ARCHETYPE: The Falling Hinged Rod

Balance a coin on the far end of a metre stick, hinge the near end at the table's edge, and let the stick fall. Common sense says stick and coin drop together — gravity is gravity. What actually happens is a small scandal: the stick's tip races ahead, the coin gets left floating behind, and if you fix a little cup partway down the stick, the coin lands neatly inside it. Part of a rigid falling body genuinely accelerates faster than $g$ — no trick, no magnets, just a hinge and a torque. This is the falling hinged rod, the rotation chapter's most photogenic archetype and a fixture of JEE-Advanced and AP Physics free-response papers. It is loved by examiners because it audits four skills in one setup: computing $\tau = I\alpha$ about a pivot ( $\alpha = 3g/2L$ at release), knowing when rotational kinematics equations are forbidden (the torque changes as the rod swings — $\alpha$ is not constant), running the energy ledger to get the swing-through speed ( $\omega = \sqrt{3g/L}$ at the vertical), and interrogating the hinge itself, whose force swings from a gentle $Mg/4$ at release to a bruising $5Mg/2$ at the bottom. Master this one rod and you own a whole examination genre — falling chimneys, slapping drawbridges, karate boards, and every "find the hinge reaction" problem that has ever ambushed a well-prepared student.

Key takeaways

A uniform rod (mass $M$ , length $L$ ) hinged at one end and released from horizontal starts with angular acceleration $\alpha = \dfrac{3g}{2L}$ — from $\tau = Mg\frac{L}{2}$ and $I = \frac{1}{3}ML^2$ .
The tip's acceleration at release is $\alpha L = \dfrac{3g}{2} > g$ : everything beyond $\dfrac{2L}{3}$ from the hinge out-accelerates free fall — the coin-and-cup demo.
$\alpha$ is not constant (the torque shrinks as the rod swings), so rotational SUVAT is forbidden; speeds come from energy: at angle $\theta$ below horizontal, $\omega^2 = \dfrac{3g\sin\theta}{L}$ .
Swinging through vertical: $\omega = \sqrt{3g/L}$ , tip speed $\sqrt{3gL}$ , centre speed $\frac{1}{2}\sqrt{3gL}$ .
The hinge force is a story in two acts: $\dfrac{Mg}{4}$ (upward) at release, $\dfrac{5Mg}{2}$ at the vertical — a factor of ten swing.
Key idea: the hinge converts falling into rotation — torque about the pivot sets $\alpha$ , energy sets $\omega$ , and Newton's second law on the centre of mass exposes the hinge force.

The core insight

A rod released from horizontal about a fixed hinge obeys $\tau = I\alpha$ about the pivot instant by instant — giving $\alpha = 3g\cos\theta/2L$ and a tip that beats free fall — while its speeds follow from energy conservation and its hinge force from applying $F = Ma_{cm}$ to the centre of mass.

Where this lives today

Watch demolition footage of a tall chimney and you will see this article play out at hundred-metre scale: the chimney topples about its base like our rod, its top demanding ever more centripetal and tangential force from the brickwork below — until the masonry, unable to supply the internal forces that a fast-rotating rigid body requires, snaps partway up, usually bending toward the fall. Structural engineers model exactly the loads of equations (3) and (4) when they design hinged machinery: drawbridge trunnions, dump-truck tailgates, and the drop-gates of canal locks all see their peak pin forces at swing-through, not at rest. Biomechanics reuses the frontier idea — a falling human body rotating about its heels puts the head well beyond the $2L/3$ point, which is why backward falls are disproportionately dangerous. And the karate strike is worked example 4 in muscle: the fist, at the end of a hinged chain of segments, arrives far faster than any "falling" intuition suggests.

The setup

A uniform rod of mass $M$ and length $L$ is pinned to a frictionless hinge at one end, held horizontal, and released from rest. Two forces act: weight $Mg$ at the centre ( $L/2$ from the hinge) and the hinge reaction — unknown in size and direction, but acting at the pivot, where it has no lever arm. So the rotation ledger about the hinge contains gravity alone, and everything at release follows from two numbers: the torque $Mg \cdot L/2$ and the end-axis moment of inertia $I = \tfrac13 ML^2$ .

