ARCHETYPE: Pulley With Mass

The Atwood machine you mastered in the Laws of Motion chapter ran on a polite fiction: a massless pulley, spinning up for free, transmitting one tension unchanged around its rim. Now the fiction is cancelled. Give the pulley mass and it becomes a third dynamical body — a disc that must be torqued into spinning — and the machine's most sacred rule dies on the spot: the two sides of the string now carry different tensions. They must: their difference, times the radius, is the only torque available to spin the pulley up. This archetype is the standard first encounter between Newton's laws and $\tau = I\alpha$ in one coupled system, and every competitive board mines it: NEET poses the single mass unwinding from a cylinder (answer: $a = g/(1 + I/mR^2)$ , a formula worth owning outright); JEE-Main and Advanced run the full massive-pulley Atwood and its table variants, with friction and multi-phase twists; AP Physics 1 and IB-HL both name pulleys explicitly in their rotation units. The master pattern is always the same four lines — one $F = ma$ per hanging body, one $\tau = I\alpha$ for the pulley, and the no-slip bridge $a = R\alpha$ welding them — and it produces one unforgettable simplification: a uniform disc pulley behaves exactly like an extra half its mass of ordinary inertia hidden in the string. Learn the pattern once, with its energy-method twin and its checkable limits (massless pulley recovers the old machine; infinite pulley freezes everything), and an entire family of exam problems — Atwoods, unwinding spools' cousins, table systems, flywheel hoists — becomes fill-in-the-blanks.

Key takeaways

A pulley with mass is a third body: it needs net torque to spin up, so the string tensions on its two sides must differ — $(T_1 - T_2)R = I\alpha$ .
Master pattern, always: one $F = ma$ per hanging mass + $\tau = I\alpha$ for the pulley + the no-slip bridge $a = R\alpha$ ; add the equations and the tensions telescope away.
Massive-pulley Atwood: $a = \dfrac{(m_2 - m_1)\,g}{m_1 + m_2 + I/R^2}$ — the pulley contributes $I/R^2$ of "effective mass" to the denominator, never to the driving force.
Uniform disc pulley: $I/R^2 = \tfrac12 M$ — the pulley acts like an extra $M/2$ riding in the string; the radius cancels entirely.
Single mass unwinding from a wheel: $a = \dfrac{g}{1 + I/mR^2}$ — NEET's favourite; a hollow cylinder of mass equal to the hanging mass gives exactly $a = g/2$ .
Tension order: light side $T > mg$ (it accelerates up), heavy side $T < mg$ (it accelerates down), and heavy-side $T$ > light-side $T$ — the difference spins the pulley.
Key idea: mass on the pulley = one more $\tau = I\alpha$ equation, one broken tension, and $I/R^2$ of stowaway inertia in the denominator — everything else is the old Atwood.

The core insight

When the pulley has moment of inertia $I$ , the cord's two sides pull with different tensions whose difference supplies $(T_1 - T_2)R = I\alpha$ ; with the no-slip bridge $a = R\alpha$ , every hanging-mass system solves by adding one force equation per body to the pulley's torque equation — yielding the massless answers with an extra $I/R^2$ of effective mass in the denominator.

Where this lives today

Every traction lift you ride is this archetype with a motor at the axle: car and counterweight hang over a driven sheave whose own rotational inertia — plus the motor armature's, geared up — is a genuine term in the drive calculations, exactly the $I/R^2$ stowaway. Winches, cranes, cable cars and mine hoists all size their motors against effective inertia (payload plus drum plus gearing), and flywheel-assisted hoists deliberately add drum inertia to smooth load surges. In the lab, the photogate "smart pulley" turns this module's equations into the standard undergraduate measurement of $I$ : hang a known mass, time the unwinding, and solve equation (1) backwards for the wheel's moment of inertia — Atwood's own 1784 correction, now the experiment itself. Even cassette tapes and film reels lived here: the changing spool radius and inertia as tape winds is the unwinding-mass problem with $R$ and $I$ slowly drifting, which is why precision decks drove the capstan, not the reels.

