Relation to linear quantities - v = rω, a_t = rα, a_c = rω²: Worked examples

1.5.3 · D3 · Physics › Rotational Mechanics › Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

Yeh parent topic ka ek practice companion hai. Wahan humne teen bridge formulas derive kiye the. Yahan hum har tarah ki situation dhundte hain jinmein unhe daala ja sakta hai — har sign, har zero, har degenerate case, aur exam ke sneaky twists — aur har ek ke liye ek example solve karte hain.

Prerequisites jo tum open rakhna chahoge: Angular velocity and angular acceleration, Radian measure and arc length, Uniform vs non-uniform circular motion, Centripetal force and circular motion.

Definition Is page ke symbols ka quick recall

Koi bhi number se pehle, yahan exactly woh hai jo har letter ka matlab hai (fuller versions Angular velocity and angular acceleration mein hain):

$r$ — radius: spin axis se kisi point ki doori, metres mein. Ek picture: bug record ke middle se kitna door baitha hai.
$θ$ — ghuma hua angle, yahan hamesha radians mein measure kiya jaata hai (ek full turn $= 2 π$ rad).
$t$ — time seconds mein; $Δ t$ ka matlab hai time mein ek badlaav (ek elapsed interval), $α = Δ ω /Δ t$ ke andar use hota hai.
$ω$ (Greek "omega") — angular velocity: body per second kitne radians sweep karti hai, yaani $ω = \frac{Δ θ}{Δ t}$ , unit rad/s. Ek rigid body ke har point ke liye same hota hai. Iska sign spin ki direction dikhata hai: positive = counter-clockwise, negative = clockwise.
$α$ (Greek "alpha") — angular acceleration: $ω$ khud kitni tezi se badal raha hai, $α = \frac{Δ ω}{Δ t}$ , unit rad/s². Positive = tez hona (in the $+$ sense), negative = dheema hona.
$v$ — linear (tangential) speed, m/s, circle ke saath. Jab hum $ω$ ka sign rakhte hain, $v = r ω$ tangent ke saath ek signed velocity hai; iska magnitude (kitna tez, koi direction nahi) $∣ v ∣ = r ∣ ω ∣$ hai.
$a_{t}$ — tangential acceleration (speed kitni tezi se badlti hai), m/s²; $a_{c}$ — centripetal acceleration (centre ki taraf point karta hai, direction badalta hai), m/s².
$ϕ$ (Greek "phi") — ek angle jo hum Ex 6 mein compute karenge: total acceleration vector ka woh chhota sa tilt jo pure inward (radial) direction se door hai.

Scenario matrix

Is topic ke har problem inhi cells mein se ek hai. Columns hain kya badal raha hai; rows hain kaunsi special value involved hai.

Cell	Kya special hai	Kaunsa formula stressed hai	Example
A	Plain uniform spin ( $α = 0$ )	$v = r ω$ , $a_{c} = r ω^{2}$	Ex 1
B	Same $ω$ , alag $r$	$v = r ω$ scaling	Ex 2
C	Degenerate: axis par point, $r = 0$	all $\to 0$	Ex 3
D	Speeding up ( $α > 0$ )	$a_{t} = r α$	Ex 4
E	Slowing down ( $α < 0$ , sign matters)	$a_{t} = r α$ negative	Ex 5
F	Both accelerations at once (vector total)	$a = a_{t}^{2} + a_{c}^{2}$ + angle	Ex 6
G	Unit trap: degrees / rpm given, must convert	radian discipline	Ex 7
H	Limiting behaviour: $ω \to$ large, what dominates?	$a_{c} \sim ω^{2}$ growth	Ex 8
I	Real-world word problem	pick the right bridge	Ex 9
J	Exam twist: work backwards (given $v$ , find $ω$ or $r$ )	invert the formula	Ex 10
K	Negative $ω$ (clockwise spin): sign of $v$	signed $v = r ω$	Ex 11

Har cell neeche fill ki gayi hai. Chalo shuru karte hain.

Ex 1 — Cell A: plain uniform spin

Worked example Ek grinding wheel steadily

$ω = 8 rad/s$ par spin karta hai. Ek point $r = 0.15 m$ par baitha hai. Uski speed $v$ aur acceleration nikalo.