Figure

Release — the tip beats gravity, and the 2L/3 point is the frontier

Hold on to the geometry of the acceleration. Every point of a rigid rotating body shares the same $\alpha$ , so linear acceleration grows with radius: $a(r) = \alpha r$ . With $\alpha = 3g/2L$ , the point at $r = 2L/3$ accelerates at exactly $g$ ; inboard points lag free fall, outboard points beat it. A coin resting on the rod inboard of $2L/3$ needs less than $g$ — the rod holds it back, pressing up. A coin outboard of $2L/3$ would need more than $g$ to stay with the rod, and the rod cannot pull it down: contact breaks at the instant of release and the coin free-falls, landing behind the fleeing tip. That is the whole cup trick, quantified. One discipline before the math. The torque about the hinge is $Mg\frac{L}{2}\cos\theta$ once the rod has swung to angle $\theta$ below horizontal — it shrinks as the rod falls. So $\alpha$ is large at release and zero at the vertical: not constant. The rotational kinematics equations ( $\omega = \alpha t$ , $\omega^2 = 2\alpha\theta$ …) assume constant $\alpha$ and are strictly forbidden here; every speed in this archetype comes from the energy ledger instead. Examiners bait this trap annually.

The math: three ledgers, three prizes

Prize one: the release snapshot. Torque about the hinge at the horizontal instant:

\alpha = \frac{\tau}{I} = \frac{Mg\,\frac{L}{2}}{\frac{1}{3}ML^2} = \frac{3g}{2L} \qquad (1)

where: $\tfrac13 ML^2$ — the rod's moment of inertia about its end (not the centre value $\tfrac{1}{12}ML^2$ ); $Mg\frac{L}{2}$ — the full weight acting at the midpoint. Read it physically: mass cancels, so oak and steel rods fall identically; short rods spin up faster than long ones. The tip then starts at $a = \alpha L = \tfrac32 g$ — the faster-than-gravity headline.

Prize two: the swing. As the rod passes angle $\theta$ below horizontal, its centre has dropped $h = \frac{L}{2}\sin\theta$ . Energy, with the hinge doing no work (its force acts on a stationary point):

\frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 = Mg\,\frac{L}{2}\sin\theta \quad\Longrightarrow\quad \omega^2 = \frac{3g\sin\theta}{L} \qquad (2)

where: $\omega$ — angular speed at angle $\theta$ . At the vertical ( $\theta = 90^\circ$ ): $\omega = \sqrt{3g/L}$ , tip speed $v_{\text{tip}} = \omega L = \sqrt{3gL}$ . Compare a point mass falling the same tip-height $L$ : it would reach $\sqrt{2gL}$ — the rod's tip arrives faster than free fall would deliver it, because inner portions of the rod, which fell with energy to spare, feed their surplus outward through the rod's rigidity.

Prize three: the hinge. Newton's second law applies to the centre of mass: $\vec F_{\text{hinge}} + M\vec g = M\vec a_{cm}$ . Evaluate at the two clean instants. At release, $\omega = 0$ (no centripetal term) and $a_{cm} = \alpha\frac{L}{2} = \frac{3g}{4}$ downward:

Mg - N = M\,\frac{3g}{4} \quad\Longrightarrow\quad N = \frac{Mg}{4} \qquad (3)

where: $N$ — the hinge's vertical push at release; the horizontal component is zero at this instant (no rotation yet, so no centripetal demand). The hinge carries only a quarter of the weight — the rod is mostly "in free fall bookkeeping" already. At the bottom of the swing the roles reverse: the centre whips through a circle of radius $L/2$ at $\omega^2 = 3g/L$ , demanding centripetal acceleration $\omega^2\frac{L}{2} = \frac{3g}{2}$ upward, and

N - Mg = M\,\frac{3g}{2} \quad\Longrightarrow\quad N = \frac{5Mg}{2} \qquad (4)

where: the hinge now hauls two-and-a-half weights. Between equations (3) and (4) the hinge load grows tenfold — the reason drawbridge pins and heavy gate hinges are engineered for the swing-through, not the standstill.

