ARCHETYPE: Yo-Yos & Unwinding Spools

Drop a ball and it falls at $g$ — story over. Drop a yo-yo holding its string and something odd happens: it falls slowly, gently, as if gravity had been turned down, while spinning faster and faster. A cheap toy is performing a serious piece of physics: the string's tension is fighting gravity, and the price of that fight is spin. The yo-yo is an Atwood machine folded into a single body — one object that is both the falling mass and the pulley. Examiners adore this setup because it forces you to run translation and rotation simultaneously: Newton's second law down the page, $\tau = I\alpha$ across it, and one string constraint stitching them together. JEE-Main asks for the acceleration and tension of a falling spool; AP Physics 1 poses the unwinding disc as a free-response staple; JEE-Advanced plays its favourite trick — a thread spool on a table that rolls toward the hand pulling it, a result so counter-intuitive that physicists use it as a party bet. This article treats the setup itself as the concept: the master solution for anything falling on its own unwinding string, the energy shortcut, the spool-on-a-table direction rule with its beautiful critical angle, and the traps that separate the prepared from the surprised. Master one toy, and a whole family of rotation problems unwinds for you.

Key takeaways

A yo-yo or spool falling on its own string obeys $mg - T = ma$ (translation) and $Tr = I\alpha$ (rotation), tied by the string constraint $a = r\alpha$ — $r$ is the radius where the string leaves the body.
The master result: $a = \dfrac{g}{1 + I/mr^2}$ and $T = \dfrac{mg\,(I/mr^2)}{1 + I/mr^2}$ — always $a < g$ , always $T < mg$ .
A uniform disc unwinding from its rim falls at $a = \tfrac{2}{3}g$ with $T = \tfrac{1}{3}mg$ ; a real yo-yo (big disc, thin axle) falls far slower because $I/mr^2$ is huge.
Energy route for speeds: $v = \sqrt{\dfrac{2gh}{1 + I/mr^2}}$ — gravity's work splits between moving and spinning.
Spool pulled on a table: it rolls toward the pull when the string's line of action passes in front of the contact point; the flip happens at the critical angle $\cos\theta_c = r/R$ .
Key idea: the string radius $r$ sets the torque and the constraint — one body, two of Newton's laws, one shared $a$ .

The core insight

A body unwinding from its own string is solved by writing Newton's second law for the fall and $\tau = I\alpha$ for the spin about the centre, then locking them together with the string constraint $a = r\alpha$ taken at the radius where the string actually unwinds.

Where this lives today

The yo-yo's two-ledger physics rides on real rockets. A spin-stabilized launch stage must stop spinning before releasing its satellite, and the standard fix is the yo-yo de-spin: two masses on cables wound around the stage are released, unwind exactly like this article's string, and are then cut loose, carrying nearly all the angular momentum away with them. The design equation is the same interplay of $I$ , $r$ and unwinding constraint you used in the worked examples — sized so that the stage's spin hits zero just as the cables release. Meanwhile the spool-direction rule earns its keep in cable-laying: a ship paying out undersea cable must respect the geometry of where the line of action meets the drum, or the reel surges. And every "gravity-powered" toy — the falling-disc lamp winch, the pull-string flywheel — is a yo-yo ledger with a product design wrapped around it.

The setup

Geometry first, because in an archetype the geometry is the physics. A body of mass $m$ — disc, spool, cylinder — has a string wound around it at radius $r$ from the centre. The free end is held fixed; the body is released. Gravity pulls the centre straight down; the string, anchored above, pulls up with tension $T$ at the winding radius. As the body falls a distance $x$ , a length $x$ of string must unwind, so the body must rotate by $x/r$ radians: fall and spin are chained together.

Figure

The falling yo-yo — one body, two laws, one constraint

Notice what makes this different from the pulley problems you have already mastered. In an Atwood machine the string's tension acts on separate bodies — blocks translate, the pulley spins. Here one body does both jobs. That means the same tension $T$ appears twice in your own equations for one object: once as an upward force fighting the fall, once as a torque $Tr$ driving the spin. Gravity, acting at the centre of mass, has zero lever arm about the centre and contributes no torque at all — the spin-up is entirely the string's doing. And keep your eye on one detail that decides every mark in this archetype: the torque and the constraint both use $r$ , the radius where the string leaves the body — the hub of a yo-yo, the drum of a spool — which is generally not the outer radius $R$ that appears inside the moment of inertia. Mixing them up is the single most common way to lose this problem.

