TECHNIQUE: The 1 + k²/R² Toolkit

Here is an open secret about rotation questions in JEE-Main and NEET: a huge fraction of them — the rolling races, the "fraction of kinetic energy in rotation," the "which reaches the bottom first," the ratio-of-speeds MCQs — are all the same question. Behind every one sits a single dimensionless number, $\beta = k^2/R^2 = I/MR^2$ , and a single magic denominator, $1 + \beta$ . Know four values of $\beta$ by heart and carry the $(1+\beta)$ reflex, and problems that cost your neighbours three minutes of energy conservation collapse for you into fifteen seconds of ratio arithmetic. This is a technique module, not a concept module. The physics was built in the rolling-motion and rotational-energy articles; here we forge it into speed craft: five named shortcut moves, the memory table, a drill ladder that runs each move against the clock, and — just as important — the honest map of where the toolkit backfires. Because the second skill of the shortcut artist is recognizing the question that looks like a rolling race but isn't: the frictionless incline, the slipping sphere, the yo-yo whose string leaves a hub rather than the rim. One number, one denominator, four shapes. Load them once, and the rotation section of every mock paper starts paying you back immediately.

Key takeaways

Every rolling body is summarized by one number: $\beta = \dfrac{I}{MR^2} = \dfrac{k^2}{R^2}$ — hoop $1$ , hollow sphere $\tfrac{2}{3}$ , disc/solid cylinder $\tfrac{1}{2}$ , solid sphere $\tfrac{2}{5}$ .
One denominator answers everything: $a = \dfrac{g\sin\theta}{1+\beta}$ , $v = \sqrt{\dfrac{2gh}{1+\beta}}$ , $t = \sqrt{\dfrac{2L(1+\beta)}{g\sin\theta}}$ .
The kinetic-energy split needs no calculation: $KE_{\text{rot}} : KE_{\text{trans}} = \beta : 1$ , so the rotational fraction is $\dfrac{\beta}{1+\beta}$ .
Race order = $\beta$ order: solid sphere beats disc beats hollow sphere beats hoop — always, independent of mass and radius.
Mass and radius are decoys in every ratio MCQ: they cancel from $a$ , $v$ , $t$ and the energy split.
Friction needed to roll: $f = \dfrac{\beta}{1+\beta}\,Mg\sin\theta$ , so $\mu_{\min} = \dfrac{\beta}{1+\beta}\tan\theta$ — the one formula the ratios can't replace.
Key idea: translate the object into $\beta$ , write $(1+\beta)$ , and answer rolling MCQs by ratio — not by re-running energy conservation.

The core insight

Every rolling-without-slipping result — acceleration, bottom speed, descent time, energy split, friction demand — depends on the body only through $\beta = I/MR^2$ , so memorizing four values of $\beta$ and the master denominator $1+\beta$ converts the entire rolling-race genre into fifteen-second ratio arithmetic.

Where this lives today

The $(1+\beta)$ tax is quietly present in the founding experiment of mechanics: Galileo's inclined-plane balls rolled, so the acceleration he measured was $\frac57 g\sin\theta$ , not $g\sin\theta$ — his conclusions about uniform acceleration survived because the tax is a constant factor, but any modern replication that compares his numbers to $g$ must apply this module's correction. Engineers meet $\beta$ as the "equivalent inertia" of drivetrains: a car's effective mass during acceleration includes its spinning wheels and flywheel through exactly $m(1+\beta_{\text{eff}})$ bookkeeping, and flywheel energy storage is the split of Move 4 built into infrastructure — energy parked deliberately in the $\beta$ share. Sports equipment designers tune $\beta$ on purpose: a golf ball's layered core, a bowling ball's asymmetric weight block, and bicycle "lightweight rims matter more than lightweight frames" folklore are all $(1+\beta)$ statements — rim mass is the most expensive mass on the bike because it pays the tax twice.