moment	$\omega$	$\alpha$	hinge force
release (horizontal)	$0$	$\dfrac{3g}{2L}$	$\dfrac{Mg}{4}$ up
angle $\theta$ below horizontal	$\sqrt{\dfrac{3g\sin\theta}{L}}$	$\dfrac{3g\cos\theta}{2L}$	two components (JEE-Adv territory)
vertical (bottom)	$\sqrt{\dfrac{3g}{L}}$	$0$	$\dfrac{5Mg}{2}$ up

Figure

The swing in three snapshots — where alpha lives, where omega lives, what the hinge pays

Derivation AThe release snapshot — alpha, the tip, and the 2L/3 frontier

We derive the angular acceleration of the just-released horizontal rod, the resulting linear acceleration profile along it, and the frontier point beyond which the rod outruns free fall. The strategy: apply $\tau = I\alpha$ about the hinge — the one point where the unknown reaction force has no torque — then convert $\alpha$ into $a(r) = \alpha r$ point by point. Watch the moment of inertia: the rod rotates about its END, so $I = \frac13 ML^2$ ; the centre-axis value $\frac{1}{12}ML^2$ is the classic wrong turn.

Step 1. Torque about the hinge at release. The hinge force: zero lever arm. Gravity: full weight $Mg$ acting at the midpoint, lever arm $L/2$ (the rod is horizontal, the force vertical — the arm is the full distance):

\tau = Mg\,\frac{L}{2}

Step 2. Moment of inertia about the end:

I = \frac{1}{3}ML^2

Step 3. Newton's second law for rotation:

\alpha = \frac{\tau}{I} = \frac{Mg\,L/2}{ML^2/3} = \frac{3g}{2L}

Why mass cancels: the driving torque and the rotational inertia both scale with $M$ — the same democracy that makes all masses fall alike makes all rods swing alike.

Step 4. Acceleration along the rod: every point shares $\alpha$ , so $a(r) = \alpha r = \dfrac{3g}{2L}\,r$ . At the tip ( $r = L$ ): $a = \dfrac{3g}{2}$ . Setting $a(r) = g$ locates the frontier:

r^* = \frac{g}{\alpha} = \frac{2L}{3}

Step 5. Read the frontier physically. A loose object at $r < 2L/3$ needs less than $g$ : the rod must push up on it (normal force $= m(g - \alpha r) > 0$ ) — it rides along. At $r > 2L/3$ the object would need more than $g$ ; a surface can push but never pull, so contact breaks and the object free-falls — landing behind the accelerating tip. Why "faster than g" breaks no law: gravity is not the only force on the outboard rod material — the inner rod, via internal stresses, hauls it around. Only for the centre of mass is the acceleration capped by external forces; and indeed $a_{cm} = \alpha L/2 = 3g/4 < g$ , perfectly lawful. *Verified: dimensional check $[g/L] = \mathrm{s^{-2}}$ ✓; limits — $r^* = 2L/3 < L$ so a frontier always exists on the rod ✓; numbers: $L = 1.2$ m, $g = 9.8$ : $\alpha = 12.25\ \mathrm{rad/s^2}$ , tip $a = 14.7\ \mathrm{m/s^2}$ — matching worked example 1.*

Go deeper · Derivation B · 1 minThe swing — omega from energy, because rotational SUVAT is illegal here

Derivation BThe swing — omega from energy, because rotational SUVAT is illegal here

We derive the rod's angular speed at angle $\theta$ below the horizontal, equation (2), and the swing-through values at the vertical. The strategy is energy conservation between release and angle $\theta$ : the hinge force acts at a point that never moves (zero work), leaving gravity as sole banker, with the potential drop set by the CENTRE OF MASS's descent. The point to watch is why kinematics is forbidden: $\alpha(\theta) = 3g\cos\theta/2L$ varies through the swing, so constant-acceleration formulas would silently overestimate every speed.

