Acceleration of rolling objects on inclines — comparison: Question bank

1.5.15 · D5 · Physics › Rotational Mechanics › Acceleration of rolling objects on inclines — comparison

Yeh page parent topic ke liye ek trap-spotting drill hai. Neeche har item kisi specific misconception ya boundary case ko pakadne ke liye design ki gayi hai. Pehle answer cover karo, ek reason decide karo (sirf "true"/"false" nahi), phir reveal karo.

Shuru karne se pehle, is page par use hone wale har symbol ko pin down karte hain — yeh traps tab tak samajh nahi aate jab tak tum har letter ke peeche ki picture nahi dekh lete. Hum inhe us order mein introduce karte hain jisme inki zaroorat padti hai, taaki koi bhi letter use hone se pehle build ho chuka ho.

Figure — Acceleration of rolling objects on inclines — comparison

Definition Forces

$m g sin θ$ , normal force, aur friction $f$ Rolling body par teen forces act karti hain (Figure 2). $m g sin θ$ gravity ka along-slope pull hai (hill ke neeche), jo hum ne abhi define kiye mass $m$ se bana hai. Normal force $N$ surface ka straight incline se bahar push hai, jiska size $m g cos θ$ hai (gravity ka "slope mein jaata hua" part jo surface ko balance karna padta hai) — yeh kabhi slope ke along point nahi karti, isliye speed up ya spin karne mein koi role nahi hai. $f$ static friction hai: surface aur object ke contact point ke beech ki grip. "Static" matlab contact point slide nahi kar raha — friction yahan koi work nahi karta, sirf ek twisting effect deta hai. Yeh slope ke upar point karta hai, kyunki yeh woh force hai jo neeche ke part ke slip karne ki tendency ko resist karta hai.

True or false — justify

Har item: pehle decide karo, phir reason do. Sirf "true/false" ka score zero hai — reasoning hi point hai.

Ek heavier solid sphere usi incline par ek lighter solid sphere se faster accelerate karti hai.

False.

a = \frac{g s i n θ}{1 + β}

mein mass

m

cancel ho chuka hai; heavy sphere ko zyada gravity milti hai lekin proportionally zyada inertia bhi, isliye dono identical

a

ke saath roll karte hain.

Same mass lekin alag radii wali do solid spheres alag accelerate karti hain.

False.

R

bhi cancel hota hai — yeh torque (

f R

) aur inertia (

m R^{2}

) ko saath scale karta hai, isliye ek marble aur ek bowling ball same acceleration share karte hain.

Ek rolling solid sphere usi incline par ek frictionless sliding block se pehle bottom par pahunchti hai.

False. Sliding block ko koi rotation feed nahi karni padti, isliye woh poora

a = g sin θ

use karta hai, jabki sphere spin ke liye kuch acceleration khoti hai (

a = \frac{5}{7} g sin θ

). Block jeet jaata hai.

Bilkul smooth (frictionless) incline par, top par rakha hua sphere roll karke neeche aayega.

False. Friction ke bina centre ke baare mein koi torque nahi hai, isliye yeh spin shuru nahi kar sakta — yeh sirf rolling ke bina slide karta hai. Friction hi translation ko rotation se couple karta hai.

Rolling without slipping object par static friction zero work karta hai.

True. Contact point instantaneously rest par hota hai, isliye friction zero displacement se act karta hai — uska work zero hai, isliye energy conserve hoti hai.

Ek ring aur ek solid sphere same height se release hoke bottom par same speed ke saath pahunchte hain.

False. Final speed

v = \frac{2 g h}{1 + β}

β

par depend karti hai; sphere (

β = 2/5

) bottom par ring (

β = 1

) se faster hai.

Incline angle ko

30°

60°

par double karne se acceleration double ho jaata hai.

False.

a \propto sin θ

hai,

θ

nahi.

sin 60°/ sin 30° = (3 /2) / (1/2) = 3 \approx 1.73

, isliye yeh

\approx 1.73 \times

badhta hai,

2 \times

nahi.

Ranking (sphere fastest, ring slowest)

10°

aur

70°

dono inclines par same hai.

True. Angle har object ki acceleration ko same

g sin θ

se multiply karta hai; sirf

β

inhe order karta hai, isliye ranking angle-independent hai (agar har object abhi bhi rolling without slipping karta ho).

Rolling without slipping incline chahe kitna bhi steep ho jaaye, chalti rehti hai.

False. Yeh tab tak hold karta hai jab tak

tan θ \leq \frac{1 + β}{β} μ_{s}

; uske baad required friction

μ_{s} m g cos θ

se zyada ho jaati hai aur object slip karne lagta hai.

