Acceleration of rolling objects on inclines — comparison: Worked examples

This page is the drill ground for the parent topic. The parent gave us one master formula. Here we throw every kind of input at it — steep angles, flat angles, weird shapes, sliding blocks, energy, time, friction — and solve each one from the ground up.

Before any symbol appears, here is the one tool we reuse everywhere. It was earned in the parent note; we restate it in full plain words so nothing is assumed.

Formula The master result (from the parent)

For a round object rolling without slipping (the contact patch never skids) down a slope tilted at angle $θ$ : $a = \frac{g s i n θ}{1 + β}, β = \frac{I}{m R ^{2}} .$

$a$ = how fast the centre of the object speeds up along the slope (metres per second, every second).
$g \approx 9.8$ = free-fall acceleration.
$θ$ = tilt of the slope measured from flat ground.
$sin θ$ = the fraction of gravity that pulls along the slope (0 when flat, 1 when vertical).
$m$ = mass of the object; $R$ = its radius.
$I$ = the object's moment of inertia — a number measuring how hard it is to spin, i.e. how much rotational "stubbornness" the mass gives (see Moment of Inertia). Bigger $I$ = mass sits farther from the spin axis.
$β$ = the shape factor $I / (m R^{2})$ : a pure number stripping out mass and radius so only the shape remains. Small = bunched near the axis (easy to spin), large = far from axis (hard to spin).

Keep the shape table pinned in your mind:

Object	$β$	one-word memory
Solid sphere	$2/5 = 0.40$	marble
Solid cylinder / disc	$1/2 = 0.50$	coin
Hollow sphere (shell)	$2/3 \approx 0.67$	tennis ball
Ring / hoop / pipe	$1$	bangle
Sliding block (no rolling)	$0$	ice cube

The scenario matrix

Every question this topic can ask lives in one of these cells. The worked examples below each carry a tag like (Cell A2) so you can see the coverage is complete.

#	Cell class	What makes it special	Covered by
A1	Standard angle ( $0 < θ < 90°$ )	plain plug-in	Ex 1
A2	Compare two shapes	ranking / who wins	Ex 2
B1	$θ \to 0$ (flat) degenerate	$sin θ \to 0$ , $a \to 0$	Ex 3
B2	$θ \to 90°$ (vertical) limit	$sin θ \to 1$ , max $a$	Ex 3
C1	$β = 0$ (sliding block) degenerate	rolling term vanishes	Ex 4
C2	$β \to \infty$ (mass far out) limit	$a \to 0$	Ex 4
D1	Energy / final speed	uses $v = 2 g h / (1 + β)$	Ex 5
D2	Time to descend	uses $t = 2 L / a$	Ex 6
E1	Friction required (find $f$ )	is the slope rough enough?	Ex 7
E2	Slip threshold (exam twist)	when rolling breaks	Ex 8
F1	Real-world word problem	translate words → symbols	Ex 9
F2	Unknown shape (back-solve $β$ )	measure $a$ , find shape	Ex 10

Worked examples

Ex 1 — Standard angle, single object (Cell A1)

Worked example Solid cylinder on a 20° slope

A solid cylinder is released from rest on a $θ = 20°$ incline. Find its acceleration. Use $g = 9.8$ .

Forecast: guess — will $a$ be more or less than $g sin 20° \approx 3.35$ ? (A rolling body is always slower than a sliding one, so expect less.)

Identify $β$ . Solid cylinder $\Rightarrow β = \frac{1}{2}$ . Why this step? $β$ is the only shape-dependent input; get it first.
Write the formula. $a = \frac{g sin θ}{1 + β} = \frac{g sin 20°}{1 + \frac{1}{2}} = \frac{g sin 20°}{1.5}$ . Why this step? Direct substitution — nothing else varies.
Compute $sin 20° = 0.342$ . So $g sin 20° = 9.8 \times 0.342 = 3.352$ . Why this step? $sin θ$ is the fraction of gravity along the slope; we need its number before dividing.
Divide. $a = 3.352/1.5 = 2.23 m/s^{2}$ . Why this step? Dividing by $1 + β$ applies the "spin tax" that slows the body.