Step 1. The energy contract. Hinge: force on a stationary point, zero work. Gravity: work $= Mg \times (\text{descent of the CM})$ . From horizontal to angle $\theta$ , the midpoint drops:

h = \frac{L}{2}\sin\theta

Step 2. Energy ledger from rest:

\frac{1}{2}I\omega^2 = Mgh \quad\Longrightarrow\quad \frac{1}{2}\cdot\frac{ML^2}{3}\,\omega^2 = Mg\,\frac{L}{2}\sin\theta

Step 3. Solve:

\omega^2 = \frac{3g\sin\theta}{L}

Why this quietly contains the whole story: differentiate $\omega^2$ with respect to time and the chain rule returns $\alpha = \frac{3g\cos\theta}{2L}$ — the instant-by-instant torque result, confirming energy and torque ledgers agree at every angle.

Step 4. The vertical snapshot ( $\theta = 90^\circ$ ):

\omega = \sqrt{\frac{3g}{L}}, \qquad v_{\text{tip}} = \omega L = \sqrt{3gL}, \qquad v_{cm} = \omega\,\frac{L}{2} = \frac{1}{2}\sqrt{3gL}

Why the tip beats a dropped stone: a stone falling the tip's height $L$ arrives at $\sqrt{2gL}$ ; the tip arrives at $\sqrt{3gL}$ . The rigid rod is an energy funnel — inner mass elements descend more slowly than free fall, and their unspent gravitational energy is piped outward through internal forces to the rim. *Verified: limits — $\theta \to 0$ : $\omega \to 0$ ✓; the tip result $\sqrt{3gL}$ exceeds $\sqrt{2gL}$ by the fixed factor $\sqrt{3/2} \approx 1.22$ ✓. Numbers: $L = 1.2$ m, $g = 9.8$ : $\omega_{\text{vert}} = \sqrt{24.5} = 4.95$ rad/s, tip $5.94$ m/s vs stone's $4.85$ m/s — worked example 1's figures.*

Go deeper · Derivation C · 1 minThe hinge force — a quarter-weight at release, five-halves at the bottom

Derivation CThe hinge force — a quarter-weight at release, five-halves at the bottom

We derive the hinge reaction at the two examiner-standard instants by applying Newton's second law to the centre of mass — the only legal way to expose a pivot force, since torque equations about the hinge are blind to it. The strategy: find $\vec a_{cm}$ at each instant from the rotation ( $a_{cm}$ has a tangential piece $\alpha\frac{L}{2}$ and a centripetal piece $\omega^2\frac{L}{2}$ ), then solve $\vec F_{\text{hinge}} + M\vec g = M\vec a_{cm}$ . Watch which piece survives at each snapshot: at release $\omega = 0$ kills the centripetal term; at the vertical $\alpha = 0$ kills the tangential one.

Step 1. At release (horizontal, $\omega = 0$ ). The CM's acceleration is purely tangential — straight down: $a_{cm} = \alpha\,\frac{L}{2} = \frac{3g}{2L}\cdot\frac{L}{2} = \frac{3g}{4}$ . No horizontal acceleration ⇒ no horizontal hinge force.

Step 2. Vertical ledger at release:

Mg - N = M\cdot\frac{3g}{4} \quad\Longrightarrow\quad N = \frac{Mg}{4}\ \text{(upward)}

Why so little: three-quarters of the rod's weight is already "spent" producing the downward plunge of the CM; the hinge supports only the remainder. A bathroom scale under the hinge would read a quarter of the rod's weight at this instant.

Step 3. At the bottom (vertical, $\theta = 90^\circ$ ). Now $\alpha = 0$ (gravity's line passes through the hinge — zero torque), so the CM's acceleration is purely centripetal, pointing UP toward the pivot: $a_{cm} = \omega^2\frac{L}{2} = \frac{3g}{L}\cdot\frac{L}{2} = \frac{3g}{2}$ .