The setup

Geometry first, because in an archetype the geometry is the physics. A cord passes over a pulley of mass $M$ , radius $R$ — a uniform disc unless stated — carrying $m_1$ (lighter) and $m_2$ (heavier) on its two sides. The cord is light and inextensible and does not slip on the rim: released from rest, $m_2$ descends, $m_1$ rises, the pulley turns, and all three accelerate together — the two masses at $a$ , the rim at the matching $a = R\alpha$ .

Figure

The Atwood machine grows a real pulley — three bodies, two tensions, one bridge

Read the new physics off the figure. In the old machine, "massless pulley" bought the rule one tension throughout: a body with $I = 0$ needs zero net torque for any $\alpha$ , so the tensions matched. Here the disc has genuine $I = \tfrac12 MR^2$ , and spinning it up requires net torque — which only a tension difference can supply. So the heavy side pulls with $T_2$ , the light side with $T_1$ , and $T_2 > T_1$ by exactly the pulley's fee. Everything else survives from the old machine: the cord's inextensibility still gives one acceleration magnitude, and the no-slip condition converts it to the pulley's $\alpha = a/R$ . One more structural reading before the algebra: where can the pulley's inertia appear in the answer? Not in the driving force — the pulley hangs at the axle, its weight supported, contributing no drive. Only in the resistance: it is one more thing that must be accelerated. Expect the massless formula with an extra inertia term in the denominator — and that is exactly what the master derivation delivers.

The master solution pattern

Derivation AThe massive-pulley Atwood, solved once and for all

We derive the acceleration and both tensions for the Atwood machine with a pulley of moment of inertia $I$ and radius $R$ — the master pattern of the whole archetype. Strategy: one Newton equation per hanging mass (each along its own direction of motion, exactly as in the massless machine), one torque equation for the pulley, and the no-slip bridge; then add everything so the internal forces telescope. Watch two things: how the bridge converts the pulley's torque equation into "effective mass," and how the radius quietly cancels for a uniform disc.

Step 1. The light mass $m_1$ , up positive: $T_1 - m_1 g = m_1 a$ .

Step 2. The heavy mass $m_2$ , down positive: $m_2 g - T_2 = m_2 a$ .

Step 3. The pulley. The two tensions act tangentially at radius $R$ with opposite turning senses; taking the heavy side's sense as positive: $(T_2 - T_1)R = I\alpha$ . The axle's reaction and the pulley's weight act through the axis — zero lever arm, out of the ledger.

Step 4. The bridge. No slipping: $\alpha = a/R$ . Substitute into Step 3 and divide by $R$ :

T_2 - T_1 = \frac{I}{R^2}\,a

Read it before moving on: the tension difference is (stowaway mass) × (acceleration), where the stowaway is $I/R^2$ . For a uniform disc, $I/R^2 = \tfrac12 M$ — the radius is gone, and the pulley has become an extra half-its-mass riding in the cord.

Step 5. Add Steps 1, 2 and 4 (as +Step 2 − Step 1 arranged so tensions cancel — writing Step 1 as $-T_1 + m_1 g = -m_1 a$ and summing all three left and right sides):

(m_2 - m_1)g = \left(m_1 + m_2 + \frac{I}{R^2}\right)a \qquad \Longrightarrow \qquad a = \frac{(m_2 - m_1)\,g}{m_1 + m_2 + I/R^2}

The old answer with one new tenant in the denominator. Tensions by substitution: $T_1 = m_1(g + a)$ , $T_2 = m_2(g - a)$ — same expressions as ever, but now $T_1 \neq T_2$ .

