Acceleration of rolling objects on inclines — comparison

WHAT are we comparing?

The only thing that differs between them is the moment of inertia, written as $I = \beta\,mR^2,$ where $\beta$ is a pure number (the "shape factor"). Bigger $\beta$ = mass farther from axis.

Object	$\beta = I/mR^2$
Solid sphere	$2/5 = 0.40$
Solid cylinder / disc	$1/2 = 0.50$
Hollow sphere (thin shell)	$2/3 \approx 0.67$
Ring / hoop / thin pipe	$1$

HOW: derive the acceleration from first principles

Set up. Take an object of mass $m$ , radius $R$ , on an incline of angle $\theta$ . Forces along the incline: gravity component $mg\sin\theta$ (down the slope) and friction $f$ (up the slope — it's what provides the torque to spin the object).

Step 1 — Translation (Newton's 2nd law along incline). $mg\sin\theta - f = ma \qquad (1)$ Why this step? Net force = mass × acceleration of the centre of mass.

Step 2 — Rotation about the centre (torque = $I\alpha$ ). Only friction has a torque about the centre (gravity and normal pass through the centre): $f R = I\alpha \qquad (2)$ Why this step? Torque spins the object; only the friction force has a lever arm $R$ .

Step 3 — Rolling-without-slipping condition. The contact point doesn't slide, so $v = \omega R$ and differentiating: $a = \alpha R \quad\Rightarrow\quad \alpha = \frac{a}{R} \qquad (3)$ Why this step? This is the glue: it links linear $a$ to angular $\alpha$ .

Step 4 — Eliminate $f$ and $\alpha$ . Put $I = \beta mR^2$ and $\alpha = a/R$ into (2): $f = \frac{I\alpha}{R} = \frac{\beta mR^2 \cdot (a/R)}{R} = \beta m a$ Why this step? Express friction purely in terms of $a$ , so we can substitute into (1).

Substitute into (1): $mg\sin\theta - \beta m a = ma$ $g\sin\theta = a(1+\beta)$

Figure — Acceleration of rolling objects on inclines — comparison

The ranking (fastest → slowest)

Plug each $\beta$ into $a = \dfrac{g\sin\theta}{1+\beta}$ :

Object	$\beta$	$a$ (in units of $g\sin\theta$ )
Solid sphere	$2/5$	$a = \frac{5}{7}g\sin\theta \approx 0.714$
Solid cylinder	$1/2$	$a = \frac{2}{3}g\sin\theta \approx 0.667$
Hollow sphere	$2/3$	$a = \frac{3}{5}g\sin\theta = 0.600$
Ring/hoop	$1$	$a = \frac{1}{2}g\sin\theta = 0.500$

Worked examples

Worked example Example 2 — Final speed (energy method, cross-check)

A solid cylinder rolls from rest down a height $h$ . Find its speed at the bottom.

Energy: $mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2$ . With $I=\tfrac12 mR^2$ , $\omega=v/R$ : $mgh = \tfrac12 mv^2 + \tfrac12\cdot\tfrac12 mR^2\cdot\frac{v^2}{R^2} = \tfrac12 mv^2(1+\tfrac12)$ Why this step? The $\tfrac12 I\omega^2$ term adds the rotational share $\beta$ . $gh = \tfrac12 v^2 (1+\beta) \Rightarrow v = \sqrt{\frac{2gh}{1+\beta}} = \sqrt{\frac{2gh}{1.5}} = \sqrt{\tfrac{4gh}{3}}$ Same $\frac{1}{1+\beta}$ factor appears — consistent with the acceleration result. ✔

Common mistakes

Active recall

Recall Quick self-test (cover the answers)

Formula for rolling acceleration? → $a=\dfrac{g\sin\theta}{1+\beta}$
Which is fastest? → Solid sphere ( $\beta=2/5$ )
Which is slowest? → Ring ( $\beta=1$ )
Does mass matter? → No, it cancels.
Does friction do work here? → No (static, contact point at rest).