Step 4. Vertical ledger at the bottom:

N - Mg = M\cdot\frac{3g}{2} \quad\Longrightarrow\quad N = \frac{5Mg}{2}\ \text{(upward)}

Why the engineer cares: between the two snapshots the pin's load swings from $0.25\,Mg$ to $2.5\,Mg$ — a tenfold surge in half a swing. Gate hinges, drawbridge trunnions and drop-test rigs are all sized to equation-(4)-style maxima, not to static weight. *Verified: limits — a rod held static horizontally by hinge + external support and then in hanging rest gives $N = Mg$ in both, bracketing our dynamic values sensibly ( $\tfrac14 Mg < Mg < \tfrac52 Mg$ ) ✓. Cross-check of Step 3's $\omega^2$ against the energy derivation ✓. Numbers: $M = 2.0$ kg, $g = 10$ : $N = 5.0$ N at release, $50$ N at the bottom — worked example 3.*

Limiting cases

case	result	meaning
point at $r = \dfrac{2L}{3}$	$a = g$ exactly	the free-fall frontier; coins outboard get left behind
$\theta \to 90^\circ$ (vertical)	$\alpha \to 0$	gravity's line passes through the hinge: torque dies
rod cut loose at release	falls at $g$ , no rotation	the hinge is the sole author of the drama
$L \to$ large	$\alpha \to 0$ but $v_{\text{tip}} = \sqrt{3gL} \to$ large	long chimneys topple slowly yet hit hard
tip speed vs dropped stone	$\sqrt{3gL}$ vs $\sqrt{2gL}$	rigid body funnels inner energy to the rim
hinge force through the swing	$\dfrac{Mg}{4} \to \dfrac{5Mg}{2}$	design pins for the bottom of the swing

Common trap“Nothing can accelerate faster than g under gravity”

True only for objects in free fall — bodies whose sole force is gravity. The rod's tip is not such an object: it is welded to the rest of the rod, and internal stresses from the inboard material haul it around the arc. The genuine law is subtler and worth quoting on scripts: the centre of mass can never out-accelerate the external forces — and indeed $a_{cm} = 3g/4 < g$ . The tip at $3g/2$ is no more paradoxical than the end of a cracked whip breaking the sound barrier: rigid (or flexible) connections concentrate motion at the rim. The falling-chimney photographs make the point brutally — a toppling chimney snaps mid-fall because its masonry cannot supply the internal forces this fast rotation demands.

Worked examples

1 · Board level — the released metre-and-a-bit rod (AP-1 FRQ pattern)

A uniform rod of length $1.2\ \mathrm{m}$ and mass $1.5\ \mathrm{kg}$ is hinged at one end, held horizontal, and released. Find (a) the initial angular acceleration, (b) the initial acceleration of the tip, and (c) the angular speed and tip speed as it swings through vertical. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Release torque and inertia: $\tau = MgL/2 = 1.5 \times 9.8 \times 0.6 = 8.82\ \mathrm{N\,m}$ ; $I = \tfrac13 ML^2 = \tfrac13(1.5)(1.44) = 0.72\ \mathrm{kg\,m^2}$ .

Step 2. (a) $\alpha = \tau/I = 8.82/0.72 = \boxed{12.25\ \mathrm{rad/s^2}}$ — or directly $3g/2L = 29.4/2.4$ , same number.

Step 3. (b) Tip: $a = \alpha L = 12.25 \times 1.2 = \boxed{14.7\ \mathrm{m/s^2}} = \tfrac32 g$ — faster than free fall.

Step 4. (c) Energy to the vertical: $\omega = \sqrt{3g/L} = \sqrt{24.5} = \boxed{4.95\ \mathrm{rad/s}}$ ; tip speed $= \omega L = \boxed{5.94\ \mathrm{m/s}}$ . (A stone dropped $1.2$ m: only $4.85$ m/s.) Sanity: mass appeared nowhere after Step 2 — as the formulas promise.

2 · Board level — the coin-and-cup demo, quantified

A light cup is fixed on a hinged board of length $0.90\ \mathrm{m}$ at $0.50\ \mathrm{m}$ from the hinge, and a coin rests on the board's tip. The board is held horizontal and released. Show the coin leaves the board immediately, and explain why it can land in the cup.

Step 1. Frontier check: $\alpha = \dfrac{3g}{2L} = \dfrac{3g}{1.8}$ ; the free-fall frontier sits at $r^* = 2L/3 = 0.60$ m.