Step 6. Audit the limits. $I \to 0$ : the massless machine returns, $a = (m_2-m_1)g/(m_1+m_2)$ and $T_1 = T_2$ ✓. $I \to \infty$ : $a \to 0$ , $T_1 \to m_1 g$ , $T_2 \to m_2 g$ — an immovable pulley, everything just hangs ✓. And the tension ordering: $T_2 - T_1 = (I/R^2)a > 0$ — the heavy side pulls harder, as the spin-up demands. Verified: numbers for $m_2 = 4.0$ , $m_1 = 2.0\ \mathrm{kg}$ over a disc pulley $M = 2.0\ \mathrm{kg}$ , $R = 0.25\ \mathrm{m}$ ( $g = 9.8$ ): $a = 2 \times 9.8/(4 + 2 + 1) = 2.8\ \mathrm{m/s^2}$ ; $T_1 = 2(12.6) = 25.2\ \mathrm{N}$ , $T_2 = 4(7.0) = 28.0\ \mathrm{N}$ ; pulley check $(28.0 - 25.2)(0.25) = 0.70\ \mathrm{N\,m}$ against $I\alpha = 0.0625 \times 11.2 = 0.70$ ✓. Matches the standard treatment (HC Verma Ch. 10; Halliday Ch. 10).

Go deeper · Derivation B · 1 minThe energy route — and when it beats the force route

Derivation BThe energy route — and when it beats the force route

We re-derive the machine's speed by energy conservation, obtaining the velocity after the heavy side falls a height $h$ — the method that wins whenever a question asks "how fast after falling..." rather than "how long." Strategy: one energy ledger for the whole machine, with the pulley's spin pocket included; the key point to watch is why conservation is legal — and which single term students forget.

Step 1. The budget. As $m_2$ falls $h$ , $m_1$ rises $h$ (inextensible cord): net potential energy released $= (m_2 - m_1)gh$ .

Step 2. The accounts it fills, at common speed $v$ (cord speed = rim speed, no slip): the two masses' translation, and the pulley's spin with $\omega = v/R$ :

K = \tfrac12 m_1 v^2 + \tfrac12 m_2 v^2 + \tfrac12 I\left(\frac{v}{R}\right)^2 = \tfrac12\left(m_1 + m_2 + \frac{I}{R^2}\right)v^2

The same stowaway $I/R^2$ , now in the energy books — one derivation's effective mass is the other's third pocket.

Step 3. Why conservation is legal: the cord does not slip on the rim, so the static grip there does no net work (it transfers energy to the pulley, exactly the $\tfrac12 I\omega^2$ term, without dissipating any); the axle is frictionless; tensions are internal to the system. Setting release from rest:

(m_2 - m_1)gh = \tfrac12\left(m_1 + m_2 + \frac{I}{R^2}\right)v^2 \qquad \Longrightarrow \qquad v = \sqrt{\frac{2(m_2 - m_1)gh}{m_1 + m_2 + I/R^2}}

Step 4. Cross-check against the dynamics: constant acceleration from rest gives $v^2 = 2ah$ , and inserting Derivation A's $a$ reproduces Step 3 exactly ✓. Triage rule for the exam: time asked → force route (energy is time-blind); speed after a distance → energy route (one line); tensions → force route only (energy methods are blind to internal forces). Verified: for the anchor machine ( $a = 2.8\ \mathrm{m/s^2}$ ) after $h = 2.0$ m: dynamics $v = \sqrt{2 \times 2.8 \times 2} = 3.35\ \mathrm{m/s}$ ; energy $\sqrt{2 \times 2 \times 9.8 \times 2/7} = \sqrt{11.2} = 3.35$ ✓.

Figure

The variant family — one pattern, three exam costumes

The variants examiners actually pose

The single unwinding mass (NEET and IB-HL's favourite). One mass $m$ hangs from a cord wound on a wheel of moment of inertia $I$ , radius $R$ (a flywheel, a cylinder on an axle). Same pattern minus one body: $mg - T = ma$ and $TR = I\alpha$ with $a = R\alpha$ give

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

where: the memorable specials — hollow cylinder ( $I = MR^2$ ) with $M = m$ : $a = g/2$ exactly; uniform disc with $M = m$ : $a = 2g/3$ . This is also the "pull versus hang" preview from the dynamics module, now solved in general.