The master solution pattern

Every falling-on-its-own-string problem begins identically. Take down as positive for the fall, and the unwinding sense as positive for the spin. Newton's second law for the centre of mass:

mg - T = ma \qquad (1)

where: $m$ — the body's mass; $T$ — the string tension; $a$ — the downward acceleration of the centre. Read it physically: the body falls slower than free fall exactly because $T$ back-pulls. If your final answer has $a \ge g$ or $T \le 0$ , the physics has gone wrong somewhere. The rotation ledger, torques about the centre of mass. Gravity acts at the centre — zero lever arm, zero torque. The tension acts at the winding radius $r$ , tangentially:

T\,r = I\,\alpha \qquad (2)

where: $I$ — moment of inertia about the central axis; $\alpha$ — angular acceleration; $r$ — the string radius. This is the engine of the problem: tension is the only torque, so a yo-yo with a slack string cannot spin up at all. The bigger the tension, the faster the spin builds. The string cannot stretch, and it unwinds without slipping on the drum, so a fall of $x$ means a turn of $x/r$ . Differentiate twice:

a = r\,\alpha \qquad (3)

where: the constraint uses the same $r$ as the torque equation — string radius, not outer radius. Substitute $\alpha = a/r$ into (2) to get $T = Ia/r^2$ , feed that into (1), and the machine solves:

a = \frac{g}{1 + \dfrac{I}{mr^2}}, \qquad T = mg\,\frac{I/mr^2}{1 + I/mr^2} \qquad (4)

where: the dimensionless ratio $\beta = I/mr^2$ is the whole personality of the body — how spin-hungry it is at the radius the string works. Big $\beta$ (fat disc, thin axle): the body falls at a crawl with $T$ nearly $mg$ . Small $\beta$ : nearly free fall, nearly slack string. Everything in this archetype is contained in that one ratio.

body and string position	$\beta = I/mr^2$	acceleration	tension
uniform disc, string at rim ( $r=R$ , $I = \tfrac12 mR^2$ )	$\tfrac12$	$\tfrac{2}{3}g$	$\tfrac{1}{3}mg$
hoop or ring, string at rim	$1$	$\tfrac{1}{2}g$	$\tfrac{1}{2}mg$
disc of radius $R$ , string on thin hub $r$	$\tfrac{R^2}{2r^2}$	$\dfrac{g}{1 + R^2/2r^2}$	close to $mg$ when $r \ll R$
any body, given $I$ , string at $r$	$\dfrac{I}{mr^2}$	$\dfrac{g}{1+\beta}$	$\dfrac{mg\,\beta}{1+\beta}$

Derivation AThe falling yo-yo — master result by the two-ledger method

We derive equation (4) — the acceleration and tension of any body of mass $m$ and moment of inertia $I$ falling as its string unwinds from radius $r$ — by the method every marking scheme expects: one Newton's-law equation for the translation of the centre of mass, one torque equation about the centre, and the no-slip string constraint to weld them together. The strategy is to use the constraint to eliminate $\alpha$ , express $T$ in terms of $a$ , and substitute. Watch two things: gravity exerts no torque about the centre of mass, and the radius in both the torque and the constraint is the string radius $r$ , not the body's outer radius.

Step 1. Translation, down positive:

mg - T = ma

Step 2. Rotation about the centre of mass, unwinding sense positive. Gravity acts at the centre (zero lever arm); the normal-size player is the tension at radius $r$ :

T r = I \alpha

Why about the centre of mass: for a body that is both accelerating and spinning, the centre of mass is the one point about which $\tau = I\alpha$ can be used without any correction terms — the same licence that lets us treat rolling bodies this way.