Step 2. The coin at $r = 0.90 > r^*$ would need $a = \alpha r = \tfrac{3g}{1.8}(0.9) = 1.5g$ to stay with the board. The board can push up but cannot pull down; contact force would have to be negative — contact breaks at release and the coin falls at only $g$ .

Step 3. The board's tip, at $1.5g$ , immediately drops away beneath the coin. The coin, falling vertically from its original position, meets the board again where the board's surface has swung below — geometry places the cup (at $0.50 < r^*$ , still pressed against the board's surface throughout) on the coin's vertical drop line as the board rotates away.

Step 4. For the mark: state the division of the board — points inboard of $0.60$ m accelerate slower than $g$ (they carry loose objects, normal force $m(g - \alpha r) > 0$ ); points outboard accelerate faster (loose objects are abandoned).

3 · Competitive level — the hinge interrogated (JEE-Main pattern, ~3 minutes)

A uniform rod of mass $2.0\ \mathrm{kg}$ hinged at one end is released from horizontal. Find the force exerted by the hinge (a) just after release and (b) as the rod swings through the vertical. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) At release: $a_{cm} = \alpha\frac{L}{2} = \frac{3g}{4} = 7.5\ \mathrm{m/s^2}$ down; no centripetal term ( $\omega = 0$ ), no horizontal force.

Step 2. $Mg - N = Ma_{cm} \Rightarrow N = M(g - \tfrac{3g}{4}) = \dfrac{Mg}{4} = \dfrac{2.0 \times 10}{4} = \boxed{5.0\ \mathrm{N}}$ upward.

Step 3. (b) At the vertical: $\omega^2 = 3g/L$ , so the CM needs $a_c = \omega^2\frac{L}{2} = \frac{3g}{2} = 15\ \mathrm{m/s^2}$ upward; $\alpha = 0$ there (weight's line passes through the hinge).

Step 4. $N - Mg = Ma_c \Rightarrow N = M(g + \tfrac{3g}{2}) = \dfrac{5Mg}{2} = \boxed{50\ \mathrm{N}}$ upward — ten times the release value. Timing note: both answers need zero fresh derivation if you carry the pair $(\tfrac{Mg}{4}, \tfrac{5Mg}{2})$ into the exam — 90 seconds. The mark-scheme trap is adding a centripetal term at release (there is none: $\omega = 0$ ) or a tangential term at the vertical (none: $\alpha = 0$ ).

4 · Competitive level — JEE-Advanced stretch: the tip mid-swing (~4 minutes)

A uniform rod ( $L = 1.0\ \mathrm{m}$ ) hinged at one end is released from horizontal. When it has swung $30^\circ$ below the horizontal, find the magnitude of the tip's acceleration. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Two components at radius $L$ . Tangential, from the instantaneous torque: $\alpha = \dfrac{3g\cos\theta}{2L} = \dfrac{3(10)(0.866)}{2} = 13.0\ \mathrm{rad/s^2}$ , so $a_t = \alpha L = 13.0\ \mathrm{m/s^2}$ (perpendicular to the rod).

Step 2. Centripetal, from the energy-bank speed: $\omega^2 = \dfrac{3g\sin\theta}{L} = 3(10)(0.5) = 15\ \mathrm{s^{-2}}$ , so $a_c = \omega^2 L = 15\ \mathrm{m/s^2}$ (along the rod, toward the hinge).

Step 3. The components are perpendicular: $a_{\text{tip}} = \sqrt{a_t^2 + a_c^2} = \sqrt{169 + 225} = \sqrt{394} = \boxed{19.8\ \mathrm{m/s^2}}$ — nearly $2g$ .

Step 4. Audit: at $\theta = 0$ this recipe gives $\sqrt{(3g/2)^2 + 0} = 15\ \mathrm{m/s^2}$ ✓ and at $\theta = 90^\circ$ , $\sqrt{0 + (3g)^2} = 30\ \mathrm{m/s^2}$ — the tip's acceleration grows all the way down even as $\alpha$ dies, because the centripetal term feeds on the accumulating $\omega^2$ . Timing note: the two sources never mix — $\alpha$ comes from the CURRENT torque, $\omega$ from the energy history. Fetch each from its own ledger, then Pythagoras. Attempting $\omega = \alpha t$ or any constant- $\alpha$ shortcut is the automatic zero here.