The table variant (AP and JEE staple). Block $m_t$ on a table (friction coefficient $\mu$ if stated), cord over a massive pulley at the edge, hanging mass $m_h$ : the driving force is $m_h g$ minus the friction $\mu m_t g$ , and the resistance is everything that moves:

a = \frac{m_h g - \mu\,m_t g}{m_t + m_h + I/R^2} \qquad (2)

where: the two tensions (table side and hanging side) again differ by $(I/R^2)a$ , and each follows from its own body's $F = ma$ after $a$ is known.

The axle's load (the one-mark twist). The axle supports the pulley's weight and both cord pulls: $F_{\text{axle}} = Mg + T_1 + T_2$ — no longer the massless machine's clean $2T$ , and less than $Mg + (m_1 + m_2)g$ while the system accelerates.

Limiting cases

case	result	meaning
$I \to 0$	massless Atwood: $T_1 = T_2$ , old $a$	the idealization recovered as a limit
$I \to \infty$	$a \to 0$ , $T_i \to m_i g$	immovable pulley — statics returns
disc pulley	denominator gains $\tfrac12 M$ ; $R$ cancels	the stowaway half-mass
hoop pulley	gains full $M$	rim-heavy pulleys tax double
$m_1 = m_2$ , any $I$	$a = 0$ , $T_1 = T_2 = mg$	balanced machine: no spin-up, no fee
single mass, $I = mR^2$ (hollow, $M = m$ )	$a = g/2$ , $T = mg/2$	the NEET special
single mass, $m \gg I/R^2$	$a \to g$ , $T \to 0$	free fall; the wheel spins helplessly along

Common trap"One string, one tension — and the pulley's mass just adds to the total like everything else"

Both reflexes are imported from the massless machine, and both are precisely what this archetype exists to break. First: with a massive pulley, writing $T_1 = T_2$ makes the spin-up torque $(T_1 - T_2)R$ vanish — your equations then assert the disc spins up for free, and the system is unsolvable or wrong. The stated pulley mass is the examiner's declaration that the tensions differ; conversely, "light pulley" is the licence for one $T$ . Second: the pulley's inertia enters the denominator as $I/R^2$ — for a disc, half its mass, not all of it — and never enters the driving force (the axle carries its weight; gravity on the pulley drives nothing). Options built on $m_1 + m_2 + M$ in the denominator, or on the pulley's weight joining the drive, are pricing exactly these two slips. The three-second audit: set $I = 0$ in your final expression — if the massless Atwood does not reappear, the algebra has already confessed.

Worked examples

1 · Board level — the unwinding cylinder (NEET/IB-HL pattern, 3 marks)

A mass of $3.0\ \mathrm{kg}$ hangs from a light cord wound around a hollow cylinder (mass $3.0\ \mathrm{kg}$ , radius $0.20\ \mathrm{m}$ ) that can rotate freely about its fixed horizontal axis. The cord does not slip. Find (a) the acceleration of the falling mass, (b) the cord's tension. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. The pattern, two bodies: falling mass, $mg - T = ma$ ; cylinder, $TR = I\alpha$ with $I = MR^2$ (hollow) and the bridge $\alpha = a/R$ , so $T = (I/R^2)a = Ma$ .

Step 2. (a) Substitute: $mg - Ma = ma$ with $M = m$ : $a = \dfrac{g}{2} = \boxed{4.9\ \mathrm{m/s^2}}$ — the hollow-cylinder special: equal masses split gravity fifty-fifty between falling and spinning.

Step 3. (b) $T = Ma = 3.0 \times 4.9 = \boxed{14.7\ \mathrm{N}}$ — exactly half the weight, and comfortingly less than $mg = 29.4\ \mathrm{N}$ : the mass genuinely accelerates down.

Step 4. The formula worth owning: $a = g/(1 + I/mR^2)$ — every unwinding-mass MCQ is this line with a different $I$ . (Disc instead of hollow cylinder, same masses: $a = g/1.5 = 6.5\ \mathrm{m/s^2}$ .)