Verify: $2.23 < 3.35$ ✔ (rolling is slower than pure sliding, as forecast). Units: $\frac{m/s ^{2}}{pure number} = m/s^{2}$ ✔.

Ex 2 — Race between shapes (Cell A2)

Worked example Marble vs bangle on the same 30° slope

A solid sphere and a ring start together from the top of a $30°$ incline. By what factor is the sphere's acceleration larger?

Forecast: the ring hides its mass at radius $R$ , so it should be markedly slower — guess a ratio around $1.3$ – $1.5$ .

Sphere: $β = \frac{2}{5}$ , so $a_{sph} = \frac{g sin 30°}{1 + 2/5} = \frac{g sin 30°}{7/5} = \frac{5}{7} g sin 30°$ . Why this step? Plug the sphere's shape factor.
Ring: $β = 1$ , so $a_{ring} = \frac{g sin 30°}{2} = \frac{1}{2} g sin 30°$ . Why this step? Same slope, different $β$ — the only change.
Ratio. $\frac{a _{sph}}{a _{ring}} = \frac{5/7}{1/2} = \frac{10}{7} = 1.4286$ . Why this step? $g sin 30°$ cancels — the race depends only on $β$ , so the answer is slope-independent.

Verify: $10/7 \approx 1.43$ , inside the forecast window ✔. The sphere accelerates $43%$ faster ⇒ wins the race, matching the parent's ranking.

Ex 3 — The angle extremes (Cells B1 & B2)

Here we stress-test $θ$ itself. The figure below is the geometric heart of this example: it draws four slopes of increasing tilt and shows, at each, the full weight (plum, always straight down) and its along-slope component $g sin θ$ (orange). Watch the orange arrow grow from a stub at $10°$ to nearly the full weight at $85°$ — that growing arrow is the driving push in our formula. Steps 1 and 2 below simply read off the two endpoints of that growth.

Figure — Acceleration of rolling objects on inclines — comparison

Figure: the along-slope drive $g sin θ$ (orange) grows from $0$ at flat ground to the full weight at vertical, while the total weight (plum) never changes.

Worked example Flat slope and near-vertical slope (solid sphere)

For a solid sphere ( $β = 2/5$ ), what is $a$ when (a) $θ = 0°$ and (b) $θ = 90°$ ? Use $g = 9.8$ .

Forecast: flat ground = nothing rolls, so $a = 0$ . Straight down = the sphere is basically in free-fall but slowed by spin — expect $a = \frac{5}{7} g \approx 7$ .

Case $θ = 0°$ (Cell B1). $sin 0° = 0$ , so $a = \frac{g \cdot 0}{7/5} = 0 m/s^{2}$ . Why this step? On flat ground the orange arrow in the figure shrinks to zero length — gravity has no component along the surface, so nothing pushes it forward.
Case $θ = 90°$ (Cell B2). $sin 90° = 1$ , so $a = \frac{g}{7/5} = \frac{5}{7} g = \frac{5}{7} (9.8) = 7.0 m/s^{2}$ . Why this step? At vertical the orange arrow reaches the full weight (its longest in the figure); $sin θ$ maxes out at 1. This is idealised — a truly vertical surface can't press back to give friction, so rolling would actually break here (that's Cell E2's job). We take it as the mathematical limit.

Verify: $a (0°) = 0$ ✔ (sensible — flat ground). $a (90°) = 7.0 < 9.8 = g$ ✔ (still slower than pure free-fall because energy goes into spin). Both extremes bracket every real answer $0 \leq a \leq 7.0$ for a solid sphere.

Ex 4 — The shape extremes (Cells C1 & C2)

Worked example The frictionless block and the mass-flung-out wheel

On a $45°$ slope: (a) a frictionless sliding block ( $β = 0$ ); (b) a hypothetical wheel whose entire mass sits on the rim at ever-larger radius, driving $β \to \infty$ . Compare accelerations. Use $g = 9.8$ , $sin 45° = 0.707$ .

