you into the road. Velodrome walls at 45° live in this regime.

Step 4. Repeat with friction reversed (up-slope) for the slow limit: $v_{\min}^2 = g\,r\,\dfrac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}$ , meaningful only when $\tan\theta > \mu_s$ — on gentler banks, even a stationary car sits happily. Setting $\mu_s = 0$ in either result recovers the design speed (3): $v^2 = gr\tan\theta$ , the unique friction-free speed. Why a minimum speed exists at all on steep ice: below it the car slides down-and-inward off the banking — the same physics that keeps a bobsled high on a wall only while it is fast. *As a self-consistency check: for $r = 200$ m, $\theta = 17.7°$ , $\mu_s = 0.5$ the design speed is 90 km/h and $v_{\max} = 157$ km/h, with no minimum speed since $\tan\theta < \mu_s$ . The limiting cases also hold: setting $\theta \to 0$ in (5) recovers the flat-road result (2), and setting $\mu_s \to 0$ recovers the design-speed relation (3).*

Go deeper · Derivation B · 2 minThe vertical circle — minimum speed, the 6mg rule, and the 5R/2 release height

Derivation BThe vertical circle — minimum speed, the 6mg rule, and the 5R/2 release height

We are deriving the three classics of vertical-circle dynamics for a body on the inside of a loop (or a ball on a string): the minimum top speed (4), the theorem that the bottom-of-loop force exceeds the top-of-loop force by exactly $6mg$ whenever energy is conserved, and the minimum release height $h = \tfrac52 R$ for a frictionless coaster to complete a loop. The strategy: apply the centripetal requirement (1) at the top and bottom points separately — where all forces are conveniently radial — then let energy conservation connect the two speeds. Watch the signs: gravity is a payer at the top and a burden at the bottom, and that asymmetry is the whole story.

Step 1. Top of the loop: both the normal force $N_{\rm top}$ (track pushes down, toward the centre) and gravity point inward, so $N_{\rm top} + mg = \dfrac{mv_{\rm top}^2}{R}$ . Why the crisis is here: the track can only push. If $v_{\rm top}^2 < gR$ , the required inward force is less than $mg$ alone supplies — the equation would demand $N_{\rm top} < 0$ , a pull the track cannot deliver. The body leaves the track and becomes a projectile before reaching the crest. Hence the minimum condition $v_{\rm top}^2 = gR$ with $N_{\rm top} = 0$ : pure free-fall curvature, momentary weightlessness.

Step 2. Bottom of the loop: the normal force pushes up (inward) and gravity pulls down (outward), so $N_{\rm bot} - mg = \dfrac{mv_{\rm bot}^2}{R}$ . Why riders feel heavy here: $N_{\rm bot}$ must cover the weight AND the full invoice — the apparent weight is $N_{\rm bot} = mg + mv_{\rm bot}^2/R$ , always more than $mg$ .

Step 3. Connect the speeds by energy conservation across the height gain $2R$ (valid for a frictionless track or an inextensible string, neither of which does work): $\tfrac12 mv_{\rm bot}^2 = \tfrac12 mv_{\rm top}^2 + mg(2R)$ , i.e. $v_{\rm bot}^2 = v_{\rm top}^2 + 4gR$ . Substitute into the Step-2 and Step-1 equations and subtract:

N_{\rm bot} - N_{\rm top} = \frac{m(v_{\rm top}^2 + 4gR)}{R} + mg - \frac{mv_{\rm top}^2}{R} + mg = 6mg \qquad (6)

Why the speed cancelled: the $6mg$ gap is universal — any ball whirled on a string in a vertical circle, however fast, has bottom tension exceeding top tension by six weights (2mg from the gravity sign flip, 4mg from the energy-mandated speed difference). A string rated at less than $6mg$ plus the top's requirement WILL snap at the bottom.