2 · Board level — the anchor machine (AP/JEE-Main pattern)

Masses of $4.0$ and $2.0\ \mathrm{kg}$ hang from a cord over a uniform disc pulley of mass $2.0\ \mathrm{kg}$ and radius $0.25\ \mathrm{m}$ ; the cord does not slip. Find the acceleration and both tensions, and verify the pulley's torque equation. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Effective machine: driving force $(4.0 - 2.0)g = 19.6\ \mathrm{N}$ ; total inertia $4.0 + 2.0 + \tfrac12(2.0) = 7.0\ \mathrm{kg}$ (the disc's stowaway half-mass — note $R$ never enters).

Step 2. $a = 19.6/7.0 = \boxed{2.8\ \mathrm{m/s^2}}$ .

Step 3. Tensions from each body: light side $T_1 = 2.0(9.8 + 2.8) = \boxed{25.2\ \mathrm{N}}$ ; heavy side $T_2 = 4.0(9.8 - 2.8) = \boxed{28.0\ \mathrm{N}}$ . Order check: $28.0 > 25.2$ — heavy side harder, as the spin-up requires; both between the two weights' demands.

Step 4. Pulley audit: $(T_2 - T_1)R = 2.8 \times 0.25 = 0.70\ \mathrm{N\,m}$ ; $I\alpha = \left(\tfrac12 \times 2.0 \times 0.0625\right)\times(2.8/0.25) = 0.0625 \times 11.2 = 0.70\ \mathrm{N\,m}$ ✓. The tension difference $2.8\ \mathrm{N}$ is $\tfrac12 Ma = 1.0 \times 2.8$ — the stowaway's bill, itemized.

3 · Competitive level — table, friction, massive pulley (JEE-Main pattern, ~3 minutes)

A $5.0\ \mathrm{kg}$ block on a table ( $\mu_k = 0.20$ ) is joined by a cord over a disc pulley ( $M = 2.0\ \mathrm{kg}$ ) at the table's edge to a hanging $3.0\ \mathrm{kg}$ mass. The cord does not slip; take $g = 10\ \mathrm{m/s^2}$ . Find the acceleration and both tensions.

Step 1. Drive and drag: hanging weight $30\ \mathrm{N}$ drives; table friction $\mu m_t g = 0.20 \times 50 = 10\ \mathrm{N}$ drags. Net drive $20\ \mathrm{N}$ .

Step 2. Total inertia: $5.0 + 3.0 + \tfrac12(2.0) = 9.0\ \mathrm{kg}$ . So $a = 20/9.0 = \boxed{2.2\ \mathrm{m/s^2}}$ .

Step 3. Hanging side: $T_h = m_h(g - a) = 3.0(10 - 2.22) = \boxed{23.3\ \mathrm{N}}$ . Table side: $T_t = m_t a + f = 5.0(2.22) + 10 = \boxed{21.1\ \mathrm{N}}$ .

Step 4. Audit: $T_h - T_t = 2.2\ \mathrm{N} = \tfrac12 Ma = 1.0 \times 2.22$ ✓ — the difference is exactly the pulley's fee, friction notwithstanding. Timing note: ~2.5 minutes with the effective-machine habit (drive over total inertia), then one substitution per tension. The planted errors: forgetting friction in the drive (gives 3.33), forgetting the stowaway (gives 2.5), and the reflex $T_h = T_t$ (kills part two outright).

4 · Competitive level — the energy route and the massless comparison (JEE-Advanced stretch, ~4 minutes)

In Example 2's machine, the heavy mass is released from rest $2.0\ \mathrm{m}$ above the floor. Find (a) its landing speed, by energy methods; (b) the landing speed had the pulley been massless; (c) what the pulley is doing at the instant of landing.

Step 1. (a) Energy ledger over the 2.0 m drop: $(m_2 - m_1)gh = \tfrac12(m_1 + m_2 + \tfrac12 M)v^2$ , so $39.2 = \tfrac12(7.0)v^2$ , giving $v = \sqrt{11.2} = \boxed{3.35\ \mathrm{m/s}}$ . (Dynamics check: $v = \sqrt{2ah} = \sqrt{2 \times 2.8 \times 2.0} = 3.35$ ✓.)