The toolkit: β values you must read at sight

The whole module rests on translating "solid sphere" or "ring" instantly into a number. The number is $\beta = I/MR^2$ — equivalently $k^2/R^2$ , where $k$ is the radius of gyration. It measures one thing: what fraction of the body's "falling budget" rotation will tax.

body	$\beta = \dfrac{k^2}{R^2}$	$1+\beta$	rotational KE fraction $\dfrac{\beta}{1+\beta}$	incline $a$
solid sphere	$\dfrac{2}{5}$	$\dfrac{7}{5}$	$\dfrac{2}{7} \approx 29\%$	$\dfrac{5}{7}g\sin\theta$
disc / solid cylinder	$\dfrac{1}{2}$	$\dfrac{3}{2}$	$\dfrac{1}{3} \approx 33\%$	$\dfrac{2}{3}g\sin\theta$
hollow sphere (shell)	$\dfrac{2}{3}$	$\dfrac{5}{3}$	$\dfrac{2}{5} = 40\%$	$\dfrac{3}{5}g\sin\theta$
hoop / ring / hollow cylinder	$1$	$2$	$\dfrac{1}{2} = 50\%$	$\dfrac{1}{2}g\sin\theta$
sliding block (no rotation)	$0$	$1$	$0$	$g\sin\theta$

Read the table's logic once and it becomes memory-proof: $\beta$ grows as mass moves outward. A solid sphere hides mass near its axis ( $\beta = 2/5$ ); a hoop parks everything at the rim ( $\beta = 1$ ). The sliding block is the $\beta = 0$ base case — the table's top row in disguise — and every rolling body is a block plus rotational tax.

Figure

The beta ladder — mass distribution sets the race, mass and radius are spectators

The five shortcut moves

Move 1 — Translate the body into β. Never carry "hollow sphere" through a calculation; carry $2/3$ . The moment you read a rolling MCQ, annotate each body with its $\beta$ . Composite or unusual bodies: the question must hand you $I$ ; compute $\beta = I/MR^2$ once and re-enter the toolkit.

Move 2 — The (1+β) reflex. All four standard answers are one denominator away: acceleration $\frac{g\sin\theta}{1+\beta}$ , bottom speed $\sqrt{\frac{2gh}{1+\beta}}$ , descent time $\sqrt{\frac{2L(1+\beta)}{g\sin\theta}}$ , translational share $\frac{1}{1+\beta}$ . No free-body diagram, no energy equation — those were done once, in the derivation below, for all bodies simultaneously.

Move 3 — Race ordering without arithmetic. "Which arrives first/fastest?" is answered by $\beta$ alone: smaller $\beta$ wins. Solid sphere > disc > shell > hoop, forever. If two bodies share $\beta$ (a disc and a solid cylinder), they tie exactly — a favourite trick option.

Move 4 — The energy split by inspection. Rolling KE divides as $KE_{\text{rot}} : KE_{\text{trans}} = \beta : 1$ . Fractions follow instantly: a hoop is 50–50; a disc carries $1/3$ of its KE in spin; a solid sphere $2/7$ . Any question phrased "what fraction of the energy…" is Move 4 — zero writing required.

Move 5 — Ratio surgery. Two bodies compared: the answer is a ratio of $(1+\beta)$ 's, usually under a square root. $\dfrac{v_1}{v_2} = \sqrt{\dfrac{1+\beta_2}{1+\beta_1}}$ , $\dfrac{t_1}{t_2} = \sqrt{\dfrac{1+\beta_1}{1+\beta_2}}$ , $\dfrac{a_1}{a_2} = \dfrac{1+\beta_2}{1+\beta_1}$ . Every mention of mass or radius in the stem is a decoy — cross it out physically. The master results themselves deserve one honest derivation each — run once, remembered forever.

a = \frac{g\sin\theta}{1+\beta} \qquad (1)

where: $\theta$ — the incline angle; $\beta$ — the body's ratio. Read it as a taxed block: a frictionless block would do $g\sin\theta$ ; rolling claims the fraction $\beta/(1+\beta)$ of the driving force to feed the spin, through friction's torque. At $\beta = 0$ the block returns; at $\beta = 1$ (hoop) exactly half the acceleration survives.