Spot the error

Har statement mein reasoning ki ek galat line hai. Flaw ko naam do.

"Kyunki

m g h

poori tarah

\frac{1}{2} m v^{2}

mein jaati hai, ek rolling cylinder

v = 2 g h

par end hota hai."

Error yeh hai ki rotational energy

\frac{1}{2} I ω^{2}

drop kar di gayi. Real balance

m g h = \frac{1}{2} m v^{2} (1 + β)

hai, jo smaller

v = \frac{2 g h}{1 + β}

deta hai.

"Friction motion ko oppose karta hai, isliye yeh sphere ko slow karta hai aur energy heat ke roop mein waste karta hai."

Rolling without slipping mein friction static hai, kinetic nahi; yeh koi work nahi karta aur koi heat generate nahi karta — yeh sirf woh torque supply karta hai jo spin create karta hai.

"Ek hollow sphere ka mass solid sphere se zyada hota hai, isliye uska

β

bada hota hai."

Error yeh hai ki mass aur distribution ko confuse kiya ja raha hai.

β

depend karta hai ki mass kahan baitha hai, kitna nahi; ek thin-shell hollow sphere (

β = 2/3

) ka

β

solid sphere (

β = 2/5

) se bada hota hai kyunki uska mass axis se zyada door baitha hai, total mass ki parwah kiye bina.

"Kyunki sphere ke liye

a = \frac{5}{7} g sin θ

hai, uski acceleration

g

se kam hai, jo gravity ke under impossible hai."

Koi rule nahi kehta ki acceleration

g

ke equal honi chahiye; sirf free-fall mein

g

milta hai. Incline par along-slope pull already

g sin θ < g

hai, aur rolling ise aur reduce karti hai — yeh perfectly consistent hai.

"Hum torque equation skip kar sakte hain aur sirf

F = ma

use kar sakte hain

F = m g sin θ

ke saath."

Yeh Newton's law mein friction

f

aur poore rotational half ko ignore karta hai. Tumhe dono

m g sin θ - f = ma

aur

f R = I α

a = α R

se tie karna hai; torque drop karne par galat (sliding) answer milta hai.

"Kyunki

a = α R

hai, bada radius matlab bada linear acceleration

a

hoga."

a = α R

ek constraint hai, cause nahi; bada

R

same

a

ke liye smaller

α

bhi mean karta hai. Final formula dikhata hai ki

R

poori tarah cancel ho jaata hai.

"Normal force ek aisa torque contribute karta hai jo sphere ko spin karne mein help karta hai."

Normal force aur gravity dono centre of mass se act karte hain (ya uski taraf point karte hain), isliye centre ke baare mein unka lever arm zero hai — sirf friction centre ke baare mein torque produce karta hai.

"Rougher surface (bada

μ_{s}

) object ko jaldi roll karne mein madad karta hai."

Jab tak yeh already rolling without slipping kar raha hai,

a = \frac{g s i n θ}{1 + β}

mein koi

μ_{s}

nahi hai; extra grip sirf steeper slopes par rolling enable karta hai, yeh already rolling object ko speed up nahi karta.

Why questions

"Why" ka jawab do, sirf "what" ka nahi.

m

aur

R

dono acceleration se kyun vanish ho jaate hain?

Gravity ka pull

m

ke saath scale karta hai aur inertia

m

ke saath (cancel ho jaate hain); torque

R

ke saath scale karta hai aur rotational inertia

R^{2}

ke saath, aur rolling constraint

α = a / R

bacha hua

R

hata deta hai — isliye sirf dimensionless

β

bachta hai.

Bada

β

matlab slower object kyun hota hai?

Bada

β

matlab gravity ke energy budget ka zyada hissa axis se door baitha mass ko spin karne mein spend karna padta hai, forward motion ke liye kam bachta hai, isliye

a = \frac{g s i n θ}{1 + β}

shrink karta hai.

Humein ek ki jagah do equations (force aur torque) ki zaroorat kyun hai?

Ek rolling body do simultaneous motions undergo karta hai — apne centre ka translation aur uske baare mein rotation — isliye Newton's law for forces aur torque law dono alag alag ek ek ko govern karte hain, aur rolling condition unhe glue karta hai.

Friction rolling ke liye zaroori kyun hai lekin no work karta hai?

Yeh akela woh force hai jo centre ke baare mein torque produce kar sakta hai (isliye spin shuru karta hai), phir bhi kyunki contact point kabhi slide nahi karta, woh force zero displacement ke through act karta hai — zaroori lekin work-free.