Step 2. (b) Massless pulley: total inertia $6.0\ \mathrm{kg}$ , $a = 19.6/6.0 = 3.27\ \mathrm{m/s^2}$ , $v = \sqrt{2 \times 3.27 \times 2.0} = \boxed{3.61\ \mathrm{m/s}}$ . The real pulley cost $0.26\ \mathrm{m/s}$ — its spin pocket holds $\tfrac12 \times 1.0 \times (3.35)^2 = 5.6\ \mathrm{J}$ of the released 39.2 J at landing.

Step 3. (c) At landing, the cord on the heavy side goes slack. The pulley (and the risen light mass) continue: the light mass moves up at $3.35\ \mathrm{m/s}$ , now decelerating at $g$ as a free projectile if the cord tension dies — but the still-spinning pulley keeps the cord taut on the light side, so pulley + light mass continue as a new two-body system: $T_1' - m_1g = -m_1 a'$ and $T_1'R = -I\alpha'$ ... solving, they decelerate together at $a' = \dfrac{m_1 g}{m_1 + I/R^2} = \dfrac{19.6}{3.0} = 6.5\ \mathrm{m/s^2}$ — slower than $g$ , because the spinning disc now feeds energy back through the cord. Timing note: (a)–(b) are ~2 minutes with the two routes; (c) is the JEE-Advanced discriminator — recognizing that landing re-partitions the system and the pulley's stored spin becomes a player. Multi-phase machine problems always re-ask: who is still connected, and what does each phase conserve?

Exam traps

Stated pulley mass = broken tension. Write $T_1$ and $T_2$ from the first line. The single-tension reflex is this archetype's defining wound.

The stowaway is $I/R^2$ , not $M$ . Disc: $M/2$ . Hoop: $M$ . It joins the denominator only — the pulley's weight never drives (the axle carries it).

Radius almost always cancels. For a uniform disc, $R$ vanishes from $a$ and the tensions; a question that supplies $R$ is often testing whether you know you don't need it (you do need it for $\alpha = a/R$ and torque audits).

Tension order tells the turning. The side the pulley accelerates toward (the heavy side's sense) must pull harder: $T_{\text{heavy}} > T_{\text{light}}$ . An answer with the order reversed fails the audit $(T_2 - T_1) = (I/R^2)a$ .

Energy ledgers include the spin pocket. $\tfrac12 I\omega^2$ with $\omega = v/R$ — dropping it inflates landing speeds by exactly the massless-machine answer.

No slip is the licence for $a = R\alpha$ and for work-free grip. "The cord slips on the pulley" changes the problem species entirely (kinetic friction at the rim, tensions related by slip physics — beyond this archetype).

Axle force: $Mg + T_1 + T_2$ . Not $2T$ (no single $T$ exists), and less than the total hanging weight plus $Mg$ while accelerating.

Which $g$ : JEE/NEET papers almost always declare $g = 10$ ; AP and IB use $9.8$ . The stated value, always.

Origins · 1 minGeorge Atwood and how this was discovered

George Atwood

1745–1807 · Trinity College, Cambridge

Look at the full title of Atwood's 1784 book: A Treatise on the Rectilinear Motion <strong>and Rotation</strong> of Bodies. The rotation was never an afterthought — Atwood knew perfectly well that his machine's wheel had inertia, and his treatise both computes its effect and shows how to correct measurements for it, half a century before anyone wrote $\tau = I\alpha$ in modern dress. His practical fix was elegant: he mounted his pulley on friction wheels — each axle resting on the rims of four larger wheels — to make the bearing losses negligible, leaving the wheel's inertia as a small, calculable correction to the gravity-diluting formula. So the "massive pulley problem" of this archetype is not a modern complication bolted onto Atwood's toy: it is the fine print of the original instrument, written by the original author. Instrument makers sold Atwood machines across Europe with the friction wheels included; the Copley Medal came in 1796. The lecture demonstrator's instinct — quantify your apparatus's own physics before trusting it about nature's — is the whole discipline of experimental error analysis, in period costume.

