Rolling without slipping — v = Rω condition: Worked examples

1.5.13 · D3 · Physics › Rotational Mechanics › Rolling without slipping — v = Rω condition

Yeh page ek drill hai har us case ke liye jo rolling constraint $v_{c m} = R ω$ tumhare saamne rakh sakta hai. Pehle hum saare scenario types ko ek matrix mein map karte hain, phir har cell ke liye ek example karte hain. Kuch bhi naya assume nahi kiya gaya hai — agar koi symbol aata hai, woh parent parent note mein ya neeche build kiya gaya tha.

Recall Teen constraint forms (parent se)

$x_{c m} = R θ v_{c m} = R ω a_{c m} = R α$ Yahan $R$ = radius (metres), $θ$ = spin angle radians mein, $ω$ = spin rate (rad/s), $α$ = spin acceleration (rad/s²), aur "cm" subscript ka matlab hai "centre of mass ka". Yeh sirf rolling without slipping ke liye hold karte hain.

Scenario matrix

Is topic ka har problem inhi cells mein se ek hota hai. Neeche diye examples mein se har ek ke saath (Cell B) jaisa ek tag lagaa hai taaki tum dekh sako ki poora space cover ho raha hai.

Cell	Case class	Kya cheez ise special banati hai
A	$v$ se $ω$ dhundho (ya ulta)	Simple constraint, ek unknown
B	Acceleration form $a = R α$	Differentiated constraint
C	Point velocities (top / bottom / side)	$v_{p o in t} = v_{c m} + ω \times r$ use karta hai
D	Spin ka sign / direction	$ω$ $+$ ya $-$ ho sakta hai; forward vs backward roll
E	Zero / degenerate input	$ω = 0$ , $v = 0$ , ya $R \to$ bahut chhota
F	Slipping — constraint FAIL ho jaata hai	$v \neq = R ω$ ; formula bilkul use mat karo
G	Rim par limiting / arbitrary point	Kisi bhi general point ki speed angle $ϕ$ par
H	Real-world word problem	Odometer, gear, distance travelled
I	Exam twist	Do wheels / relative spin / mixed units

Example set

Worked example Example 1 — Plain constraint

(Cell A) Ek wheel jiska radius $R = 0.5 m$ hai, $ω = 4 rad/s$ par spin kar raha hai rolling without slipping ke saath. Centre speed $v$ dhundho.

Forecast: Bada radius ya tez spin ⇒ tez centre. Aage padhne se pehle ek number guess karo.

Step 1 — Sahi form chuno. Hume $ω$ aur $R$ pata hai, hum $v$ chahte hain, toh $v = R ω$ use karo. Yeh step kyun? Yeh ek maatra form hai jo spin ko directly centre speed se link karta hai bina kisi time-derivative ke.

Step 2 — Substitute karo. $v = (0.5) (4) = 2 m/s$ . Yeh step kyun? $R$ metres mein hai, $ω$ rad/s mein; radians dimensionless hote hain, toh product m/s hoga — speed ke liye correct.

Verify: Units: $m \times rad/s = m/s$ ✓. Sanity: 4 rad/s spin lagbhag 0.64 turns per second hai; ek ~1 m circumference wala rim bahar glide karta hai — ek couple of m/s bilkul reasonable lagta hai. $v = 2 m/s$

Worked example Example 4 — Rim par koi bhi point

(Cell C, G) Ek ball $v = 3 m/s$ par roll kar rahi hai, radius $R = 0.2 m$ . Wheel ke side par point ki speed dhundho (3-o'clock point, centre ke level par).

Forecast: Yeh top ( $2 v$ ) nahi hai aur bottom ( $0$ ) nahi hai. Guess karo: kahin beech mein hoga.

Step 1 — Do velocities mein split karo. Translation har point ko $v = 3 m/s$ forward (horizontal) deta hai. Rotation side point ko $R ω = v = 3 m/s$ deta hai — lekin rotation velocity spoke ke perpendicular hai, toh 3-o'clock point par yeh seedha upar (ya neeche) point karti hai. Yeh step kyun? $v_{p o in t} = v_{c m} + ω \times r$ : cross-product $ω \times r$ hamesha radius $r$ ke perpendicular hota hai. Figure dekho — dono arrows right angles par hain.

Step 2 — Pythagoras se combine karo. Woh perpendicular hain, toh $v_{s i d e} = v^{2} + (R ω)^{2} = 3^{2} + 3^{2} = 18 = 4.24 m/s .$ Yeh step kyun? Perpendicular vectors ek right triangle ke do legs ki tarah add hote hain.

Verify: Value $4.24$ , $0$ (bottom) aur $6$ (top) ke beech mein hai ✓. Kyunki $R ω = v$ yahan, $v_{s i d e} = v 2 \approx 1.41 v$ — ek clean check. $v_{s i d e} \approx 4.24 m/s$

Worked example Example 5 — General angle aur do extreme points

(Cell G) Ek wheel jo $v$ par roll kar rahi hai (toh $R ω = v$ ), rim par ek point bottom contact point se $ϕ$ angle banata hai (ghumte hue). Dikhao ki speed $v_{ϕ} = 2 v sin (ϕ /2)$ hai, phir bottom, side, aur top par check karo.