v = \sqrt{\frac{2gh}{1+\beta}}, \qquad \frac{KE_{\text{rot}}}{KE_{\text{trans}}} = \beta \qquad (2)

where: $h$ — the vertical drop. The speed is the free-fall/sliding result $\sqrt{2gh}$ shrunk by the same tax; the split is the deeper fact behind it — with $v = \omega R$ , $\frac12 I\omega^2 = \frac12 (\beta MR^2)(v/R)^2 = \beta \cdot \frac12 Mv^2$ : spin energy is $\beta$ times translational energy, always, at every instant of rolling.

f = \frac{\beta}{1+\beta}\,Mg\sin\theta \quad\Longrightarrow\quad \mu_{\min} = \frac{\beta}{1+\beta}\tan\theta \qquad (3)

where: $f$ — the static friction that spins the body up. This is the toolkit's one formula that keeps $M$ : friction is a real force, not a ratio. It also carries the surprise MCQ answer: rolling friction demand grows with $\beta$ — the hoop needs the grippiest incline ( $\mu \ge \tfrac12\tan\theta$ ), the solid sphere the least ( $\mu \ge \tfrac27\tan\theta$ ).

Derivation AThe master denominator — derived once, for every body at once

We derive the universal rolling results (1) and (2) for a body of mass $M$ , radius $R$ and moment of inertia $I = \beta MR^2$ rolling without slipping down an incline of angle $\theta$ — by the energy route for the speed and the force route for the acceleration, so both appearances of the magic denominator are seen to be the same physics. The strategy: parameterize the body by $\beta$ alone, and watch $M$ and $R$ cancel — the cancellation IS the technique, because it licenses every ratio shortcut in this module.

Step 1. Energy route. Rolling from rest through vertical drop $h$ : static friction acts at the momentarily-stationary contact point (zero work), so gravity funds everything:

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

Step 2. Impose rolling, $\omega = v/R$ , and the body's identity, $I = \beta MR^2$ :

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\beta MR^2\frac{v^2}{R^2} = \frac{1}{2}Mv^2(1+\beta)

Why the split falls out here: the two kinetic terms stand in the ratio $1 : \beta$ before any solving — equation (2)'s energy split is a structural fact of rolling, not a result about inclines.

Step 3. Solve for the speed — $M$ and $R$ both gone:

v = \sqrt{\frac{2gh}{1+\beta}}

Step 4. Force route for the acceleration. Along the incline: $Mg\sin\theta - f = Ma$ . Torque about the centre: $fR = I\alpha = \beta MR^2 \cdot \frac{a}{R}$ , so $f = \beta Ma$ . Substitute:

Mg\sin\theta - \beta Ma = Ma \quad\Longrightarrow\quad a = \frac{g\sin\theta}{1+\beta}, \qquad f = \frac{\beta}{1+\beta}Mg\sin\theta

Why the two routes must agree: constant acceleration down slope length $L = h/\sin\theta$ gives $v^2 = 2aL = \frac{2gh}{1+\beta}$ — the energy answer. One denominator, two derivations, zero exceptions. *Verified: limits — $\beta = 0$ : $a = g\sin\theta$ , $f = 0$ , the frictionless block ✓; $\beta = 1$ (hoop): $a = \frac12 g\sin\theta$ , half of everything ✓. Friction check: $f \le \mu N = \mu Mg\cos\theta$ reproduces equation (3)'s $\mu_{\min}$ ✓.*

Go deeper · Derivation B · 1 minRatio surgery — why every comparison MCQ is a square root of (1+β)s

Derivation BRatio surgery — why every comparison MCQ is a square root of (1+β)s

We derive the three comparison formulas of Move 5, because using a shortcut whose origin you cannot reproduce is how shortcuts betray you. The strategy: write each quantity's $(1+\beta)$ dependence from the master results, take the ratio for two bodies on the same incline with the same release, and watch everything shared — $g$ , $\theta$ , $h$ , $L$ , masses, radii — cancel wholesale. The point to watch: whether $(1+\beta)$ lands in the numerator or denominator, which is exactly what the exam options test.