History

1784

— Atwood's Treatise on the Rectilinear Motion and Rotation of Bodies publishes the machine — with the wheel's inertia analysed and corrected for.

1790

s — instrument makers ship Atwood machines with friction-wheel bearings across Europe: the first mass-produced physics apparatus.

1880

s — traction elevators hang car and counterweight over a driven sheave: the massive-pulley Atwood becomes urban infrastructure.

1900

s — machine dynamics matures: every winch, hoist and cable drum sized by effective-inertia bookkeeping.

2000

s — "smart pulleys" with photogates make the massive-pulley Atwood the standard first lab of rotational dynamics.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 10 (massive-pulley systems) · Halliday–Resnick–Walker Ch. 10 · NCERT Physics Class 11, Ch. 7. The parent archetype: the Atwood Machine & Variants module (massless case, sign discipline, lift variants). Full rigor: Torque & Fixed-Axis Dynamics (UG deep dive).

Provenance. All results are standard. A verifier should recompute: the master formula's limits ( $I \to 0$ , $I \to \infty$ ); the anchor machine ( $a = 19.6/7 = 2.8\ \mathrm{m/s^2}$ , $T_1 = 25.2$ , $T_2 = 28.0\ \mathrm{N}$ , torque audit $0.70 = 0.70\ \mathrm{N\,m}$ ); Example 1's $g/2$ and $T = 14.7\ \mathrm{N}$ ; Example 3 ( $a = 20/9 = 2.22$ , $T_h = 23.3$ , $T_t = 21.1$ , difference $= \tfrac12 Ma$ ✓); Example 4 ( $v = \sqrt{11.2} = 3.35$ vs massless $3.61\ \mathrm{m/s}$ ; phase-two deceleration $19.6/3.0 = 6.5\ \mathrm{m/s^2}$ ); the axle load $72.8 < 78.4\ \mathrm{N}$ with shortfall $(m_2 - m_1)a = 5.6\ \mathrm{N}$ .

Prerequisites. The Atwood Machine & Variants · Newton's Second Law for Rotation · Tension & Ideal Strings. Leads to. Yo-Yos & Unwinding Spools (archetype) · The 1 + k²/R² Toolkit (technique) · Torque & Fixed-Axis Dynamics (UG deep dive).

Where does the pulley's radius go? It appears in the setup but not in the answers.

It cancels, in a way worth watching once. The torque equation $(T_2 - T_1)R = I\alpha$ , divided by $R$ and fed the bridge $\alpha = a/R$ , becomes $T_2 - T_1 = (I/R^2)\,a$ — and for a uniform disc $I = \tfrac12 MR^2$ , so $I/R^2 = \tfrac12 M$ : pure mass, no radius. Physically: a bigger pulley has more inertia ( $\propto R^2$ ) but also more leverage for the same tension difference and less spin for the same cord speed — and the three effects cancel exactly. The radius does matter for the angular quantities themselves ( $\alpha$ , $\omega$ , torque audits) and for non-uniform pulleys specified by $I$ directly, which is why questions still supply it.

Why is the heavy side's tension larger? I expected the rising side to be pulled harder.

Follow the pulley's needs. It must angularly accelerate in the heavy side's sense, so the net torque must point that way: the heavy-side cord must win the tug on the rim, $T_2 > T_1$ . Now check each mass agrees. The heavy mass accelerates down, so its tension must lose to its weight: $T_2 < m_2 g$ — large, but not too large. The light mass accelerates up, so its tension beats its weight: $T_1 > m_1 g$ — but by less than $T_2$ 's margin below $m_2g$ ... the consistent solution threads both needles, and the algebra confirms $m_1 g < T_1 < T_2 < m_2 g$ . In the anchor machine: $19.6 < 25.2 < 28.0 < 39.2$ (newtons) — a chain worth checking in any answer, since a single inverted inequality flags an algebra slip immediately.