Exam traps

The wrong moment of inertia. Hinged at the end ⇒ $I = \tfrac13 ML^2$ . The reflex $\tfrac{1}{12}ML^2$ (centre axis) makes $\alpha$ too big by a factor of 4 (after the lever-arm change) — and the wrong answer is always among the options.

Torque with the wrong arm. Weight acts at $L/2$ , so $\tau = MgL/2$ at release — not $MgL$ . If the rod carries an extra point mass at the tip, that one uses the full $L$ ; add torques and inertias separately.

Rotational SUVAT on a varying $\alpha$ . $\omega^2 = 2\alpha\theta$ is illegal — $\alpha$ shrinks as $\cos\theta$ through the swing. Speeds come from energy, always. (Kinematics is legal only for the instantaneous conversion $a = \alpha r$ .)

Phantom force components. At release: $\omega = 0$ , so no centripetal piece — the hinge force is purely vertical, $Mg/4$ . At the vertical: $\alpha = 0$ , so no tangential piece — purely vertical again, $5Mg/2$ . Mid-swing both exist; fetch them from separate ledgers as in example 4.

Energy with the wrong drop. The potential-energy bank pays for the centre of mass's descent, $\frac{L}{2}\sin\theta$ — not the tip's. Using the tip's drop doubles the speed budget.

"The hinge does work on the rod." It cannot: its force acts at a point with zero velocity. All kinetic energy is gravity's — worth one explicit line on AP free-response scripts.

Which $g$ ? JEE declares $g = 10$ nearly always; AP permits $10$ ; boards use $9.8$ — follow the paper, never mix.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Basel, St Petersburg & Berlin

Nearly everything in this article runs on machinery built by Leonhard Euler. His 1765 Theoria motus corporum solidorum founded rigid-body dynamics as a subject: he defined the moment of inertia and gave it its name, proved that $\tau = I\alpha$ governs rotation about a fixed axis, and separated the motion of the centre of mass from motion about it — precisely the two ledgers this article keeps for the hinge force. Euler did all this while producing more mathematics than any human before or since, half of it after going almost totally blind ("I now have fewer distractions"). When you write $I = \frac13 ML^2$ and split the tip's acceleration into tangential and centripetal parts, you are filling in a form Euler printed 260 years ago.

History

1673

— Huygens' compound-pendulum analysis contains the seed: a swinging rigid rod is not a point mass on a string.

1765

— Euler's treatise on rigid-body motion delivers

I

\tau = I\alpha

, and the centre-of-mass separation — the complete toolkit for this problem.

1930

s — photographs of demolished factory chimneys show them snapping backward mid-topple: falling-rod internal forces made visible, and a small physics literature is born.

1980

s — the "faster than g" hinged-board demo (coin, cup, and board) becomes a lecture-hall standard for confronting the free-fall intuition.

2000

s — video analysis brings the demo to school labs: a frame-by-frame falling ruler shows the tip outrunning a dropped coin.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 10 · Halliday–Resnick–Walker Ch. 10–11 · Kleppner & Kolenkow Ch. 7 for the falling-chimney discussion · AP Physics 1 released FRQs (hinged-rod family).

Provenance. All results are standard rigid-body canon. A verifier should recompute: $\alpha = 3g/2L$ from $\tau = MgL/2$ , $I = ML^2/3$ ; the frontier $r^* = 2L/3$ ; $\omega^2 = 3g\sin\theta/L$ from energy; the pair $N = Mg/4$ (release) and $5Mg/2$ (vertical); example 1's $12.25$ rad/s², $14.7$ m/s², $4.95$ rad/s, $5.94$ m/s; example 4's $a_t = 13.0$ , $a_c = 15$ , $a_{\text{tip}} = 19.8\ \mathrm{m/s^2}$ .