Step 1. Accelerations, from (1): $a \propto \dfrac{1}{1+\beta}$ , so

\frac{a_1}{a_2} = \frac{1+\beta_2}{1+\beta_1}

Step 2. Bottom speeds, from (2): $v \propto \dfrac{1}{\sqrt{1+\beta}}$ , so

\frac{v_1}{v_2} = \sqrt{\frac{1+\beta_2}{1+\beta_1}}

Step 3. Descent times: from rest, $L = \frac12 a t^2 \Rightarrow t \propto 1/\sqrt{a} \propto \sqrt{1+\beta}$ , so

\frac{t_1}{t_2} = \sqrt{\frac{1+\beta_1}{1+\beta_2}}

Why the swap between Steps 2 and 3 matters: speed ratios carry the OTHER body's $(1+\beta)$ on top; time ratios carry your OWN. Options in ratio MCQs are built precisely from this inversion — $\sqrt{15/14}$ and $\sqrt{14/15}$ will both be on offer. *Verified: sphere vs disc: $v_s/v_d = \sqrt{(3/2)/(7/5)} = \sqrt{15/14} \approx 1.035$ — the sphere arrives about 3.5% faster, matching drill 4's arithmetic; and $t_s/t_d = \sqrt{14/15} < 1$ , earlier arrival ✓. Degenerate check: $\beta_1 = \beta_2$ gives all ratios 1 — disc ties solid cylinder exactly ✓.*

When it backfires

The toolkit has a strict licence: rolling without slipping under gravity-driven descent. Four situations void it, and paper-setters plant all four.

The frictionless incline. No friction, no torque about the centre, no rotation: every body slides down at $g\sin\theta$ and arrives together, spinning not at all (or at whatever spin it started with). A "rolling race" on a smooth incline is a dead heat — the option "all reach together" exists precisely for this.

Insufficient friction. If $\mu < \frac{\beta}{1+\beta}\tan\theta$ , the body rolls and slips: kinetic friction takes over, $v \ne \omega R$ , and every formula in this module dies. The question will signal this by giving you a specific $\mu$ — that is your cue to check equation (3) before reaching for ratios.

The string radius is not R. Yo-yos and spools falling on strings obey the same-shaped formula $a = g/(1 + I/mr^2)$ — but with the string radius $r$ , not the rolling radius. A yo-yo's effective " $\beta$ " can be 18, not $1/2$ . Reach for the yo-yo archetype, not this table.

Rolling uphill, or driven wheels. A ball rolled up an incline still obeys $a = \frac{g\sin\theta}{1+\beta}$ (deceleration), and friction still points uphill — but torque-driven wheels (cars, motors) reverse friction's role entirely. The toolkit covers freely rolling bodies only.

Figure

The routing decision — fifteen seconds that pick the right weapon

Limiting cases

case	result	meaning
$\beta = 0$ (block, smooth)	$a = g\sin\theta$ , no spin share	the untaxed baseline
$\beta = 1$ (hoop)	$a = \tfrac12 g\sin\theta$ , KE split 50–50	the maximal standard tax
$\beta_1 = \beta_2$	all ratios $= 1$	disc ties solid cylinder — a planted trick
$\mu < \dfrac{\beta}{1+\beta}\tan\theta$	slipping: toolkit void	check (3) whenever $\mu$ is quoted
smooth incline	everything slides at $g\sin\theta$	"all arrive together" is the right option
same body, any $M$ , any $R$	$a, v, t$ identical	mass/radius mentions are decoys

Worked examples — the drill ladder

Drill 1 · Board level — the two-body race, timed

A disc and a ring are released together from rest at the top of a $30^\circ$ incline of slope length $2.1\ \mathrm{m}$ . Find each descent time and the winner. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Move 1: disc $\beta = \tfrac12$ , ring $\beta = 1$ . Move 3 already names the winner: the disc.