Step 4. The release-height classic: a frictionless coaster starts from rest at height $h$ above the loop's bottom and must reach the top ( $height 2R$ ) with $v_{\rm top}^2 \ge gR$ . Energy: $mgh = mg(2R) + \tfrac12 mv_{\rm top}^2 \ge 2mgR + \tfrac12 mgR$ , so

h_{\min} = \tfrac52\,R \qquad (7)

Why more than the loop's height: the extra half-radius of drop finances the speed the crest demands — arriving with zero speed at the top is not an option, because the circle must be flown, not merely reached. At the corresponding bottom speed, $v_{\rm bot}^2 = 5gR$ , the apparent weight is $N_{\rm bot} = 6mg$ : the $\tfrac52 R$ coaster briefly loads its riders at six gravities, which is why real coaster loops are teardrop-shaped (clothoid) rather than circular — larger radius at the bottom, smaller at the top, taming both ends of the invoice. *As a self-consistency check: for $R = 7$ m the results are $v_{\rm top,\min} = 8.3$ m/s, $h_{\min} = 17.5$ m, and 6g at the bottom. The limiting behaviour is also correct: $R \to 0$ sends all thresholds to zero (a tiny loop is trivially easy), and the $6mg$ gap is confirmed by direct substitution into the force equations.*

Go deeper · Derivation C · 2 minThe conical pendulum — tilt, speed, and the impossible horizontal string

Derivation CThe conical pendulum — tilt, speed, and the impossible horizontal string

We are deriving the equilibrium tilt and period of a conical pendulum — a bob of mass $m$ on a string of length $L$ , circling steadily so the string sweeps a cone of half-angle $\theta$ from the vertical — and, as the payoff, explaining why the string can never be whirled fully horizontal. The strategy is the same two-axis routine: tension is the only force besides gravity, its vertical part must balance $mg$ , and its horizontal part must pay the entire invoice. The radius is set by geometry, $r = L\sin\theta$ — keep that substitution for last to get the cleanest period formula.

Step 1. Resolve the tension $T$ : vertical $T\cos\theta = mg$ (no vertical acceleration), horizontal $T\sin\theta = \dfrac{mv^2}{r}$ . Why tension must do both jobs: nothing else is available — and the division of labour fixes the geometry: the faster the circle, the larger the inward share must be, so the string swings outward and $\theta$ grows. The tilt is not chosen; it is negotiated.

Step 2. Divide the two equations: $\tan\theta = \dfrac{v^2}{rg}$ — precisely the banked-curve design relation (3), with the string's tilt playing the bank angle. Why the coincidence is exact: both systems have one tilted push (tension/normal force) sharing itself between holding weight and funding the circle; the geometry of that split is identical.

Step 3. Extract the period. With $r = L\sin\theta$ and $v = 2\pi r/T_{\rm period}$ , substitute into $\tan\theta = v^2/rg$ : $\ \tan\theta = \dfrac{4\pi^2 r}{g\,T_{\rm period}^2} = \dfrac{4\pi^2 L\sin\theta}{g\,T_{\rm period}^2}$ , and since $\tan\theta = \sin\theta/\cos\theta$ ,

T_{\rm period} = 2\pi\sqrt{\frac{L\cos\theta}{g}} \qquad (8)

Why the period shrinks as the cone opens: $\cos\theta$ falls with $\theta$ — a faster whirl both widens the cone and quickens the circuit. In the small-angle limit $\cos\theta \to 1$ the formula becomes $2\pi\sqrt{L/g}$ : the simple pendulum's period, no accident — a conical pendulum at small angle is two perpendicular small oscillations in quadrature.

Step 4. Take the horizontal limit $\theta \to 90°$ : the vertical equation demands $T\cos\theta = mg$ with $\cos\theta \to 0$ , so $T \to \infty$ , and (8) gives $T_{\rm period} \to 0$ , i.e. $v \to \infty$ . A truly horizontal string would leave gravity unbalanced — no finite speed achieves it. Why the hammer throw still looks horizontal: at large $v$ the tilt needed to hold the weight becomes small ( $\cos\theta \approx mg/T$ ); a 7.26-kg hammer whirled with 2 kN of pull hangs only ~2° below horizontal — close enough to fool the eye, never exact. *As a self-consistency check: for $L = 1.5$ m and $\theta = 30°$ , equation (8) gives $T_{\rm period} = 2\pi\sqrt{1.5 \times 0.866/9.8} = 2\pi\sqrt{0.1326} = 2.29$ s, and the speed from the force equations is $v = \sqrt{rg\tan\theta} = \sqrt{0.75 \times 9.8 \times 0.577} = 2.06$ m/s — in agreement with the independent check $v = 2\pi r/T_{\rm period} = 4.71/2.29 = 2.06$ m/s.*

Which equation answers which question

you want	use	notes
the master routine	inward net $= mv^2/r$ , eq. (1)	real forces only; no centripetal/centrifugal arrows
flat-curve speed limit	$v_{\max} = \sqrt{\mu_s gr}$ , eq. (2)	mass-free; static friction
bank angle for a road/track	$\tan\theta = v^2/rg$ , eq. (3)	zero-friction design speed
safe range on a banked curve	eq. (5) and its $v_{\min}$ twin	friction widens the design speed into a band
loop-the-loop survival	$v_{\rm top}^2 \ge gR$ , eq. (4)	N = 0 at the minimum: weightless crest
string tension budget	$N_{\rm bot} - N_{\rm top} = 6mg$ , eq. (6)	universal, speed-independent gap
coaster release height	$h \ge \tfrac52 R$ , eq. (7)	and 6g at the bottom if taken at minimum
conical pendulum	$\tan\theta = v^2/rg$ , $T = 2\pi\sqrt{L\cos\theta/g}$ , eq. (8)	horizontal string impossible

Limiting cases

Case	Result	Meaning
$r \to \infty$ at fixed $v$	required force $\to 0$	straight-line motion needs no payer — Newton's first law recovered
$\mu_s \to 0$ , flat road	$v_{\max} \to 0$	no friction, no cornering: the ice-rink limit
$\mu_s\tan\theta \ge 1$	$v_{\max} \to \infty$ in (5)	steep + grippy: geometry locks the car to the wall
$\tan\theta \le \mu_s$	no $v_{\min}$	gentle banks hold parked cars; steep icy ones don't
$v_{\rm top}^2 = gR$ exactly	$N_{\rm top} = 0$	free-fall curvature: the weightless crest
$\theta \to 0$ , conical pendulum	$T \to 2\pi\sqrt{L/g}$	simple pendulum recovered — two SHMs in quadrature
$\theta \to 90°$ , conical pendulum	$T \to \infty$ , $v \to \infty$	the horizontal whirl is unreachable
satellite orbit	gravity is the sole payer: $mg' = mv^2/r$	circular orbits are this article with gravity cast as payer

Common trapCentrifugal force pushed the car off the road

In an inertial frame — the frame of every free-body diagram in this article — there is no outward force on the circling body, and adding one is the most damaging habit in circular dynamics. The evidence is what happens when the payer quits: cut the string, hit the ice patch, and the body moves off along the tangent, in a straight line, exactly as a force-free body must — not radially outward, as a real outward force would dictate. What your body reports in a turning car is not an outward push but the absence of an inward one: you continue straight while the car curves beneath you, until the door (or seatbelt) supplies your share of $mv^2/r$ . The related error is drawing a "centripetal force" arrow alongside gravity, normal force, and friction: centripetal force is the name of the total the real forces must reach, not an extra participant — double-booking it breaks every equation downstream. (The word "centrifugal" does have a legitimate home: in a rotating reference frame, a frame-correction term by that name appears — that machinery arrives with the non-inertial-frames articles. Until then: real forces, inertial frame, tangent-line escapes.)

Worked examples

1 · The highway exit ramp (confidence builder)

An exit ramp curve has radius $r = 200$ m and is designed for $v = 25$ m/s (90 km/h). (a) Find the design bank angle. (b) With $\mu_s = 0.5$ (wet but decent tyres), find the actual maximum speed. (c) Is there a minimum speed on ice ( $\mu_s \approx 0$ )? (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Design angle from (3): $\tan\theta = \dfrac{25^2}{200 \times 9.8} = \dfrac{625}{1960} = 0.319$ , so $\theta = \boxed{17.7°}$ — steep by road standards (real highways use gentler banks plus friction; racing ovals use up to 30°).

Step 2. Maximum speed from (5): numerator $\tan\theta + \mu_s = 0.819$ ; denominator $1 - \mu_s\tan\theta = 1 - 0.159 = 0.841$ . $v_{\max}^2 = 200 \times 9.8 \times \dfrac{0.819}{0.841} = 1960 \times 0.974 = 1909$ , so $v_{\max} = \boxed{43.7\ \mathrm{m/s} \approx 157\ \mathrm{km/h}}$ — friction plus banking nearly doubles the design speed. (Same curve, flat: $v_{\max} = \sqrt{0.5 \times 9.8 \times 200} = 31.3$ m/s — banking buys 12 m/s.)

Step 3. Minimum speed: exists only if $\tan\theta > \mu_s$ ; on ice, $\mu_s \approx 0 < 0.319$ , so yes — $v_{\min}^2 = gr\tan\theta = 1960 \times 0.319$ , giving $v_{\min} = 25$ m/s: on frictionless ice the only safe speed is the design speed, and a car slower than 90 km/h slides down the banking. Ordinary conditions ( $\mu_s = 0.5 > \tan\theta$ ): no minimum — you may park on the ramp.

Step 4. Read the design philosophy: engineers bank for a typical speed so that friction is held in reserve for braking, gusts, and the unplanned — the bank pays the routine invoice, friction handles the exceptions. That division of labour is equation (5)'s entire purpose in civil engineering.

2 · Loop-the-loop, with a safety audit (exam level)

A coaster's circular loop has radius $R = 7.0$ m. For the frictionless ideal: (a) the minimum speed at the top; (b) the minimum release height above the loop's bottom; (c) the riders' apparent weight at the bottom if released from exactly that height; (d) the same at the top. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Top minimum from (4): $v_{\rm top} = \sqrt{gR} = \sqrt{9.8 \times 7.0} = \sqrt{68.6} = \boxed{8.3\ \mathrm{m/s}}$ (30 km/h). At this speed the track force is zero: riders are weightless for the moment of crest — gravity alone is curving them.

Step 2. Release height from (7): $h_{\min} = \tfrac52 R = \boxed{17.5\ \mathrm{m}}$ — a full 3.5 m above the loop's 14-m top, the premium that finances the crest speed.

Step 3. Bottom of the loop: $v_{\rm bot}^2 = 2gh_{\min} = 5gR = 343\ \mathrm{m^2/s^2}$ ( $v_{\rm bot} = 18.5$ m/s). Apparent weight $N_{\rm bot} = mg + \dfrac{mv_{\rm bot}^2}{R} = mg + 5mg = \boxed{6mg}$ — six gravities, at the edge of blackout territory for unprepared riders. Cross-check with the $6mg$ theorem (6): $N_{\rm top} = 0$ , so $N_{\rm bot}$ must be $6mg$ ✓.

Step 4. Top apparent weight: $N_{\rm top} = \boxed{0}$ by construction — and this is why real coasters never build circular loops at minimum energy. The clothoid (teardrop) profile enlarges $R$ at the bottom (shrinking $v^2/R$ there, cutting the 6g spike to ~4g) and tightens it at the top (so the crest is flown comfortably above the weightless threshold). The 1901 Flip-Flap Railway's circular loop delivered ~12g and neck injuries; the modern teardrop is equation (1) applied with mercy.

Figure

Around the vertical circle — the invoice is fixed, the payers rotate through the roles

Origins · 1 minChristiaan Huygens and how this was discovered

Christiaan Huygens

1629–1695 · The Hague & Paris

Huygens got to the mathematics of circular motion first — before Newton's Principia existed. In 1659, analysing why a stone in a sling pulls outward on the hand, he derived that the "centrifugal tendency" grows as $v^2/r$ , publishing the result as thirteen theorems in Horologium Oscillatorium (1673), the same masterwork that gave the pendulum clock its theory, the cycloid its isochronous pendulum, and physics its first precision standard of time. Huygens' conical pendulum — this article's third derivation — was his own proposal for a clock escapement, its steady sweep converting circular dynamics into timekeeping. Newton, who possessed the $v^2/r$ result independently, seized on Huygens' quantitative law: applied to the Moon, whose orbital "fall" toward Earth is a centripetal acceleration, it became a decisive rung in the ladder to universal gravitation — the famous Moon test comparing $v^2/r$ at the lunar orbit with $g$ diluted by the inverse square. Huygens himself never accepted action at a distance, and spent his last years disputing gravity's mechanism with Newton's followers — the man who armed the Principia with its circular-motion mathematics, politely declining its metaphysics.

History

1659

— Huygens derives the

v^2/r

law of circular motion ("centrifugal force" in his vocabulary), thirteen theorems he will publish in 1673.

1687

— Newton's Principia re-derives the law as centripetal — centre-seeking — and uses the Moon's

v^2/r

to weld terrestrial gravity to celestial motion.

1901

— Coney Island's Flip-Flap Railway loops a circular loop at ~12g; the injuries teach engineering the hard version of this article's bottom-of-loop equation.

1907

— Brooklands motor circuit opens with 9-m-high bankings, the first purpose-built demonstration of design-speed engineering, eq. (3) poured in concrete.

1976

— Revolution at Six Flags debuts the modern clothoid (teardrop) vertical loop — variable radius taming the invoice at both crest and trough.

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 6 (uniform circular motion & applications) · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 2–3 (with the conical pendulum and rotor problems) · Morin, Introduction to Classical Mechanics, Ch. 3 · Huygens, Horologium Oscillatorium (1673), for the original thirteen theorems on centrifugal tendency. Key numerical results: exit-ramp set — $\tan\theta = 625/1960 = 0.319 \Rightarrow 17.7°$ ; $v_{\max}^2 = 1960 \times 0.819/0.841 \Rightarrow 43.7$ m/s; icy $v_{\min} = 25$ m/s. Loop set — $v_{\rm top} = \sqrt{68.6} = 8.3$ m/s, $h = 17.5$ m, $N_{\rm bot} = 6mg$ , with the $6mg$ gap speed-independent. Conical pendulum — $T = 2\pi\sqrt{1.5\cos30°/9.8} = 2.29$ s, $v = 2.06$ m/s, cross-checked via $2\pi r/T$ . Flat-curve limit — $\sqrt{0.9 \times 9.8 \times 50} = 21$ m/s.

Builds on: Circular Motion and Centripetal Acceleration · Newton's Second Law · Friction Forces · Conservation of Mechanical Energy. Leads to: Non-inertial Reference Frames · Fictitious (Inertial) Forces · Central Force Fundamentals.

Is centripetal force a real force or not? Different books seem to disagree.

The force is perfectly real — it is whichever ordinary forces happen to be paying: tension in the string, friction from the road, gravity on a satellite, the normal force from a banked wall. What is not real is centripetal force as a separate species that appears because motion is circular. "Centripetal" is a job description, not a kind of force — it names the inward role, the requirement $mv^2/r$ that the real forces must total. Books that say "the centripetal force on the car is 4000 N" mean "the inward components of the real forces sum to 4000 N," which is fine; books that draw a centripetal arrow in addition to friction and gravity are double-counting, and every equation after that diagram is wrong. The safe habit: free-body diagrams carry only forces with an identifiable agent (Earth pulls, road pushes, string pulls); then Newton's second law aims the net at the centre. If you cannot name what is physically pulling inward, the body cannot move in a circle — that observation solves more circular-dynamics puzzles than any formula.

If there's no outward force, why does the water stay in a bucket swung overhead?

Ask instead why the water would leave — things need a force to stop falling, not to keep moving. At the top of the swing the water is moving horizontally at speed $v$ , and gravity pulls it down at $g$ . Falling is exactly what it does — but the bucket is also accelerating downward, at $v^2/R$ . As long as $v^2/R \ge g$ , the bucket floor curves away from underneath the water at least as fast as gravity can drop it, and the floor must actually push down on the water (its share of $mv^2/R$ beyond gravity's) to keep it on the circular path. The water presses back on the floor — it feels "held in" — with no outward force anywhere in the story. Swing too slowly ( $v^2/R < g$ ), and gravity now out-accelerates the circle: the water leaves the floor and parts company — you get wet not because an outward force won, but because the inward requirement went unmet. Same equation, kitchen edition: $v_{\rm top} = \sqrt{gR}$ with $R \approx 1$ m says about 3.1 m/s, one revolution per two seconds. Comfortably achievable, as generations of physics teachers have gambled.

Why do I lean inward on a bicycle but a car stays upright?

Because a bicycle must satisfy a torque balance as well as the force equation, and leaning is how it pays that second bill. The friction force at the tyres supplies the inward $mv^2/r$ , but it acts at the ground — below the centre of mass — so it exerts a torque that would flip the bike outward if the rider stayed vertical. Leaning tilts the gravity lever arm to cancel that torque: the balance works out at $\tan\phi = v^2/rg$ — the same magic angle as the banked curve and the conical pendulum, now appearing as a lean angle (25 m/s around 200 m: about 18°; MotoGP riders exceed 60° with the help of wide tyres and knee-out geometry). A car doesn't lean because its four-wheel base absorbs the torque as a load shift — outer tyres press harder, inner tyres lighten — with no tilt needed until the shift exhausts itself and the car rolls over (the SUV rollover threshold from the statics article: the same $v^2/rg$ compared against the track-width-to-height ratio $w/2h$ ). One requirement, three body plans.