Recall Feynman: explain to a 12-year-old

Imagine rolling a marble and a metal ring down a slide. Gravity is "pocket money" given to each. But the ring keeps most of its mass far from the centre, so it's hard to get spinning — it spends a lot of its pocket money just on spinning and has little left for going forward, so it's slow. The marble has its weight bunched in the middle, easy to spin, so it saves most of its money for going fast — it wins the race! And funnily, it doesn't matter if the marble is big or small, light or heavy — same speed.

Connections

Moment of Inertia — where $\beta = I/mR^2$ comes from.
Rolling Without Slipping — the condition $v=\omega R$ , $a=\alpha R$ .
Energy Conservation in Rolling — the $v=\sqrt{2gh/(1+\beta)}$ derivation.
Torque and Angular Acceleration — equation $fR = I\alpha$ .
Friction in Rolling — why static friction does no work.
Newton's Second Law on Inclines — the sliding-block limit ( $\beta=0$ ).

What is the general acceleration of an object rolling without slipping down an incline?

a = \dfrac{g\sin\theta}{1+I/mR^2} = \dfrac{g\sin\theta}{1+\beta}

Define the shape factor

\beta

\beta = I/(mR^2)

, a pure number; larger means mass farther from the axis.

Why do mass and radius not affect the acceleration?

They appear in both gravity/torque and inertia in the same proportion, so they cancel.

Rank solid sphere, cylinder, hollow sphere, ring by acceleration (fastest first).

Solid sphere (

5/7

) > cylinder (

2/3

) > hollow sphere (

3/5

) > ring (

1/2

), all ×

g\sin\theta

Acceleration of a solid sphere down a 30° incline (g=9.8)?

\frac{5}{7}(9.8)(0.5)=3.5\ \text{m/s}^2

What provides the torque that makes the object spin while rolling?

Static friction at the contact point (lever arm

R

about the centre).

Does friction do work during rolling without slipping?

No — the contact point is instantaneously at rest, so static friction does zero work.

Final speed at bottom from height

h

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

(less than the sliding value

\sqrt{2gh}

How does descent time depend on

\beta

t \propto \sqrt{1+\beta}

, so larger

\beta

means slower descent.

The two equations used to derive rolling acceleration?

Translation

mg\sin\theta - f = ma

; Rotation

fR = I\alpha

; linked by

a=\alpha R

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi round object incline par bina slip kiye roll karta hai, to gravity ka kaam sirf usse aage push karna nahi hota — usse ghumana (spin) bhi padta hai. Matlab gravity ki energy do jagah baant jaati hai: translation (aage badhna) aur rotation (ghoomna). Jis object ka mass axis se jitna door hoga, uska moment of inertia $I$ utna bada (yaani $\beta = I/mR^2$ bada), aur utni hi zyada energy spinning me chali jaati hai — isliye woh slow chalega.

Formula simple hai: $a = \dfrac{g\sin\theta}{1+\beta}$ . Iski derivation me bas teen cheezein chahiye — translation ke liye $mg\sin\theta - f = ma$ , rotation ke liye $fR = I\alpha$ , aur rolling condition $a = \alpha R$ . Inhe mila do, $f$ aur $\alpha$ hata do, bas formula aa jaata hai. Sabse important baat: mass $m$ aur radius $R$ cancel ho jaate hain! Isliye chhota marble aur bada solid sphere dono same acceleration se aate hain — yeh exam me favourite trick hai.

Ranking yaad rakho: solid sphere sabse fast ( $\beta=2/5$ , factor $5/7$ ), phir cylinder, phir hollow sphere, aur ring sabse slow ( $\beta=1$ , factor $1/2$ ). Logic: jiska $\beta$ chhota, woh swift. Aur ek common galti — yeh mat sochna ki friction yahan energy waste karta hai. Rolling without slipping me contact point ruka hua hota hai, isliye static friction ka work zero hai; woh sirf energy ko rotation me bhejta hai. Energy method se check karo: $v=\sqrt{2gh/(1+\beta)}$ , jo sliding block ke $\sqrt{2gh}$ se hamesha kam aata hai. Yeh consistency exam me confidence deti hai.

Acceleration of rolling objects on inclines — comparison

WHAT are we comparing?

HOW: derive the acceleration from first principles

The ranking (fastest → slowest)

Worked examples

Common mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections