Rolling without slipping — v = Rω condition: Question bank

1.5.13 · D5 · Physics › Rotational Mechanics › Rolling without slipping — v = Rω condition

Shuru karne se pehle, ek vocabulary reminder jo hum baar baar use karenge:

== $v_{c m}$ == woh speed hai jis par body ka center ground ke saath saath chalta hai.
== $ω$ == (omega) spin rate hai — radians per second mein.
== $R$ radius hai; $R ω$ == woh speed hai jis par ek rim point center ke relative purely spinning se move karta hai.
No-slip ka matlab hai contact point (ground ko touch karne wala hissa) ka zero velocity ground ke relative us instant par.

True or false — justify

Rolling without slipping ka matlab hai wheel contact point par kabhi move nahi karta.

True, lekin sirf instantaneously — har moment ek alag material point contact point banta hai, aur har ek momentarily rest mein hota hai jab wo touch karta hai. Koi ek chalk mark zyada der tak frozen nahi rehta. Dekho Instantaneous axis of rotation.

Agar ek wheel

v = R ω

satisfy karta hai, toh us par koi friction act nahi karta.

False. Level ground par constant speed mein friction zero ho sakta hai — lekin no-slip ki possibility static friction se enforce hoti hai, jo nonzero ho sakti hai (jaise incline par) jabki phir bhi zero work karti hai. Dekho Static vs kinetic friction.

Ek rolling wheel ka sabse upar wala point

2 v

par move karta hai, isliye uski kinetic energy center se double hai.

"Double" energy wale part par False — kinetic energy speed squared ke saath scale karti hai, isliye

2 v

par move karne wala point center par same-mass chunk ke

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

se chaar guna zyada hai. Speed double hoti hai, energy chaar guna ho jaati hai.

Ek slipping tyre phir bhi

a = R α

satisfy karta hai chahe

v = R ω

violate kare.

False. Velocity aur acceleration dono forms usi no-slip geometry se aate hain. Agar

v \neq = R ω

(slipping), toh generally

a \neq = R α

bhi; kinetic friction accelerate kar raha hota hai instead.

Rolling without slipping ke liye, contact point ki velocity ki direction horizontal hoti hai.

False — uski velocity zero hai, aur zero ki koi direction nahi hoti. Jo actually horizontal hai woh uska acceleration hai (jo centripetally center ki taraf upar point karta hai, ek baar jab aap isko account karo — edge-case section dekho).

Ek heavier wheel ko

v = R ω

satisfy karne ke liye same speed par bada

ω

chahiye.

False. Constraint

v = R ω

mein koi mass nahi hai — sirf geometry hai. Mass dynamics ko affect karta hai (us

ω

tak pahunchna kitna mushkil hai), constraint ko nahi. Dekho Moment of inertia.

Spot the error

"Wheel

v

par roll karta hai, toh rim par har point

v

par move karta hai."

Galat. Sirf center

v

par move karta hai. Rim points

0

(bottom) se

2 v

(top) tak range karte hain; unki speeds ek smooth curve trace karti hain jaise aap around jaate ho kyunki rotation translation mein add ya subtract karta hai.

"Kyunki

v = R ω

, bada wheel hamesha slow spin karta hai."

Incomplete, universal nahi. Ek fixed

v

par, haan:

ω = v / R

shrink karta hai jaise

R

badhta hai. Lekin agar

v

choose karne ki freedom hai, toh bada wheel kisi bhi

ω

par spin kar sakta hai — relation

v

aur

ω

ko saath mein tie karta hai, wheels ko size se akela rank nahi karta.

"Friction ek rolling ball ko slow karta hai kyunki yeh hamesha motion ko oppose karta hai."

Pure rolling ke liye Galat. Contact point slide nahi kar raha, isliye static friction (kinetic nahi) act karta hai, aur kyunki woh point move nahi karta,

W = F \cdot d = 0

— zero work, koi energy loss nahi. Ball ideal ground par hamesha roll karta rehega.

"Ek ball ko rest se bina slip ke rolling shuru karne ke liye,

v = R ω

apply karo."

Category error.

v = R ω

ek constraint hai, force ya action nahi. Tum ek torque/force apply karte ho; constraint phir dictate karta hai ki

v

aur

ω

ka kya relation hona chahiye agar no-slip maintain ki jaaye. Dekho Rolling down an incline.

"Instantaneous axis wheel ka center hai."

Galat. Motion ko pure rotation treat karte hue, axis contact point hai (ek maatra point jo rest mein hai), center nahi. Center khud

v

par move karta hai, isliye woh pivot nahi ho sakta. Dekho Instantaneous axis of rotation.

"Inverted track ki ceiling par roll karne wala wheel

v = R ω

satisfy nahi kar sakta."

Galat. Unrolling arc = distance covered ki geometry orientation-independent hai. Jab tak contact point slide nahi karta,

v = R ω

hold karta hai — gravity ki direction constraint ke liye irrelevant hai.