Forecast: Steady spin ka matlab $α = 0$ hai. Toh kya point bilkul bhi accelerate kar raha hai? Padhne se pehle yes/no guess karo.

Speed. $v = r ω = 0.15 \times 8 = 1.2 m/s$ . Yeh step kyun? $ω$ already rad/s mein hai, isliye hum directly multiply karte hain — bridge $v = r ω$ ko koi conversion nahi chahiye.
Tangential acceleration. Steady spin $\Rightarrow α = 0 \Rightarrow a_{t} = r α = 0$ . Yeh step kyun? $a_{t}$ tabhi exist karta hai jab turning ki rate badlti hai. Yahan nahi badlti.
Centripetal acceleration. $a_{c} = r ω^{2} = 0.15 \times 8^{2} = 0.15 \times 64 = 9.6 m/s^{2}$ . Yeh step kyun? Constant speed par bhi velocity direction ghoomti rehti hai, isliye center ki taraf hamesha ek pull hota hai.

Verify: Cross-check $a_{c} = v^{2} / r = 1. 2^{2} /0.15 = 1.44/0.15 = 9.6 m/s^{2}$ se karo. ✓ $a_{c}$ ke dono formulas agree karte hain, isliye answer consistent hai. Units: $m \cdot (rad/s)^{2} = m/s^{2}$ (radian dimensionless hai). ✓

Ex 2 — Cell B: same $ω$ , alag radius

Picture baat bilkul clear kar deti hai: same angular speed, lekin outer bug same time mein bada circle sweep karta hai, isliye usse tez move karna padta hai.

Ex 3 — Cell C: degenerate case, axis par bilkul ek point

Worked example Grinding wheel ka bilkul center (

$r = 0$ ) same $ω = 8 rad/s$ ke saath spin karta hai. Uski speed aur dono accelerations nikalo.

Forecast: Bilkul axis par ka ek point — kya disk spin karte waqt woh bilkul bhi move karta hai?

Speed. $v = r ω = 0 \times 8 = 0 m/s$ . Yeh step kyun? Axis point kabhi circle trace nahi karta — woh bas apni jagah spin karta hai, isliye zero arc length cover karta hai.
Tangential accel. $a_{t} = r α = 0$ ( $α$ chahe kuch bhi ho).
Centripetal accel. $a_{c} = r ω^{2} = 0$ . Yeh step kyun? Koi circle nahi jis mein pull ho $\Rightarrow$ koi center-pointing acceleration nahi.

Verify: Har linear quantity $r$ times ek angular quantity hai, isliye $r = 0$ par sab vanish ho jaate hain — point angularly busy hai lekin linearly rest mein hai. Yeh woh boundary case hai jo poori "farther = faster" rule ko anchor karta hai.

Ex 4 — Cell D: speeding up ( $α > 0$ )

Worked example Ek potter's wheel ka radius

$r = 0.20 m$ hai, rest se shuru hokar $6 s$ mein $ω = 12 rad/s$ tak pahunchta hai (uniform angular acceleration). $α$ aur rim ki tangential acceleration $a_{t}$ nikalo.

Forecast: Kya spin-up ke dauran $a_{t}$ badlega, ya constant rahega?

Angular acceleration. $α = \frac{Δ ω}{Δ t} = \frac{12 - 0}{6} = 2 rad/s^{2}$ . Yeh step kyun? $α$ $ω$ ke change ki rate hai; uniform speed-up ke saath yeh total time $Δ t$ mein total change hi hai.
Tangential accel. $a_{t} = r α = 0.20 \times 2 = 0.4 m/s^{2}$ . Yeh step kyun? $a_{t}$ angular "tez hone" ko tangent ke saath linear "tez hone" mein convert karta hai.

Verify: Kyunki $r$ aur $α$ dono constant hain, $a_{t} = 0.4 m/s^{2}$ poore time constant rehta hai — yeh current $ω$ par depend nahi karta. Kinematics se sanity check: $t = 6$ par rim speed $v = r ω = 0.20 \times 12 = 2.4 m/s$ hai, aur $v = a_{t} t = 0.4 \times 6 = 2.4 m/s$ . ✓ Match karte hain.