Step 3. The constraint. The string is inextensible and leaves the drum without slipping: when the centre drops $x$ , a length $x$ has unwound, so the body has turned $\theta = x/r$ . Twice differentiated:

a = r\alpha \quad\Rightarrow\quad \alpha = \frac{a}{r}

Step 4. Substitute Step 3 into Step 2: $T = \dfrac{I a}{r^2}$ . Feed into Step 1:

mg - \frac{Ia}{r^2} = ma \;\;\Rightarrow\;\; a = \frac{g}{1 + I/mr^2}

Step 5. Back-substitute for the tension:

T = \frac{Ia}{r^2} = mg\,\frac{I/mr^2}{1 + I/mr^2}

Why the shape of the answer: the denominator $1 + I/mr^2$ is gravity's driving force divided among two appetites — inertia for falling ( $m$ ) and inertia for spinning ( $I/r^2$ , the "equivalent mass" of the rotation as seen by the string). The spin's share is what the tension carries. *Verified: limits — $I \to 0$ gives $a \to g$ , $T \to 0$ (a point mass just falls); $r \to 0$ with finite $I$ gives $a \to 0$ , $T \to mg$ (a thread on an infinitely thin axle can barely turn the body: it hangs). Disc at the rim: $\beta = \tfrac12$ , $a = \tfrac{2}{3}g$ , $T = \tfrac13 mg$ — the standard textbook result.*

Go deeper · Derivation B · 1 minThe energy route — speed after a given drop

Derivation BThe energy route — speed after a given drop

We derive the speed of the yo-yo's centre after it has fallen a height $h$ from rest, using energy conservation — the fastest tool whenever a question asks for a speed rather than a time or a tension. The strategy: gravity is the only force doing net work (the tension acts at the instantaneously-at-rest unwinding point, like static friction in rolling, and does no work on the body), so the lost potential energy $mgh$ becomes kinetic energy split between translation and spin. Watch how the constraint $v = r\omega$ converts the split into a single factor.

Step 1. Energy ledger from release to a drop of $h$ :

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

Why the tension is absent: the point where the string leaves the drum is momentarily at rest (the string above it is stationary), so the tension does zero work — exactly the argument that lets friction be ignored in ideal rolling. All of gravity's work lands in kinetic energy.

Step 2. Apply the string constraint $\omega = v/r$ :

mgh = \frac{1}{2}mv^2\left(1 + \frac{I}{mr^2}\right)

Step 3. Solve:

v = \sqrt{\frac{2gh}{1 + I/mr^2}}

Why this matches the force route: with constant acceleration $a = g/(1+\beta)$ , kinematics gives $v^2 = 2ah = 2gh/(1+\beta)$ — the same answer, confirming the two ledgers agree. *Verified: disc at the rim ( $\beta = \tfrac12$ ): $v = \sqrt{4gh/3}$ . Numbers: $h = 1.2$ m, $g = 9.8$ : force route $v^2 = 2(6.53)(1.2) = 15.7$ , energy route $4(9.8)(1.2)/3 = 15.7$ — identical, $v = 3.96\ \mathrm{m/s}$ .*

The spool on the table — which way does it roll?

Now the JEE-Advanced showpiece. A thread spool — outer radius $R$ riding on the table, thread wound on an inner drum of radius $r$ — rests on a rough surface. You pull the thread gently. Every untrained instinct says the spool must roll away, paying out thread. Try it with a real spool: pulled horizontally with the thread leaving the underside of the drum, the spool rolls toward your hand, swallowing its own thread as it comes. The clean way to see it is to take torques about the contact point $P$ where the spool touches the table. About $P$ , friction and the normal force have zero lever arm — they act at $P$ — and gravity's line passes straight through the centre above $P$ , so for a spool on level ground it has no torque either. The only torque left is the pull's, and its sense is decided by pure geometry: which side of $P$ does the string's line of action pass?

\tau_P = F \times (\text{lever arm of the pull about } P) \qquad (5)

where: if the line of action passes in front of $P$ (between $P$ and the direction you pull), the torque rolls the spool toward you; passing behind $P$ , it rolls away. A horizontal pull from the bottom of the drum has lever arm $R - r$ and rolls the spool toward the hand. And there is one magic angle where the line of action passes exactly through $P$ — zero torque about the contact point, so the spool cannot roll at all: it can only skid.