Step 2. Move 2: $a_{\text{disc}} = \dfrac{10 \times 0.5}{1.5} = 3.33\ \mathrm{m/s^2}$ ; $a_{\text{ring}} = \dfrac{5}{2} = 2.5\ \mathrm{m/s^2}$ .

Step 3. $t = \sqrt{2L/a}$ : disc $\sqrt{4.2/3.33} = \boxed{1.12\ \mathrm{s}}$ ; ring $\sqrt{4.2/2.5} = \boxed{1.30\ \mathrm{s}}$ .

Step 4. Note what was never used: the masses and radii the question would normally quote. Twenty-five seconds, most of it square roots.

Drill 2 · Board level — the energy split by inspection

What fraction of a rolling solid sphere's kinetic energy is rotational? A rolling ring's? For the ring, no calculation is permitted.

Step 1. Move 4: split is $\beta : 1$ . Solid sphere: $\dfrac{\beta}{1+\beta} = \dfrac{2/5}{7/5} = \boxed{\dfrac{2}{7}} \approx 29\%$ .

Step 2. Ring: $\beta = 1$ means spin and translation carry equal energy — $\boxed{50\%}$ , stated cold.

Step 3. For the mark scheme's benefit, one line of justification: $\tfrac12 I\omega^2 = \tfrac12(\beta MR^2)(v/R)^2 = \beta \times \tfrac12 Mv^2$ .

Drill 3 · Board level — bottom speed without energy conservation

A solid sphere rolls without slipping down from a vertical height of $1.4\ \mathrm{m}$ . Find its speed at the bottom. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Move 2: $v = \sqrt{\dfrac{2gh}{1+\beta}} = \sqrt{\dfrac{2 \times 10 \times 1.4}{7/5}} = \sqrt{\dfrac{28}{1.4}} = \sqrt{20} = \boxed{4.47\ \mathrm{m/s}}$ .

Step 2. Audit against the slide: $\sqrt{2gh} = 5.29$ m/s. Rolling arrives slower — the missing energy is spinning, not lost. If your "rolling" answer beats $\sqrt{2gh}$ , the denominator went missing.

Drill 4 · Competitive level — ratio surgery under the clock (~45 seconds)

A solid sphere and a hollow sphere of different masses and radii roll without slipping down the same incline. The ratio of their speeds at the bottom is…

Step 1. Move 5, speeds carry the other body's $(1+\beta)$ on top: $\dfrac{v_{\text{solid}}}{v_{\text{hollow}}} = \sqrt{\dfrac{1+\tfrac23}{1+\tfrac25}} = \sqrt{\dfrac{5/3}{7/5}} = \sqrt{\dfrac{25}{21}} = \boxed{\dfrac{5}{\sqrt{21}}} \approx 1.09$ .

Step 2. "Different masses and radii" — decoys, crossed out on sight. Timing note: 45 seconds, of which 30 are simplifying the fraction. The wrong option $\sqrt{21}/5$ (the inverted ratio) will be present; remember — the faster body (smaller $\beta$ ) must come out with the ratio above 1.

Drill 5 · Competitive level — the friction check that saves the question (~90 seconds)

A solid sphere is released on a $30^\circ$ incline with $\mu_s = 0.12$ . A student instantly writes $a = \frac{5}{7}g\sin\theta$ . Check the licence, and give the correct behaviour. ( $g = 10\ \mathrm{m/s^2}$ , $\tan 30^\circ = 0.577$ .)

Step 1. Licence check, equation (3): rolling demands $\mu \ge \dfrac{\beta}{1+\beta}\tan\theta = \dfrac{2}{7}(0.577) = 0.165$ .

Step 2. Supplied $\mu = 0.12 < 0.165$ : the licence is refused — the sphere rolls with slipping, and the toolkit's answer is wrong.

Step 3. Correct regime: kinetic friction $f = \mu Mg\cos\theta$ ; $a = g(\sin\theta - \mu\cos\theta) = 10(0.5 - 0.12 \times 0.866) = 10(0.5 - 0.104) = \boxed{3.96\ \mathrm{m/s^2}}$ , with $\omega$ growing separately via $\tau = fR$ until (if ever) $v = \omega R$ is reached. Timing note: the check costs 15 seconds and is the entire question — the "obvious" $\tfrac57 g\sin\theta = 3.57\ \mathrm{m/s^2}$ sits among the options as the punishment for skipping it. Whenever a rolling question quotes a numeric $\mu$ , run equation (3) first.

Exam traps

Mass and radius worship. They cancel. Every time. A stem that specifies "2 kg, radius 10 cm" versus "4 kg, radius 20 cm" for a ratio question is testing whether you know they cancel. (They return only in the friction force $f$ and in energy totals.)

Inverted ratios. $v \propto 1/\sqrt{1+\beta}$ but $t \propto \sqrt{1+\beta}$ : speed ratios carry the other body's factor on top, time ratios your own. Fix the direction with physics before algebra: smaller $\beta$ must be faster and earlier.

Rolling assumed for free. $v = \omega R$ and everything downstream requires $\mu \ge \frac{\beta}{1+\beta}\tan\theta$ . A quoted $\mu$ is an instruction to check, not decoration (Drill 5).

The smooth-incline ambush. "Frictionless" kills rotation entirely: all bodies slide at $g\sin\theta$ and tie. The trap option is the memorized race order, faithfully applied where it no longer holds.

Hollow versus solid sphere. $2/3$ versus $2/5$ — the single most-confused pair. Anchor: hollow bodies push mass outward, raising $\beta$ ; hollow sphere sits between disc and hoop.

"Friction does negative work on the rolling body." In pure rolling the contact point is momentarily at rest: static friction does zero work — that is why energy conservation held in the master derivation. Kinetic friction (slipping regime) is a different animal and does dissipate.

Which $g$ ? JEE-Main/Advanced and NEET: as stated, almost always $10$ . AP-1 permits $10$ for estimation. Ratios, mercifully, don't care.

Origins · 1 minJean le Rond d'Alembert and how this was discovered

Jean le Rond d'Alembert

1717–1783 · Paris

Abandoned as an infant on the steps of a Paris church and raised by a glazier's wife, d'Alembert rose to co-edit the Encyclopédie and to give mechanics one of its great compressions: his 1743 Traité de dynamique showed that problems of motion can be recast as problems of equilibrium — inertia treated as just another force to balance. That is the spiritual ancestor of this module's trick: the rolling body's dynamics compress into a single effective factor $(1+\beta)$ , after which "which body wins the race" becomes bookkeeping, not analysis. d'Alembert would have approved of Move 5 in particular; he spent his career arguing that the highest form of mechanics is the one that makes the answer obvious before the calculation starts.

History

1638

— Galileo times bronze balls rolling down inclined planes to establish uniform acceleration — unaware that rolling taxes his result by the factor

7/5

; his "diluted gravity" is diluted twice.

1673

— Huygens' pendulum work introduces what becomes the radius of gyration

k

— the

k^2/R^2

in this module's title.

1743

— d'Alembert's Traité de dynamique recasts dynamics as equilibrium, the philosophical licence for effective-inertia factors like

(1+\beta)

1765

— Euler's rigid-body treatise formalizes

I

and

\tau = I\alpha

; the master denominator becomes a two-line computation.

1900

s — the rolling race becomes a lecture and exam staple; JEE/NEET papers industrialize the ratio MCQ this module is built to defeat.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 10 (rolling problems) · NCERT Class 11 Ch. 7 · DC Pandey Rotational Mechanics for drill volume — apply the moves against its MCQ sets.

Provenance. All results are standard rolling canon, re-derived here: master denominator by both routes; splits $\beta/(1+\beta)$ ; $\mu_{\min} = \frac{\beta}{1+\beta}\tan\theta$ ; drill answers $1.12/1.30$ s, $2/7$ and $50\%$ , $\sqrt{20} = 4.47$ m/s, $5/\sqrt{21}$ , and drill 5's licence refusal ( $0.12 < 0.165$ ) with $a = 3.96\ \mathrm{m/s^2}$ .