Ex 5 — Cell E: slowing down ( $α < 0$ , sign matters)

Worked example Ek bicycle wheel, radius

$r = 0.35 m$ , $ω_{0} = 15 rad/s$ par spin kar raha hai aur $Δ t = 5 s$ mein brake karke ruk jaata hai. $α$ , tangential acceleration $a_{t}$ nikalo, aur sign interpret karo.

Forecast: Kya $a_{t}$ negative aayega, aur negative $a_{t}$ physically kya matlab rakhta hai?

Angular acceleration. $α = \frac{ω _{final} - ω _{0}}{Δ t} = \frac{0 - 15}{5} = - 3 rad/s^{2}$ . Yeh step kyun? Slowing down ka matlab $ω$ decrease karta hai, isliye $Δ ω < 0$ aur $α$ negative hai. Sign decoration nahi hai — yeh direction encode karta hai.
Tangential accel. $a_{t} = r α = 0.35 \times (- 3) = - 1.05 m/s^{2}$ . Yeh step kyun? Minus sign batata hai ki $a_{t}$ motion ke against point karta hai (tangent ke saath peeche) — yeh ek braking acceleration hai.

Verify: Speed loss ka magnitude: $∣ a_{t} ∣ \times Δ t = 1.05 \times 5 = 5.25 m/s$ . Initial rim speed thi $v_{0} = r ω_{0} = 0.35 \times 15 = 5.25 m/s$ , jo $0$ tak drop honi chahiye. ✓ Exactly consume ho gayi. Sign ne correctly full stop deliver kiya, naa zyada naa kam.

Recall Kya centripetal acceleration braking ke dauran bhi peeche point karta hai?

Nahi — $a_{c} = r ω^{2}$ hamesha $\geq 0$ hai aur hamesha center ki taraf point karta hai, chahe $α$ ka sign kuch bhi ho. Sirf tangential piece sign flip karta hai.

Ex 6 — Cell F: dono accelerations, vector total (geometry)

Worked example Jis instant Ex 5 ka braking wheel

$ω = 15 rad/s$ par hai, $α = - 3 rad/s^{2}$ ke saath, $r = 0.35 m$ : $a_{c}$ , $a_{t}$ , total acceleration magnitude, aur woh angle $ϕ$ nikalo jo yeh radius ke saath banata hai.

Forecast: Is instant par kaunsa bada hai — center-pull $a_{c}$ ya braking $a_{t}$ ?

Centripetal. $a_{c} = r ω^{2} = 0.35 \times 1 5^{2} = 0.35 \times 225 = 78.75 m/s^{2}$ . Yeh step kyun? $a_{c}$ motion ke direction-change part ko handle karta hai; high $ω$ par dominate karta hai.
Tangential (magnitude). $∣ a_{t} ∣ = ∣ r α ∣ = 0.35 \times 3 = 1.05 m/s^{2}$ . Yeh step kyun? Yeh speed-change part hai; yahan chhota hai kyunki braking gentle hai.
Total magnitude. Dono perpendicular hain ( $a_{c}$ radius ke saath, $a_{t}$ tangent ke saath), isliye Pythagoras use karo: $a = a_{c}^{2} + a_{t}^{2} = 78.7 5^{2} + 1.0 5^{2} \approx 78.75 m/s^{2} .$ Yeh step kyun? Perpendicular vectors right triangle ki legs ki tarah add hote hain — exactly yahi Pythagoras ke liye hai. Yahan $a_{t}$ itna tiny hai ki, $a_{c}$ jitni hi 2-decimal precision par, total $a_{c}$ se alag nahi dikh raha.
Angle $ϕ$ (inward) radius se. $ϕ = arctan (\frac{∣ a _{t} ∣}{a _{c}}) = arctan (\frac{1.05}{78.75}) \approx 0.76 4^{\circ}$ . Yeh step kyun? Hum pooch rahe hain "kis angle ka tangent-to-radial ratio yahi hai?" — yahi $arctan$ answer karta hai, aur yeh total vector ko pure inward direction se thoda sa tilt kar deta hai. $ϕ$ wahi tilt hai.