\cos\theta_c = \frac{r}{R} \qquad (6)

where: $\theta_c$ — the critical angle of the pull above the horizontal. The geometry: the string leaves the drum tangentially, and the tangent line from $P$ to the inner circle of radius $r$ makes exactly this angle. Pull shallower than $\theta_c$ : the spool rolls toward you. Pull steeper: it rolls away, unwinding. Pull at $\theta_c$ : it drags without rotating — the spool's version of a perfectly placed billiard stroke.

Figure

The spool decision — the string's line of action versus the contact point

Why is judging torques about $P$ legitimate for an accelerating spool? For rolling without slipping, the contact point is the instantaneous axis of rotation, and $\tau_P = I_P \alpha$ holds about it with $I_P = I_{cm} + mR^2$ by the parallel-axis theorem. That single equation contains friction and the normal force nowhere — which is exactly why the direction answer is pure geometry. (When the spool skids, this licence lapses and you must go back to the centre-of-mass ledgers.)

Limiting cases

case	result	meaning
$I \to 0$ (all mass at the axis)	$a \to g$ , $T \to 0$	nothing to spin up: free fall on a slack string
$r \to 0$ (string on a needle axle)	$a \to 0$ , $T \to mg$	the string can barely turn the body: it hangs almost still
disc, string at rim	$a = \tfrac23 g$ , $T = \tfrac13 mg$	the standard unwinding disc — memorize this pair
hoop, string at rim	$a = \tfrac12 g$ , $T = \tfrac12 mg$	all mass at the rim: gravity's work splits evenly
spool pulled at $\cos\theta = r/R$	zero torque about contact	cannot roll: skids in place when friction lets go
pull angle $\to 0$ (horizontal, under-drum)	rolls toward the hand	the party trick: the spool eats its own thread

Worked examples

1 · Board level — the unwinding disc (AP-1/JEE-Main entry pattern)

A uniform solid disc of mass $2.0\ \mathrm{kg}$ and radius $0.10\ \mathrm{m}$ has a light string wound around its rim. The free end is tied to the ceiling and the disc is released from rest. Find the acceleration of the centre, the string tension, and the speed of the centre after it has fallen $1.2\ \mathrm{m}$ . Take $g = 9.8\ \mathrm{m/s^2}$ .

Step 1. Identify $\beta$ : string at the rim, so $r = R$ , and $I = \tfrac12 mR^2$ gives $\beta = I/mR^2 = \tfrac12$ .

Step 2. Master result: $a = \dfrac{g}{1 + \tfrac12} = \dfrac{2}{3}(9.8) = \boxed{6.53\ \mathrm{m/s^2}}$ .

Step 3. Tension: $T = mg\,\dfrac{\beta}{1+\beta} = \dfrac{1}{3}(2.0)(9.8) = \boxed{6.53\ \mathrm{N}}$ . Cross-check with ledger (1): $mg - T = 19.6 - 6.53 = 13.07 = ma$ ✓.

Step 4. Speed after $1.2$ m, energy route: $v = \sqrt{\dfrac{2gh}{1+\beta}} = \sqrt{\dfrac{2(9.8)(1.2)}{1.5}} = \sqrt{15.7} = \boxed{3.96\ \mathrm{m/s}}$ . Sanity: $a < g$ ✓, $T < mg = 19.6$ N ✓, and a free-falling disc would have hit $\sqrt{2(9.8)(1.2)} = 4.85$ m/s — the spin ate the difference.

2 · Board level — the real yo-yo (why toys fall slowly)

A yo-yo is modelled as a uniform disc of mass $0.15\ \mathrm{kg}$ and radius $3.0\ \mathrm{cm}$ , with the string wound on a thin axle of radius $0.50\ \mathrm{cm}$ . Find its acceleration and the string tension, and the time to descend its $0.80\ \mathrm{m}$ string. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. The ratio does everything: $\beta = \dfrac{I}{mr^2} = \dfrac{\tfrac12 mR^2}{mr^2} = \dfrac{R^2}{2r^2} = \dfrac{(0.030)^2}{2(0.0050)^2} = \dfrac{9.0\times10^{-4}}{5.0\times10^{-5}} = 18$ .

Step 2. Acceleration: $a = \dfrac{9.8}{1 + 18} = \boxed{0.52\ \mathrm{m/s^2}}$ — about one-nineteenth of $g$ . This is the toy's secret: a fat disc on a thin axle is a gravity diluter, the yo-yo's own version of Atwood's machine.

