Acceleration of rolling objects on inclines — comparison

1.5.15 · Physics › Rotational Mechanics

WHAT are we comparing?

Unke beech sirf moment of inertia alag hota hai, jo likha jaata hai $I = β m R^{2},$ jahaan $β$ ek pure number hai (the "shape factor"). Bada $β$ = mass axis se zyada door.

Object	$β = I / m R^{2}$
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. Maano ek object hai jiska mass $m$ , radius $R$ hai, aur woh $θ$ angle ke incline par hai. Incline ke saath forces: gravity component $m g sin θ$ (slope ke neeche) aur friction $f$ (slope ke upar — yahi object ko spin karne ka torque provide karta hai).

Step 1 — Translation (Newton's 2nd law along incline). $m g sin θ - f = ma (1)$ Yeh step kyun? Net force = mass × acceleration of the centre of mass.

Step 2 — Rotation about the centre (torque = $I α$ ). Centre ke baare mein sirf friction ka torque hai (gravity aur normal centre se guzarte hain): $f R = I α (2)$ Yeh step kyun? Torque object ko spin karta hai; sirf friction force ka lever arm $R$ hai.

Step 3 — Rolling-without-slipping condition. Contact point nahi phislata, isliye $v = ω R$ aur differentiate karne par: $a = α R \Rightarrow α = \frac{a}{R} (3)$ Yeh step kyun? Yeh glue hai: yeh linear $a$ ko angular $α$ se jodata hai.

Step 4 — $f$ aur $α$ ko eliminate karo. $I = β m R^{2}$ aur $α = a / R$ ko (2) mein daalo: $f = \frac{I α}{R} = \frac{β m R ^{2} \cdot ( a / R )}{R} = β ma$ Yeh step kyun? Friction ko sirf $a$ ke terms mein express karo, taaki (1) mein substitute kar sakein.

(1) mein substitute karo: $m g sin θ - β ma = ma$ $g sin θ = a (1 + β)$

Figure — Acceleration of rolling objects on inclines — comparison

The ranking (fastest → slowest)

Har $β$ ko $a = \frac{g sin θ}{1 + β}$ mein daalo:

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

Worked examples

Common mistakes

Active recall

Recall Quick self-test (cover the answers)

Rolling acceleration ka formula? → $a = \frac{g sin θ}{1 + β}$
Sabse fast kaun? → Solid sphere ( $β = 2/5$ )
Sabse slow kaun? → Ring ( $β = 1$ )
Kya mass matter karta hai? → Nahi, cancel ho jaata hai.
Kya friction yahan kaam karta hai? → Nahi (static, contact point rest par).

Recall Feynman: explain to a 12-year-old

Socho ek marble aur ek metal ring ko slide par neeche roll karo. Gravity har ek ko "pocket money" deti hai. Lekin ring ka zyaadaatar mass centre se door hota hai, isliye use spin karna mushkil hota hai — woh apni pocket money ka zyada hissa sirf spinning par kharch kar deti hai aur aage jaane ke liye kam bachta hai, isliye woh slow hai. Marble ka weight beech mein ikatta hota hai, spin karna easy hai, isliye woh apna zyaadaatar paisa tezi se jaane ke liye bachata hai — woh race jeetta hai! Aur mazedaar baat yeh hai ki marble bada ho ya chota, halka ho ya bhaari — same speed.

Connections

Moment of Inertia — jahaan se $β = I / m R^{2}$ aata hai.
Rolling Without Slipping — condition $v = ω R$ , $a = α R$ .
Energy Conservation in Rolling — $v = 2 g h / (1 + β)$ ki derivation.
Torque and Angular Acceleration — equation $f R = I α$ .
Friction in Rolling — kyun static friction koi kaam nahi karta.
Newton's Second Law on Inclines — sliding-block limit ( $β = 0$ ).

Rolling without slipping ke saath incline par roll karne wale object ki general acceleration kya hoti hai?

a = \frac{g sin θ}{1 + I / m R ^{2}} = \frac{g sin θ}{1 + β}

Shape factor

β

ko define karo.

β = I / (m R^{2})

, ek pure number; bada matlab mass axis se zyada door.

Mass aur radius acceleration ko affect kyun nahi karte?

Yeh gravity/torque aur inertia dono mein same proportion mein appear hote hain, isliye cancel ho jaate hain.

Solid sphere, cylinder, hollow sphere, ring ko acceleration ke hisaab se rank karo (sabse fast pehle).

Solid sphere (

5/7

) > cylinder (

2/3

) > hollow sphere (

3/5

) > ring (

1/2

), sabhi ×

g sin θ

30° incline par solid sphere ki acceleration (g=9.8)?

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

Rolling karte waqt object ko spin karne wala torque kaun provide karta hai?

Contact point par static friction (centre ke baare mein lever arm

R

Kya rolling without slipping mein friction kaam karta hai?

Nahi — contact point instantaneously rest par hota hai, isliye static friction zero kaam karta hai.

Height

h

se neeche aane par final speed?

v = \frac{2 g h}{1 + β}

(sliding value

2 g h

se kam).

Descent time

β

par kaise depend karta hai?

t \propto 1 + β

, isliye bada

β

matlab slow descent.

Rolling acceleration derive karne ke liye use ki gayi do equations?

Translation

m g sin θ - f = ma

; Rotation

f R = I α

;

a = α R

se linked.