In the banked-curve derivation, why resolve horizontally and vertically instead of along the incline like in every ramp problem?

Align axes with the acceleration, and the algebra thanks you. In ramp problems the acceleration (if any) points along the slope, so slope-parallel axes put the whole of $ma$ on one axis. Here the acceleration is centripetal — horizontal, pointing at the circle's centre, which lies level with the car, not down the embankment. Resolve along the incline and the acceleration smears across both axes, giving two coupled equations with $ma\cos\theta$ and $ma\sin\theta$ terms; resolve horizontally/vertically and you get the clean pair — vertical balance, horizontal $= mv^2/r$ — that divides into $\tan\theta = v^2/rg$ in one line. Nothing is wrong with the other choice beyond pain: the physics is axis-independent, but the bookkeeping is not. The transferable rule: in circular-motion problems, one axis toward the centre, one along the no-acceleration direction — for the conical pendulum and the banked curve alike, that is horizontal + vertical; for the vertical loop's top and bottom, it is the vertical itself.

Does the vertical-circle minimum speed apply to a ball on a ROD, or a coaster with wheels that grip the track?

No — the $v_{\rm top}^2 \ge gR$ condition is specific to payers that can only push (an inside track) or only pull (a string). A rigid rod can do both: below the critical speed it simply pushes up on the ball at the top, supporting it like a strut, and the ball can crawl through the crest arbitrarily slowly (the true minimum becomes an energy condition — just enough to reach the top — not a force condition). Modern roller coasters are the engineering version of the rod: their wheel bogies clamp the rail from both sides (road wheels, upstop wheels, side wheels), so the track can pull as well as push, and a coaster valleying below design speed hangs safely rather than raining riders. The famous minimum-speed formula governs the genuinely one-sided cases: water in the swung bucket, a ball in a hemispherical bowl, a car cresting a humpback bridge (where leaving the surface at $v^2 = gR$ is the point of the problem), and any string-and-stone arrangement. Always ask of the payer: push, pull, or both? The answer selects the physics.

How is a satellite in orbit an example of this article? Nothing seems to hold it in a circle.

Gravity is the payer — the entire invoice, every second. A circular orbit is the case where the only force on the body happens to point at a fixed centre with exactly the right magnitude: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , which solves to the orbital speed $v = \sqrt{GM/r}$ (about 7.7 km/s at ISS altitude, one lap in 92 minutes). The astronauts' weightlessness is the loop-crest condition made permanent: like the coaster rider at $N = 0$ , they are in continuous free fall whose curvature happens to close on itself — Newton's cannonball fired sideways fast enough that the ground curves away as fast as the ball drops. What makes orbits a richer subject than exit ramps is that gravity, unlike a road, weakens with $r$ and permits non-circular paths — ellipses, escapes, transfers — which is where this article hands off to the central-force and Kepler articles. But the circular special case is nothing more than equation (1) with gravity cast as the sole payer, and orbital velocity is the design-speed concept of the banked curve, applied to a bend that goes all the way around the planet.

The car corners at constant speed, so no work is done — but the engine clearly burns fuel in a corner. What gives?

Two true statements, different subjects. The centripetal force does no work — it points at the centre, perpendicular to the velocity at every instant, so $\vec F \cdot \vec v = 0$ and the circular motion itself is energetically free (this is why a satellite orbits forever without a motor: its payer, gravity, works for nothing). The fuel bill in a real corner goes to everything else: rolling resistance and air drag (present on the straight too), plus corner-specific losses — tyres cornering at their limit run a small slip angle, so the friction force is not exactly perpendicular to the contact patch's actual motion and dissipates energy as tyre heating (the squeal, the wear, the black lines); the suspension pumps; and drivers usually brake in and accelerate out, cycling kinetic energy through the brakes. The clean statement: an ideal circular turn at constant speed costs nothing beyond straight-line losses, and the ideal is nearly achieved by trains on banked curves and satellites in vacuum. The perpendicularity of force and velocity is also why the magnetic force on a charged particle — always perpendicular to $\vec v$ — can bend paths into circles but never change a particle's speed: cyclotron physics is this FAQ with the payer swapped.