Forecast: the block has no spinning to feed, so it should hit the theoretical maximum $g sin θ$ . The $β \to \infty$ wheel wastes everything on spin — guess $a \to 0$ .

Block, $β = 0$ (Cell C1). $a = \frac{g sin 45°}{1 + 0} = g sin 45° = 9.8 \times 0.707 = 6.93 m/s^{2}$ . Why this step? With no rotational inertia the master formula collapses to plain Newton on an incline — this is the sliding-block limit. It is the fastest anything can go on this slope.
$β \to \infty$ (Cell C2). $a = \frac{g sin 45°}{1 + β} \to \frac{6.93}{\infty} = 0$ . Why this step? As mass moves infinitely far from the axis, it becomes infinitely hard to spin; gravity spends all its budget on rotation, leaving nothing for forward motion.

Verify: Every real rolling object ( $0 < β \leq 1$ ) must satisfy $0 < a < 6.93$ on this slope. Check ring: $a = 6.93/2 = 3.46$ ✔ — comfortably between the two extremes. The whole ranking is boxed in by these two cells.

Ex 5 — Energy method: final speed (Cell D1)

Worked example Hollow sphere down a 1.2 m drop

A hollow sphere ( $β = 2/3$ ) rolls from rest down an incline, dropping a vertical height $h = 1.2 m$ . Find its speed at the bottom. Use $g = 9.8$ .

Forecast: a sliding block would reach $2 g h = 2 (9.8) (1.2) \approx 4.85$ . Rolling steals some for spin, so expect less — guess ~ $3.8$ .

Energy balance. All the lost height-energy becomes motion: $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$ . Why this step? Static friction does no work in rolling (Friction in Rolling), so energy is conserved — see Energy Conservation in Rolling.
Fold spin into the shape factor. Using $I = β m R^{2}$ and $ω = v / R$ , the whole right side becomes $\frac{1}{2} m v^{2} (1 + β)$ . Cancel $m$ : $g h = \frac{1}{2} v^{2} (1 + β)$ . Why this step? Same $(1 + β)$ "spin tax" that appears in the acceleration formula; folding it in leaves a single unknown $v$ .
Solve for $v$ . Multiply both sides by $2/ (1 + β)$ then take the square root: $v = \frac{2 g h}{1 + β} = \frac{2 ( 9.8 ) ( 1.2 )}{1 + 2/3} = \frac{23.52}{1.667} = 14.11 = 3.76 m/s$ . Why this step? We want $v$ alone, so we undo the $\frac{1}{2} (1 + β)$ multiplying $v^{2}$ , then undo the square by rooting.

Verify: $3.76 < 4.85$ ✔ (slower than a sliding block, as forecast). Units: $m/s^{2} \cdot m = m^{2} / s^{2} = m/s$ ✔.

Ex 6 — Time to descend a fixed length (Cell D2)

Worked example How long to roll 3 m?

A solid sphere ( $β = 2/5$ ) rolls from rest down $L = 3 m$ of slope at $θ = 25°$ . How many seconds to reach the bottom? Use $g = 9.8$ , $sin 25° = 0.4226$ .

Forecast: it accelerates a few m/s², over 3 m — guess about 1–1.5 seconds.

Acceleration first. $a = \frac{g sin 25°}{1 + 2/5} = \frac{9.8 ( 0.4226 )}{7/5} = \frac{4.141}{1.4} = 2.958 m/s^{2}$ . Why this step? Time depends on how fast it speeds up, so we need $a$ from the master formula before anything else.
Kinematics from rest. $L = \frac{1}{2} a t^{2} \Rightarrow t = \frac{2 L}{a}$ . Why this step? Constant acceleration with zero start speed gives the standard "distance = ½at²"; we invert it to get $t$ .
Plug in. $t = \frac{2 ( 3 )}{2.958} = 2.029 = 1.424 s$ . Why this step? Substitute the known $L$ and computed $a$ to get the number.