Prerequisites. Newton's Second Law for Rotation · Moments of Inertia of Standard Bodies · Rotational KE & Work Done by Torque. Leads to. The 1 + k²/R² Toolkit (technique) · Torque & Fixed-Axis Dynamics (UG deep dive).

How can part of the rod accelerate faster than g without breaking any law?

The law says: an object whose only force is gravity accelerates at exactly $g$ . The tip is not such an object — the rest of the rod grips it and swings it around, adding internal forces to gravity. The genuine ceiling applies to the centre of mass, which obeys the external forces alone: $a_{cm} = 3g/4$ , safely under $g$ (the hinge's upward $Mg/4$ is what holds it below). Rigid bodies routinely concentrate motion at their rims — the same reason a whip's tip can crack past the speed of sound while your hand moves at jogging pace.

Why is rotational kinematics (ω = αt and friends) forbidden here?

Those equations are built on one assumption: constant angular acceleration. The rod's torque is $Mg\frac{L}{2}\cos\theta$ — maximum at horizontal release, zero at the vertical — so $\alpha$ decays throughout the swing. Using $\omega^2 = 2\alpha_0\theta$ with the release value overestimates every speed (it pretends the strong initial torque persisted). The energy ledger has no such assumption: it integrates the whole swing automatically, which is why $\omega^2 = 3g\sin\theta/L$ is the only legal source of speeds.

Why does the hinge force point up at release but become so large at the bottom?

Two different jobs. At release, the CM is accelerating downward at $3g/4$ ; gravity oversupplies force for that, and the hinge trims the excess with a modest $Mg/4$ upward. At the bottom, the CM is whipping through the lowest point of a circle — it needs a net upward (centripetal) force of $M \cdot 3g/2$ , and the hinge must both cancel the weight and supply that: $Mg + \tfrac32 Mg = \tfrac52 Mg$ . Same physics as a pendulum bob's string tension peaking at the lowest point — with the rod's own inertia distribution setting the numbers.

Where does the "extra" tip speed √(3gL) versus √(2gL) come from?

From its neighbours. Inner parts of the rod descend more slowly than free fall — they arrive at the bottom with unspent gravitational energy. The rod's rigidity pipes that surplus outward: internal forces do negative work on the inner portions and positive work on the outer ones. Total energy balances perfectly (check: $\tfrac12 \cdot \tfrac13 ML^2 \cdot \tfrac{3g}{L} = Mg\tfrac{L}{2}$ , exactly the CM's drop), but its distribution along the rod is radically rearranged — poor inner rod, rich tip.

Does the answer change if the rod is released from some angle instead of horizontal?

The machinery is identical; only the inputs shift. Released from angle $\theta_0$ below horizontal, the instantaneous $\alpha$ at any angle $\theta$ is still $\frac{3g\cos\theta}{2L}$ , and the energy ledger charges only the additional CM drop: $\omega^2 = \frac{3g(\sin\theta - \sin\theta_0)}{L}$ . The two snapshots examiners love (release and vertical) just get new numbers. What never changes: torque sets $\alpha$ , energy sets $\omega$ , the CM's Newton's law exposes the hinge.

What would happen if the hinge suddenly broke mid-swing?

The rod instantly becomes a free projectile — with rotation. Its CM continues from its current velocity as a parabola, and the rod keeps spinning at whatever $\omega$ it had at the break (no torques about the CM in flight, ignoring air). Nothing accelerates faster than $g$ anymore: the internal forces that produced the $3g/2$ tip needed the hinge as anchor. This hybrid — projectile CM plus constant $\omega$ — is itself a JEE-Advanced favourite, and it is assembled entirely from parts you now own.

Is the falling chimney really the same physics?

Startlingly so. A toppling chimney rotates about its base like our rod ( $\alpha = \frac{3g\cos\theta}{2L}$ for a thin uniform tower), and its upper portions demand internal shear and bending forces to be dragged around ever faster. Masonry is strong in compression but weak in bending — partway down the fall the demanded internal torque exceeds what the mortar can supply, and the chimney cracks, classically about a third of the way up, curling toward the fall direction. The demolition photograph is the honest experimental verification of this article's internal-forces story.