Prerequisites. Rolling With & Without Slipping · Moments of Inertia of Standard Bodies · Rotational KE & Work Done by Torque. Leads to. Time-Period Pattern Recognition (technique) · Option Elimination & Answer Checking (track capstone).

Why does everything depend only on β and not on mass or radius?

Because both the driving term and the inertia term scale identically. Gravity's pull along the incline is $Mg\sin\theta$ ; the resistance is $M(1+\beta)$ — ordinary inertia plus rotational inertia $I/R^2 = \beta M$ dressed as extra mass. The $M$ 's cancel, exactly as in free fall. Radius cancels inside $\beta$ itself: $I \propto MR^2$ for every standard body, so $I/MR^2$ is a pure shape number. What survives is the only genuinely distinguishing property: how far from the axis the body parks its mass.

How should I actually memorize the β table?

Two anchors and one principle. Anchors: hoop $= 1$ (all mass at the rim — the maximum for standard bodies) and solid sphere $= 2/5$ (the most centrally packed — the minimum). Principle: hollowing a body pushes mass outward and raises $\beta$ — so disc $\tfrac12 <$ hollow sphere $\tfrac23 <$ hoop $1$ . Check yourself with the energy splits, which are friendlier fractions: sphere $2/7$ , disc $1/3$ , shell $2/5$ , hoop $1/2$ of the total KE in spin.

Why does the descent time carry (1+β) inside a square root, not directly?

Time enters through kinematics, not dynamics: from rest, $L = \frac12 a t^2$ , so $t = \sqrt{2L/a}$ — any factor in $a$ arrives under a root. The same square-root softening applies to speeds ( $v^2 = 2aL$ ). Accelerations compare linearly in $(1+\beta)$ ; everything integrated once (time, speed) compares as the square root. Getting this hierarchy straight kills the most common wrong options instantly.

Doesn't friction slow rolling bodies? Why does energy conservation ignore it?

In rolling without slipping, the friction is static and its application point — the contact point — is momentarily at rest. Force at a stationary point does zero work. Friction's role is redirection: it converts would-be sliding into spinning, taxing the translational share but banking the difference as rotational KE. Total mechanical energy is untouched, which is exactly why the master derivation could use $Mgh = KE_{\text{total}}$ . Only when slipping starts (kinetic friction, Drill 5) does real dissipation begin — and then the toolkit is already void.

What do I do with a body that isn't in the table — a composite wheel, say?

The toolkit doesn't care where $\beta$ comes from. If the question supplies $I$ , compute $\beta = I/MR^2$ once and proceed with every move intact. A spoked wheel with $I = 0.8\,MR^2$ has $\beta = 0.8$ : slower than a shell, faster than a hoop, energy split $0.8 : 1$ , $\mu_{\min} = \frac{0.8}{1.8}\tan\theta$ . The four memorized values are conveniences; the structure — one dimensionless number, one denominator — is the actual technique.

Does the incline angle change the race order or the energy split?

Never. $\theta$ scales all accelerations equally ( $a \propto \sin\theta$ ) and the split $\beta : 1$ contains no $\theta$ at all — it is a property of rolling itself, valid on slopes, flats and loops alike. What $\theta$ does change is the friction licence: $\mu_{\min} = \frac{\beta}{1+\beta}\tan\theta$ grows with steepness, so a sphere that rolls beautifully at $20^\circ$ may slip at $50^\circ$ . Order and split: angle-proof. Licence: angle-sensitive.

Is the (1+β) factor related to the yo-yo's g/(1 + I/mr²)?

Same skeleton, different radius. Both problems tax gravity's drive with an effective rotational mass $I/(\text{radius})^2$ — but the radius is wherever the constraint lives: the rolling contact at $R$ for incline bodies, the string's unwinding radius $r$ for yo-yos. For a rim-wound disc the two coincide and the yo-yo result is literally $\beta = \tfrac12$ : $a = \tfrac23 g$ on a "vertical incline." Recognizing the shared skeleton is the mark of owning the technique rather than the table.