Step 3. Tension: $T = mg\,\dfrac{18}{19} = (0.15)(9.8)(0.947) = \boxed{1.39\ \mathrm{N}}$ , compared with the weight $1.47$ N — the string carries nearly everything.

Step 4. Time down the string: $t = \sqrt{\dfrac{2h}{a}} = \sqrt{\dfrac{2(0.80)}{0.516}} = \boxed{1.8\ \mathrm{s}}$ — a leisurely descent, exactly as the toy behaves.

3 · Competitive level — spool with given I, JEE-Main pattern (~3 minutes)

A spool of mass $2.0\ \mathrm{kg}$ has moment of inertia $0.020\ \mathrm{kg\,m^2}$ about its axis. Thread unwinds from its drum of radius $5.0\ \mathrm{cm}$ as the spool falls vertically. Take $g = 10\ \mathrm{m/s^2}$ . Find the acceleration, the tension, and the angular speed after the spool has fallen $1.0\ \mathrm{m}$ from rest.

Step 1. Compute the personality ratio: $\beta = \dfrac{I}{mr^2} = \dfrac{0.020}{(2.0)(0.050)^2} = \dfrac{0.020}{0.0050} = 4.0$ .

Step 2. Acceleration: $a = \dfrac{10}{1+4} = \boxed{2.0\ \mathrm{m/s^2}}$ . Tension: $T = m(g - a) = 2.0(8.0) = \boxed{16\ \mathrm{N}}$ .

Step 3. Speed after $1.0$ m: $v = \sqrt{2ah} = \sqrt{2(2.0)(1.0)} = 2.0\ \mathrm{m/s}$ .

Step 4. Angular speed from the constraint — at the drum radius: $\omega = \dfrac{v}{r} = \dfrac{2.0}{0.050} = \boxed{40\ \mathrm{rad/s}}$ . Timing note: with $\beta$ computed, everything else is equation (4) plus kinematics — 3 minutes flat. The mark-loser is dividing $v$ by an outer radius the question mentions as a decoy; the constraint lives at the drum, where the string is.

4 · Competitive level — JEE-Advanced stretch: which way does the spool roll? (~4 minutes)

A spool (mass $1.0\ \mathrm{kg}$ , outer radius $R = 0.20\ \mathrm{m}$ , drum radius $r = 0.10\ \mathrm{m}$ , $I_{cm} = 0.020\ \mathrm{kg\,m^2}$ ) rests on a rough table. The thread leaves the underside of the drum and is pulled horizontally with $F = 10\ \mathrm{N}$ . The spool rolls without slipping. (a) Which way does it move? (b) Find its acceleration. (c) At what pull angle would it skid in place?

Step 1. Direction by torque about the contact point $P$ . The horizontal line of action passes at height $r = 0.10$ m, above $P$ but below the centre: lever arm $R - r = 0.10$ m, and the sense rolls the spool toward the pull. (Friction and $N$ act at $P$ : no vote.)

Step 2. Rotation about $P$ (legitimate for rolling): $I_P = I_{cm} + mR^2 = 0.020 + (1.0)(0.20)^2 = 0.060\ \mathrm{kg\,m^2}$ .

Step 3. $\tau_P = F(R - r) = 10(0.10) = 1.0\ \mathrm{N\,m}$ , so $\alpha = \tau_P/I_P = 16.7\ \mathrm{rad/s^2}$ and $a = \alpha R = \boxed{3.3\ \mathrm{m/s^2}}$ toward the hand.

Step 4. Skid-in-place condition, equation (6): $\cos\theta_c = \dfrac{r}{R} = 0.50 \Rightarrow \theta_c = \boxed{60^\circ}$ . Shallower pulls drag the spool in; steeper pulls roll it away, unwinding thread. Timing note: the entire problem is won by choosing the contact point as pivot — friction (unknown in size and direction) never enters. Attempt it with centre-of-mass equations and you will spend four minutes solving simultaneous equations for a friction force the question never asked about. Budget: 90 seconds for the insight, the rest is arithmetic.