Why questions

Contact point ka rest mein hona

v = R ω

ko kyun force karta hai rather than sirf suggest karta hai?

Contact point ki velocity (forward

v

) plus (spin se backward

R ω

) hai. Us sum ko zero set karna — no-slip ki definition — equation

v - R ω = 0

hai. Constraint wahi arithmetic hai "no sliding" ka.

v = R ω

forces se nahi balki arc length se kyun derive hota hai?

Kyunki yeh ek kinematic geometry fact hai, dynamics fact nahi. Angle

θ

unroll karne par rim ka arc

R θ

lay hota hai; no-slip kehta hai woh arc ground cover kiye barabar hai,

x = R θ

deta hai aur differentiate karne par

v = R ω

— koi force nahi chahiye.

Static friction zero work kyun karta hai chahe yeh ek badi force ho?

Work = force times displacement us point ka jahan force act karta hai. Static friction contact point par act karta hai, jiska velocity zero hai, isliye zero displacement per instant — toh

W = F \cdot d = 0

chahe

F

kitna bhi bada ho.

Photos mein wheel ka top blur kyun dikhta hai jabki bottom sharp rehta hai?

Bottom momentarily rest mein hai (zero speed), isliye exposure ke dauran barely move karta hai aur crisp rehta hai; top

2 v

par move karta hai, sabse fast point, isliye sabse zyada smear karta hai. Yeh

0, v, 2 v

rule ka direct visual evidence hai.

Wahi wheel soft sand par roll karne ke liye hard road se different

ω

kyun chahta hai bina slip ke?

Agar contact point sink kare ya effective radius change ho, toh geometry

x = R θ

shift ho jaati hai — aur agar sand shear (slip) kare, toh constraint bilkul fail ho jaata hai. Firm ground par fixed

R

ke saath,

ω = v / R

fixed hai; deformable surfaces clean relation ko tod dete hain.

Edge cases

Ek wheel ice par spinning in place (

v = 0

ω \neq = 0

) — kya yeh rolling without slipping hai?

Nahi. Yahan

v = 0

lekin

R ω \neq = 0

, isliye

v \neq = R ω

— contact point

R ω

par backward slide karta hai. Yeh pure slipping hai; kinetic friction act karta hai aur energy dissipate karta hai.

Ek block sliding without rotating (

ω = 0

v \neq = 0

) — kya

v = R ω

hold karta hai?

Nahi. Yeh

v = 0

deta hai,

v \neq = 0

ko contradict karta hai. Non-spinning slide spinning-in-place case ka opposite extreme hai; dono no-slip ko opposite tareekon se violate karte hain.

ω = 0

aur

v = 0

par (wheel at rest), kya constraint satisfy hota hai?

Trivially haan —

0 = R \cdot 0

. Ground ko touch karta hua rest mein body

v = R ω

vacuously obey karta hai; yeh degenerate boundary hai jahan dono motions vanish ho jaati hain.

Kya contact point ka acceleration bhi zero hota hai jab uski velocity zero hoti hai?

Nahi — yeh classic trap hai. Contact point ki velocity zero hoti hai lekin nonzero centripetal acceleration

R ω^{2}

hoti hai jo center ki taraf upar point karti hai. Zero speed ka matlab zero acceleration nahi hai; yeh instant to instant turn around kar raha hai.

Infinitely large wheel (

R \to \infty

) ki limit mein, fixed

v

ke liye

ω

ka kya hota hai?

ω = v / R \to 0

. Ek enormous wheel ground cover karne ke liye barely rotate karta hai — ek flat surface ke limit ko approach karta hai jo bina visible spin ke translate kare. Constraint limit tak exact rehta hai.

Ek ball rolling without slipping phir ek frictionless patch se takraaye — pehle kya toot ta hai?

v = R ω

ki maintenance. Bina friction ke, koi torque

ω

ko changing

v

se match karne ke liye adjust nahi kar sakta; agar

v

ω

phir independently change ho, equality fail ho jaati hai aur ball slip karna shuru kar deta hai. Dekho Static vs kinetic friction.

Do gears apne teeth par bina slip ke mesh kar rahe hain — same constraint?

Haan, spirit mein: contact-point speeds match karti hain (

R_{1} ω_{1} = R_{2} ω_{2}

), jo "no-slip at the interface" idea hai do bodies ke beech apply kiya hua instead of body-and-ground ke. Unrolling-arc logic identical hai.

Connections

Instantaneous axis of rotation — contact point as pivot jo $2 v$ aur $0$ obvious banata hai
Static vs kinetic friction — kaun si friction no-slip enforce karti hai aur kyun zero work karti hai
Kinetic energy of rolling bodies — jahan $0, v, 2 v$ speed spread energy sum mein feed hoti hai
Moment of inertia — mass constraint mein kyun nahi aata lekin dynamics par rule karta hai
Angular velocity and angular acceleration — $ω, α$ jise yeh poora page constrain karta hai
Rolling down an incline — constraint applied jahan slipping ek real risk hai