Why is the acceleration smaller than with a massless pulley — the pulley isn't hanging on either side?

Because acceleration is drive over everything that must be accelerated, and the pulley's rim must be accelerated right along with the cord glued to it. The drive is unchanged — still the weight difference $(m_2 - m_1)g$ ; the pulley hangs at its axle and pushes nothing. But the response now includes spinning up a disc, and the no-slip weld makes that spin-up proportional to the cord's own acceleration: the pulley is dynamically indistinguishable from an extra $I/R^2$ of mass riding in the string. Same drive, more inertia, less acceleration — and correspondingly, at any later moment, less speed: the "missing" energy sits in the pulley's $\tfrac12 I\omega^2$ pocket, as Example 4 itemizes.

When should I use the energy method instead of the three-equation force method?

Triage by what is asked. Speed after a given fall or rise: energy, one line — the ledger $(m_2 - m_1)gh = \tfrac12(\text{total effective mass})v^2$ needs no $\alpha$ , no time, no tensions. Time, or acceleration itself: forces — energy methods are time-blind. Any tension: forces only — tensions are internal, and energy ledgers never see them (that blindness is exactly why they are fast). Mixed questions reward the hybrid: get $a$ from the force route's one addition, then use $v^2 = 2ah$ ; or get $v$ from energy and back out $a$ . And in every route, the pulley appears identically — as $I/R^2$ of effective mass (forces) or as the third kinetic pocket (energy). They are the same bookkeeping, as Derivation B's cross-check shows.

What changes if the pulley is a hoop, or some odd wheel specified only by its I?

Nothing structural — only the stowaway's size. The master pattern never used "disc" until the last substitution: the tension difference is always $(I/R^2)a$ , so a hoop-rimmed pulley ( $I = MR^2$ ) stows its full mass in the cord, a disc half, a lightweight spoked wheel with a heavy rim close to full. Exam questions exploit this as a one-word switch: identical numbers, "disc" changed to "ring," and every answer shifts. When the wheel is specified by $I$ directly (a flywheel of stated moment of inertia), skip the shape entirely and carry $I/R^2$ symbolically — the NEET single-mass formula $a = g/(1 + I/mR^2)$ is written that way for exactly this reason.

What force does the support (axle) feel — is it still 2T like the massless machine?

No — that formula died with the single tension. The axle now carries three loads: the pulley's own weight $Mg$ , plus the two cord pulls $T_1$ and $T_2$ (both downward for a hanging Atwood). Total: $F_{\text{axle}} = Mg + T_1 + T_2$ . For the anchor machine: $19.6 + 25.2 + 28.0 = 72.8\ \mathrm{N}$ — less than the naive "everything's weight" total $Mg + (m_1+m_2)g = 78.4\ \mathrm{N}$ , because the system's centre of mass accelerates downward and the support is excused the difference. That shortfall, $(m_1 + m_2)g - (T_1 + T_2) = 5.6\ \mathrm{N} = (m_2 - m_1)a$ , is the same "accelerating machines weigh less" bookkeeping as the massless case — only the itemization changed.

What happens physically if the cord slips on the pulley?

The weld breaks, and with it most of this archetype's machinery. Slipping means the cord and rim move at different speeds: $a \neq R\alpha$ , so the bridge equation is gone, and the grip at the rim becomes kinetic friction — which now does real (negative) work, heating the rim, so the energy method loses its licence too. The system gains a degree of freedom and the problem needs the slip physics of the friction chapter (in machines, belt slip is governed by wrap angle and the capstan effect — a rope wound around a post multiplies grip enormously, which is how sailors hold ships). Exam questions stay on the safe side of the boundary and tell you: "the cord does not slip" is the phrase that licenses everything this module did. Its absence, or "the cord slips freely over a smooth peg," is a different problem — a smooth peg spins not at all and transmits one tension, ironically restoring the massless machine's rule by removing the grip instead of the mass.