Exam traps

Torque radius versus inertia radius. The moment of inertia uses the body's mass geometry (outer radius $R$ for a disc); the torque and the constraint use the string radius $r$ . For a yo-yo, $I = \tfrac12 mR^2$ but $\tau = Tr$ and $a = r\alpha$ . Papers deliberately quote both radii — using $R$ in the constraint is the classic slip.

Gravity torque about the centre. Weight acts at the centre of mass, so it has no torque about the centre — the string alone spins the body. Writing $mgR$ as a torque in the rotation ledger doubles-counts gravity and wrecks the answer.

Energy without the spin share. $mgh = \tfrac12 mv^2$ is wrong here — a third or more of gravity's work is in $\tfrac12 I\omega^2$ . The one-line fix: divide $2gh$ by $(1 + \beta)$ before taking the square root.

Assuming the spool rolls away from the pull. Direction is not intuition, it is the line of action versus the contact point. Below the critical angle the spool comes toward you. State the torque-about- $P$ argument in one line and the direction mark is safe.

Friction's direction as a guess. On the table, friction on a rolling spool can point either way depending on geometry. Never pre-decide it: either work about the contact point (friction disappears) or leave friction as an unknown symbol and let its sign emerge.

$T$ sanity band. Always $0 < T < mg$ for a descending yo-yo. $T = mg$ only in the steady "sleeping" state or hanging at rest; $T > mg$ happens only during the jerk at the end of the string or while climbing with deceleration.

Which $g$ ? CBSE/ISC boards expect $9.8\ \mathrm{m/s^2}$ ; JEE-Main/Advanced and NEET papers almost always declare $g = 10$ ; AP Physics 1 permits $10$ for estimates. Use what the paper states, and never mix values mid-solution.

Origins · 1 minJames Clerk Maxwell and how this was discovered

James Clerk Maxwell

1831–1879 · Edinburgh & Cambridge

Long before his electromagnetic triumphs, James Clerk Maxwell was obsessed with spinning bodies. As a student he was famous for devices of "diabolical" cleverness — he built and demonstrated a "dynamical top" and wrote the 1857 essay On a Dynamical Top that turned rigid-body rotation into a subject an undergraduate could hold in the hand. Maxwell's habit of testing grand theory on toys — tops, discs, unwinding threads — became a style of physics teaching that survives in every yo-yo problem: a cheap object, two of Newton's laws, and no place for hand-waving to hide. He would go on to apply the same taste for mechanical honesty to the rings of Saturn (proving they must be swarms of particles, effectively countless tiny spinning bodies) and then to the electromagnetic field itself.

History

440

BC — Greek vase paintings show a boy playing with a yo-yo; a terracotta disc from the period survives in Athens — the physics demo is at least 2,400 years old.

1784

— George Atwood publishes his gravity-diluting machine; the yo-yo achieves the same dilution with one body instead of two, though nobody yet does the algebra.

1857

— Maxwell's essay on the dynamical top makes the rotation of rigid bodies a hands-on subject; tension-driven spin-up becomes standard lecture fare.

1928

— Pedro Flores begins manufacturing yo-yos in California; Donald Duncan buys the company and the toy — and its physics problem — go global.

1959

— the first "yo-yo de-spin" mechanism flies on a US rocket: weighted cords unwind from the spinning stage and carry away its angular momentum, the falling-spool ledger run in reverse.

Going further

Texts. NCERT Physics Class 11, Ch. 7 (Systems of Particles and Rotational Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 10 · Halliday-Resnick-Walker Ch. 11 (the yo-yo section) · Morin, Introduction to Classical Mechanics, Ch. 8 for the spool game at full depth.

Provenance. All results are standard rotation-chapter canon. A verifier should recompute: equation (4) from the three ledgers; disc-at-rim $a = 2g/3 = 6.53\ \mathrm{m/s^2}$ , $T = mg/3 = 6.53$ N for $m = 2.0$ kg; the real-yo-yo ratio $\beta = R^2/2r^2 = 18$ giving $a = 0.52\ \mathrm{m/s^2}$ , $T = 1.39$ N, $t = 1.8$ s over $0.80$ m; example 3's $\beta = 4$ , $a = 2.0$ , $T = 16$ N, $\omega = 40$ rad/s; example 4's $I_P = 0.060$ , $a = 3.3\ \mathrm{m/s^2}$ , $\theta_c = 60^\circ$ .