Ek rolling object slope se neeche hamesha ek identical sliding object se slower kyun hota hai?

Rolling wala gravity ki energy ka kuch hissa rotation mein divert karta hai (

β > 0

), isliye uska translational acceleration

\frac{g s i n θ}{1 + β}

sliding value

g sin θ

se strictly kam hai.

Incline angle har object ki speed kyun change karta hai lekin ranking kabhi nahi?

sin θ

ek common multiplier hai jo saari accelerations ko equally hit karta hai; order sirf

β

se fix hota hai, jise angle touch nahi karta.

Steeper incline eventually pure rolling kyun tod deta hai?

Zaroori friction

\frac{β}{1 + β} m g sin θ

ke roop mein badhti hai jabki maximum available grip

μ_{s} m g cos θ

steepness ke saath shrink karta hai, isliye

tan θ = \frac{1 + β}{β} μ_{s}

ke baad demand supply se zyada ho jaati hai aur yeh slip karta hai.

Edge cases

Boundary aur degenerate scenarios jinke liye tumhe ready rehna hai.

Formula

a = \frac{g s i n θ}{1 + β}

mein

β = 0

set karna kya correspond karta hai, aur

a

kya hai?

β = 0

no-rotation limit hai — poora rotational share vanish ho jaata hai, ek frictionless sliding block ko

a = g sin θ

ke saath reproduce karta hai. (Strictly,

β = I / m R^{2}

rotating bodies describe karta hai;

β = 0

ek limiting value hai jo "no spin" represent karta hai, na ki koi real solid object jiska sara mass axis par ho.)

Jab

β \to \infty

(imaginary "infinitely spread" mass) hota hai,

a

ka kya hota hai?

a = \frac{g s i n θ}{1 + β} \to 0

: ek object jiska mass axis se infinitely door baitha ho usse sirf spin karne ke liye apni saari energy chahiye hogi, isliye woh barely aage accelerate karta hai.

θ = 0°

(flat ground) par, ek rolling object ka acceleration kya hai?

sin 0° = 0

, isliye

a = 0

; koi along-surface gravity component nahi hai, isliye rest par ek rolling object rest par rehta hai aur pehle se move karta hua constant speed par coast karta hai.

θ = 90°

(vertical drop) par, kya rolling formula abhi bhi apply hoti hai?

Nahi —

90°

par incline object ko press nahi kar sakta (normal force

\to 0

), isliye rolling enforce karne ke liye koi friction nahi; object simply

g

par free-fall karta hai, aur rolling model break down kar jaata hai.

Ek solid sphere (

β = 2/5

) ke liye maximum incline angle kya hai jis par woh abhi bhi rolling without slipping karta hai?

tan θ \leq \frac{1 + β}{β} μ_{s} = \frac{7/5}{2/5} μ_{s} = \frac{7}{2} μ_{s}

se, isliye

θ_{m a x} = arctan (3.5 μ_{s})

; iske baad sphere slip karta hai.

Do objects jinka same

β

hai lekin alag shapes hain (e.g. ek solid cylinder aur koi bhi

β = 1/2

body) — woh kaise compare karte hain?

Identically: acceleration sirf

β

par depend karta hai, isliye equal

β

matlab equal

a

, equal final speed, aur equal descent time, specific geometry ki parwah kiye bina.

Agar surface itni smooth hai ki object roll karte hue slip kare, toh kya

a = \frac{g s i n θ}{1 + β}

abhi bhi hold karta hai?

Nahi — woh formula rolling without slipping (

a = α R

) assume karta hai. Ek baar slip hone par, kinetic friction (jo work karta hai) constraint ki jagah le leta hai, aur clean

β

result apply nahi hoti.

Ek ball jo incline upar roll kar rahi hai (decelerate ho rahi hai) — kya same

∣ a ∣

magnitude appear hoti hai?

Haan, rolling without slipping ke dauran deceleration magnitude abhi bhi

\frac{g s i n θ}{1 + β}

hai, kyunki friction abhi bhi sirf energy ko translation aur rotation ke beech redistribute karta hai — algebra unchanged hai.

Connections

Moment of Inertia — $β = I / m R^{2}$ ka source.
Rolling Without Slipping — woh constraint jo $R$ ko kill karta hai aur friction ko work-free banata hai.
Energy Conservation in Rolling — kyun same $1 + β$ factor final speed govern karta hai.
Torque and Angular Acceleration — in traps ke peeche doosra equation.
Friction in Rolling — "zaroori lekin no work" static friction aur slipping threshold.
Newton's Second Law on Inclines — poora derivation aur $β = 0$ sliding limit.