Verify: inside the forecast ✔. Sanity: final speed $v = a t = 2.958 \times 1.424 = 4.21 m/s$ ; also $v = 2 a L = 2 (2.958) (3) = 17.75 = 4.21$ ✔ — the two routes agree. Units: $m / (m/s^{2}) = s^{2} = s$ ✔.

Ex 7 — How much friction is actually needed? (Cell E1)

Worked example Friction force on a rolling disc

A solid cylinder of mass $m = 2 kg$ rolls without slipping down a $30°$ slope. Find the friction force $f$ that makes this happen. Use $g = 9.8$ .

Forecast: friction here isn't wasting energy — it's the torque supplier. It should be a modest fraction of the weight component.

Acceleration. $β = \frac{1}{2}$ , so $a = \frac{g sin 30°}{1.5} = \frac{9.8 ( 0.5 )}{1.5} = 3.267 m/s^{2}$ . Why this step? We need $a$ to find how much force was "used up" by rotation.
Friction from the rotation equation. In the parent derivation $f = β ma$ . Why this step? Rearranging torque $f R = I α$ with $α = a / R$ gives exactly the force that spins the body (Torque and Angular Acceleration).
Plug in. $f = \frac{1}{2} (2) (3.267) = 3.267 N$ . Why this step? Substitute $β$ , $m$ , and the computed $a$ into $f = β ma$ to get the number.

Verify: cross-check with translation law $m g sin θ - f = ma$ : $2 (9.8) (0.5) - 3.267 = 9.8 - 3.267 = 6.533$ , and $ma = 2 (3.267) = 6.533$ ✔. Units: $kg \cdot m/s^{2} = N$ ✔.

Ex 8 — When does rolling break? The slip threshold (Cell E2, exam twist)

Worked example Minimum coefficient of friction to keep a hoop rolling

A ring/hoop ( $β = 1$ ) rolls without slipping down a slope of angle $θ$ . The available static friction is $f_{m a x} = μ m g cos θ$ . Find the smallest $μ$ that prevents slipping, for $θ = 40°$ .

Forecast: steeper slopes demand more friction; a hoop needs a lot of torque, so expect $μ$ of a few tenths.

Required friction. $f_{req} = β ma$ with $a = \frac{g sin θ}{1 + β}$ , so $f_{req} = \frac{β m g sin θ}{1 + β}$ . Why this step? This is the friction the rolling motion demands (from Ex 7's logic).
Set demand ≤ supply. Rolling holds when $f_{req} \leq μ m g cos θ$ . The threshold is equality: $μ_{m i n} = \frac{β t a n θ}{1 + β} .$ Why this step? Divide both sides by $m g cos θ$ ; the $m g$ cancels and $sin / cos = tan$ . We use $tan$ because it is exactly the ratio "pull-along ÷ press-into-slope", i.e. how hard gravity tries to make it skid versus how hard it's pinned down.
Plug in $β = 1$ , $θ = 40°$ , $tan 40° = 0.8391$ : $μ_{m i n} = \frac{1 \cdot 0.8391}{2} = 0.4196$ . Why this step? Substitute the hoop's $β$ and the angle to get the numeric threshold.

Verify: dimensionless (a coefficient) ✔. Sanity on shape: a solid sphere ( $β = 2/5$ ) needs only $μ_{m i n} = \frac{( 2/5 ) ( 0.8391 )}{7/5} = 0.2397$ — less than the hoop ✔, because it needs less spin-torque. Below these $μ$ values the object would slip, and the whole $a = g sin θ / (1 + β)$ analysis stops being valid.

Ex 9 — Real-world word problem (Cell F1)

Worked example A can of soup on a ramp

"A sealed can of soup (treat as a solid cylinder, $β = 1/2$ ) is rolled from rest down a $15°$ loading ramp that is $2.5 m$ long. How fast is it going when it reaches the floor, and how long did it take?" Use $g = 9.8$ , $sin 15° = 0.2588$ .