Prerequisites. Newton's Second Law for Rotation · Rolling With & Without Slipping · Moments of Inertia of Standard Bodies. Leads to. The 1 + k²/R² Toolkit (technique) · Rolling Motion (UG deep dive).

Why doesn't the yo-yo just fall at g? Gravity is the same as for a stone.

Because gravity's work must now buy two things: downward speed and spin. The string enforces the trade — it back-pulls with tension $T$ , slowing the fall, and that same tension's torque spins the body up. The ratio $\beta = I/mr^2$ sets the exchange rate: a stone has $\beta = 0$ and falls at $g$ ; a rim-wound disc has $\beta = \tfrac12$ and falls at $\tfrac23 g$ ; a fat yo-yo on a thin axle has $\beta \approx 18$ and drifts down at $g/19$ .

Which radius do I use where? The question gives me two.

One rule covers everything: the string radius $r$ (where the thread actually leaves the body) appears in the torque $Tr$ and the constraint $a = r\alpha$ ; the body radii appear only inside the moment of inertia $I$ . A yo-yo with disc radius $R$ and axle radius $r$ : $I = \tfrac12 mR^2$ , torque $= Tr$ , constraint $a = r\alpha$ . If only one radius exists (string at the rim), they coincide and the danger evaporates.

Why does the tension do no work in the energy method? The string clearly pulls.

The tension acts at the point where the string leaves the drum, and at that instant that material point is at rest — the string above it hangs stationary, and no-slip unwinding means the contact point matches the string's zero speed. Force times zero velocity is zero power. It is exactly the argument for static friction doing no work in ideal rolling. The tension redirects energy (from fall into spin) without adding or removing any.

How can the spool roll toward the hand that pulls it? That feels like it violates something.

Nothing is violated — the net force on the spool points toward your hand (your pull beats friction), so accelerating toward you is exactly what Newton demands. Intuition goes wrong by imagining the thread must "unwind." Take torques about the contact point: a shallow pull's line of action passes on the near side of the contact point, so the only available rotation is the one that carries the spool toward you — winding thread in. Steepen the pull past $\cos\theta_c = r/R$ and the geometry, and the verdict, flips.

What happens at the bottom of the string — why does a yo-yo bounce or "sleep"?

When the string is fully unwound, the yo-yo has its maximum spin. If the string's end is tied to the axle, the wind suddenly reverses: the spin now re-winds string, converting rotational KE back into height — the yo-yo climbs, which is the "bounce." A trick yo-yo instead has a loose loop around the axle: the body keeps spinning freely at the bottom ("sleeping") with the string slack, until a flick re-engages friction between loop and axle and the spin winds it home. During the reversal jerk the tension briefly spikes above $mg$ — the one moment this article's $T < mg$ band is allowed to break.

Is the falling yo-yo the same problem as the pulley with mass?

Same physics family, different casting. In the rotational Atwood, the string translates blocks and spins a separate pulley; each body gets one ledger. In the yo-yo, a single body owns both ledgers, and the "pulley" falls with the mass. The unifying idea is identical — tension couples $ma$ ledgers to $I\alpha$ ledgers through a no-slip constraint — which is why both problems produce answers with the signature denominator $1 + I/mr^2$ .

Does the mass of the yo-yo matter for its acceleration?

Not if the shape is fixed. In $a = g/(1 + I/mr^2)$ , the moment of inertia of any body scales as $I = \beta_0 m \times (\text{size})^2$ , so mass cancels in the ratio just as it does for blocks on inclines and rolling races. What matters is the geometry: how the mass is spread ( $\beta_0$ ) and how thin the string radius is relative to the body. A lead yo-yo and a wooden one of identical proportions fall together.

In the spool problem, why are we allowed to ignore friction by choosing the contact point?

Choosing the contact point as pivot does not ignore friction — it makes friction's torque zero, because a force acting at the pivot has no lever arm. Friction still acts on the spool and still shapes the linear motion; but the direction-of-rolling question is a rotation question, and about the contact point only the pull votes. For rolling without slipping the contact point is the instantaneous axis, where $\tau_P = I_P\alpha$ is valid with $I_P = I_{cm} + mR^2$ — a standard, examiner-approved licence.