Forecast: Ek formula jo bottom par $0$ aur top par $2 v$ deta hai. Guess karo ki yeh smoothly grow karta hai ya nahi.

Step 1 — Yeh trick kyun kaam karti hai. Pure rolling mein contact point instantaneous axis hota hai (parent note). Toh har rim point momentarily bottom ke baare mein rotate karta hai speed $= ω \times (contact point se doori)$ ke saath. Yeh step kyun? Poore wheel ko fixed contact point ke baare mein spin karte hue treat karne se "translation + rotation" ek simple rotation mein badal jaata hai.

Step 2 — Contact se point tak ki doori. Rim-angle $ϕ$ par ek point, contact point se ek chord door hai; woh chord length hai $d = 2 R sin (ϕ /2)$ . Yeh step kyun? Ek circle ki chord jo central angle $ϕ$ subtend karti hai uski length $2 R sin (ϕ /2)$ hoti hai — standard circle geometry (figure ki dashed chord dekho).

Step 3 — $ω$ se multiply karo. $v_{ϕ} = ω d = ω \cdot 2 R sin (ϕ /2) = 2 (R ω) sin (ϕ /2) = 2 v sin (ϕ /2)$ . Yeh step kyun? Pivot ke baare mein speed $=$ (angular rate) $\times$ (pivot se doori), aur $R ω = v$ .

Verify (teeno cases):

Bottom $ϕ = 0$ : $2 v sin 0 = 0$ ✓ (Cell E degenerate: contact rest par).
Side $ϕ = 9 0^{\circ}$ : $2 v sin 4 5^{\circ} = 2 v \cdot 0.707 = 1.41 v$ ✓ (Example 4 se match karta hai).
Top $ϕ = 18 0^{\circ}$ : $2 v sin 9 0^{\circ} = 2 v$ ✓. $v_{ϕ} = 2 v sin (ϕ /2)$

Worked example Example 6 — Spin ka direction / sign

(Cell D) Ek wheel left ki taraf $v = 2 m/s$ par roll karta hai, $R = 0.25 m$ . $ω$ sahi sign ke saath do (counter-clockwise ko $+$ maano) aur top point ki velocity (magnitude aur direction).

Forecast: Left roll karna ⇒ wheel right roll karne ke ulti taraf spin karta hai. Guess karo ki sign flip hoga.

Step 1 — Constraint se magnitude. $∣ ω ∣ = v / R = 2/0.25 = 8 rad/s$ . Yeh step kyun? Constraint magnitudes link karta hai; sign geometry se aata hai, equation se nahi.

Step 2 — Sign fix karo. Left roll karte hue, wheel ka top left move karta hai jabki bottom (contact) ruka rehta hai — yeh clockwise hai, jo hamare convention mein negative sense hai. Toh $ω = - 8 rad/s$ . Yeh step kyun? Ek wheel apne top ki taraf roll karta hai; leftward motion ⇒ clockwise spin ⇒ hamare convention se negative.

Step 3 — Top point velocity. Top speed $= 2 v = 4 m/s$ , left ki taraf point karti hai (roll ki same direction mein). Yeh step kyun? Top hamesha travel ki direction mein double centre speed par move karta hai, sign convention se regardless.

Verify: $∣ R ω ∣ = 0.25 \times 8 = 2 = v$ ✓. Sign check: agar yeh right roll karta toh hume $ω = + 8$ milta; leftward ise $- 8$ par flip karta hai ✓. $ω = - 8 rad/s, v_{t o p} = 4 m/s left$

Worked example Example 7 — Degenerate inputs

(Cell E) Do quick sub-cases. (a) Ek wheel $ω = 0$ ke saath spin karta hai jabki uska centre frictionless ice par $v = 1 m/s$ forward glide karta hai — kya yeh rolling without slipping hai? (b) Ek wheel apne centre par pin kiya gaya hai aur $ω = 10 rad/s$ , $R = 0.1 m$ par spin kiya gaya hai, lekin centre move nahi kar sakta ( $v = 0$ ). Contact-point speed kya hai?

Forecast: Dono $v = R ω$ violate karte hain. Guess karo kaun forward slide karta hai aur kaun backward.

Step 1 — (a) ko constraint se test karo. $R ω = R \cdot 0 = 0 \neq = v = 1$ . Constraint fail ⇒ yeh sliding hai, rolling nahi. Yeh step kyun? Rolling without slipping ke liye $v = R ω$ zaroori hai; agar fail ho, toh slipping hai. Contact point $v - R ω = 1 - 0 = 1 m/s$ forward move karta hai — pure skid.

Step 2 — (b) test karo. $R ω = 0.1 \times 10 = 1 m/s$ , lekin $v = 0$ , toh $v \neq = R ω$ . Contact-point velocity $= v - R ω = 0 - 1 = - 1 m/s$ (backward). Yeh ice par spinning tyre hai: rubber ground ke peeche backward uda jaata hai. Yeh step kyun? Same test; yahan contact point backward slip karta hai — (a) ke opposite.