Forecast: gentle slope, short ramp — slow-ish, maybe 2 m/s and a couple of seconds.

Translate words to symbols. "solid cylinder" $\Rightarrow β = \frac{1}{2}$ ; "from rest" $\Rightarrow v_{0} = 0$ ; " $2.5$ m long" $\Rightarrow L = 2.5$ (distance along the slope). Why this step? Word problems are 90% translation — pin every phrase to a symbol.
Acceleration. $a = \frac{g sin 15°}{1.5} = \frac{9.8 ( 0.2588 )}{1.5} = 1.691 m/s^{2}$ . Why this step? Both the speed and the time need $a$ , so compute it first from the master formula.
Final speed. $v = 2 a L = 2 (1.691) (2.5) = 8.454 = 2.907 m/s$ . Why this step? Starting from rest, $v^{2} = 2 a L$ links speed directly to distance without needing time; $L$ is along the ramp so it pairs with $a$ .
Time. $t = \frac{v}{a} = \frac{2.907}{1.691} = 1.719 s$ . Why this step? From $v = a t$ (started at rest), invert to get $t$ from the just-found $v$ .

Verify: forecast matched (≈2.9 m/s, ≈1.7 s) ✔. Cross-check $L = \frac{1}{2} a t^{2} = \frac{1}{2} (1.691) (1.71 9^{2}) = \frac{1}{2} (1.691) (2.955) = 2.50$ ✔.

Ex 10 — Back-solve the mystery shape (Cell F2)

Worked example Identify an unknown object from its acceleration

An unknown uniform round object rolls from rest down a $30°$ incline and is measured to accelerate at $a = 3.5 m/s^{2}$ . What is its $β$ , and which standard shape is it? Use $g = 9.8$ .

Forecast: $g sin 30° = 4.9$ , and $a = 3.5$ is fairly close to it → small $β$ → probably a solid sphere.

Invert the master formula. From $a = \frac{g sin θ}{1 + β}$ , multiply both sides by $(1 + β)$ and divide by $a$ : $1 + β = \frac{g sin θ}{a}$ , so $β = \frac{g sin θ}{a} - 1$ . Why this step? We're given $a$ and want $β$ , so we algebraically isolate $β$ instead of plugging into the usual direction.
Plug in. $g sin 30° = 9.8 (0.5) = 4.9$ . $β = \frac{4.9}{3.5} - 1 = 1.4 - 1 = 0.4$ . Why this step? Substitute the known angle and measured $a$ to get the numeric shape factor.
Match the table. $β = 0.4 = 2/5$ → solid sphere. Why this step? Compare the computed $β$ against the pinned shape table to name the object.

Verify: forward-check — solid sphere gives $a = \frac{4.9}{1.4} = 3.5 m/s^{2}$ ✔, exactly the measured value. Forecast confirmed.

Coverage check

Every cell of the scenario matrix now has a worked home:

Recall Which formula for which question?

Find $a$ ::: $a = \frac{g sin θ}{1 + β}$ Find final speed ::: $v = 2 g h / (1 + β)$ or $v = 2 a L$ Find time from rest ::: $t = 2 L / a$ Find friction needed ::: $f = β ma$ Find slip threshold ::: $μ_{m i n} = \frac{β tan θ}{1 + β}$ Back-solve the shape ::: $β = \frac{g sin θ}{a} - 1$

Connections

Parent topic — the master formula these examples drill.
Moment of Inertia — where each $β$ value (and $I$ itself) comes from.
Rolling Without Slipping — the $v = ω R$ , $a = α R$ glue (breaks in Ex 8).
Energy Conservation in Rolling — behind Ex 5.
Torque and Angular Acceleration — behind $f = β ma$ (Ex 7).
Friction in Rolling — why friction does no work (Ex 5) yet is essential (Ex 7–8).
Newton's Second Law on Inclines — the $β = 0$ sliding limit (Ex 4).