Verify: (a) contact $+ 1$ m/s forward slip karta hai; (b) contact $- 1$ m/s (backward) slip karta hai. Koi bhi $0$ nahi, toh koi bhi true rolling nahi hai ✓. Yeh exactly parent ka " $v = R ω$ hamesha sach NAHI hai" warning hai. $(a) slip + 1 m/s, (b) slip - 1 m/s$

Worked example Example 8 — Real-world word problem

(Cell H) Ek bike wheel jiska radius $R = 0.34 m$ hai, slipping ke bina $N = 1000$ full turns karta hai. Bike kitni door gayi, aur agar iska time $t = 200 s$ laga, toh average $v$ aur $ω$ kya tha?

Forecast: Har turn ek circumference lay out karta hai. Low thousands of metres mein distance guess karo.

Step 1 — Angle radians mein. $θ = N \cdot 2 π = 1000 \times 2 π = 6283.2 rad$ . Yeh step kyun? Constraint $x_{c m} = R θ$ ko $θ$ radians mein chahiye; ek turn $2 π$ rad hai.

Step 2 — Distance. $x = R θ = 0.34 \times 6283.2 = 2136.3 m$ . Yeh step kyun? No-slip ka matlab hai ground distance = total rim unrolled.

Step 3 — Averages. $v = x / t = 2136.3/200 = 10.68 m/s$ ; $ω = v / R = 10.68/0.34 = 31.4 rad/s$ . Yeh step kyun? Average speed distance over time hai; phir constraint $ω$ deta hai.

Verify: Circumference se cross-check: $x = N \cdot 2 π R = 1000 \times 2 π \times 0.34 = 2136.3 m$ ✓. Aur $ω = θ / t = 6283.2/200 = 31.4 rad/s$ match karta hai ✓. $x \approx 2136 m, v \approx 10.7 m/s, ω \approx 31.4 rad/s$

Worked example Example 9 — Exam twist: do coupled wheels

(Cell I) Ek bada wheel ( $R_{1} = 0.4 m$ ) aur ek chhota wheel ( $R_{2} = 0.1 m$ ) same axle par hain aur same road par rolling without slipping kar rahe hain (ek toy-train double wheel). Agar chhota wheel actually track par hai aur axle centre $v = 1.2 m/s$ par move karta hai, toh common $ω$ aur bade wheel ke top ki speed dhundho.

Forecast: Same axle ⇒ same $ω$ . Bada rim, uss $ω$ par spin karte hue, ground se aage nikal jaayega — yeh track ke upar hona chahiye. Guess karo ki uska top $2 v$ se zyaada hai ya nahi.

Step 1 — Kaun sa wheel $ω$ set karta hai? Sirf contact wheel (radius $R_{2}$ ) ground ke saath $v = R_{2} ω$ obey karta hai. Toh $ω = v / R_{2} = 1.2/0.1 = 12 rad/s$ . Yeh step kyun? No-slip constraint uss wheel par apply hota hai jo surface ko touch kar raha hai, kisi aisi rim par nahi jo contact mein nahi hai.

Step 2 — Bade wheel ka top. Uski top speed $= v + R_{1} ω = 1.2 + 0.4 \times 12 = 1.2 + 4.8 = 6.0 m/s$ . Yeh step kyun? Top point = translation $v$ + rotation $R_{1} ω$ (dono forward). Note karo $R_{1} ω = 4.8 \neq = v$ , toh yeh simple " $2 v$ " case NAHI hai — bada rim rolling wala nahi hai.

Verify: Sanity: bade wheel ka bottom hoga $v - R_{1} ω = 1.2 - 4.8 = - 3.6 m/s$ (backward) — jo theek hai kyunki woh point track ke upar hai aur freely move kar sakta hai. $ω = 12 rad/s, v_{t o p, bi g} = 6.0 m/s$

Recall Coverage check — kya humne har cell hit ki?

Cell A: Ex 1, 2 ::: plain constraint dono directions mein Cell B: Ex 3 ::: acceleration form $a = R α$ Cell C: Ex 4 ::: side-point velocity Cell D: Ex 6 ::: $ω$ ka sign (leftward roll) Cell E: Ex 5 (bottom), Ex 7 ::: degenerate / zero inputs Cell F: Ex 7 ::: slipping, constraint fails Cell G: Ex 5 ::: general rim angle $2 v sin (ϕ /2)$ Cell H: Ex 8 ::: odometer word problem Cell I: Ex 9 ::: do coupled wheels

Connections

Instantaneous axis of rotation — pivot trick jo Example 5 mein use hua
Angular velocity and angular acceleration — Example 6 mein $ω$ ke signs
Static vs kinetic friction — slipping cases (Example 7) mein kinetic friction involve hoti hai
Kinetic energy of rolling bodies — agla step jab point speeds samajh aa jaayein
Rolling down an incline — jahan $a = R α$ (Example 3) use hota hai
Moment of inertia — in kinematics ko dynamics mein turn karne ke liye zaroori