Verify: Kyunki $a_{t} ≪ a_{c}$ hai, total $a_{c}$ se thoda sa upar hona chahiye: $78.7 5^{2} + 1.0 5^{2} = 78.757 \dots$ , jo 2 d.p. par $78.75 m/s^{2}$ round hota hai. ✓ Aur tilt angle tiny hona chahiye: $0.7 6^{\circ}$ center ki taraf almost seedha hai. ✓

Figure right triangle dikhata hai: ek lamba inward $a_{c}$ leg, ek chhota tangential $a_{t}$ leg, aur thoda sa tilted total arrow (hypotenuse).

Ex 7 — Cell G: unit trap (rpm aur degrees)

Ex 8 — Cell H: limiting behaviour, $ω$ badhne par kya dominate karta hai

Worked example

$r = 0.3 m$ wale ek rotor mein fixed spin-up $α = 4 rad/s^{2}$ hai. $ω = 2$ par aur phir $ω = 20 rad/s$ par $a_{t}$ aur $a_{c}$ compute karo. $ω$ badhne par kaunsa acceleration dominate karta hai?

Forecast: Jab $ω$ badhta hai, kya $a_{t}$ aur $a_{c}$ same pace se badhte hain?

Tangential (unchanged). $a_{t} = r α = 0.3 \times 4 = 1.2 m/s^{2}$ dono speeds par. Yeh step kyun? $a_{t}$ mein $ω$ nahi hai — chahe kitna bhi tez spin karo, yeh apni jagah rehta hai.
Centripetal at $ω = 2$ . $a_{c} = r ω^{2} = 0.3 \times 4 = 1.2 m/s^{2}$ .
Centripetal at $ω = 20$ . $a_{c} = 0.3 \times 400 = 120 m/s^{2}$ . Yeh step kyun? $a_{c}$ $ω^{2}$ ke saath badhta hai — square ke saath — isliye $10 \times$ tez spin $100 \times$ center-pull deta hai.

Verify: Ratio $a_{c} (20) / a_{c} (2) = 120/1.2 = 100 = 1 0^{2}$ . ✓ Square law confirm ho gaya. Limiting conclusion: jab $ω \to$ bada, $a_{c}$ ( $\propto ω^{2}$ ) constant $a_{t}$ ko completely overwhelm kar deta hai — yahi reason hai ki fast rotors rim par fly apart karte hain, tangential push ki wajah se nahi.

Graph ise visual banata hai: flat $a_{t}$ line versus $a_{c}$ ka parabola jo ise bahut peeche chhod deta hai.

Ex 9 — Cell I: real-world word problem

Worked example Ek car tyre ka radius

$r = 0.31 m$ hai. Car (bina skid kiye) $v = 25 m/s$ par seedhi road par chal rahi hai. Tyre kitna tez spin kar raha hai ( $ω$ ), aur tread mein fansa ek pathar kaunsi centripetal acceleration feel karta hai?

Forecast: Seedhi road par tyre mein ek pathar — kya woh accelerate kar raha hai jabki car seedhi jaa rahi hai?

Rolling se angular speed. No skidding ka matlab tread speed road speed $v$ ke barabar hai, isliye bridge ulta karo: $ω = \frac{v}{r} = \frac{25}{0.31} \approx 80.6 rad/s$ . Yeh step kyun? Rolling without slipping road speed ko same bridge $v = r ω$ ke zariye spin se link karta hai, bas $ω$ ke liye solve kiya. Dekho Rolling motion.
Pathar par centripetal accel. Wheel ke frame mein pathar axle ke around circle karta hai: $a_{c} = r ω^{2} = 0.31 \times 80. 6^{2} \approx 0.31 \times 6503 \approx 2016 m/s^{2}$ . Yeh step kyun? Axle ke relative pathar abhi bhi circle trace karta hai, isliye woh center-pull feel karta hai chahe car seedhi line mein chal rahi ho.

Verify: Cross-check $a_{c} = v^{2} / r = 2 5^{2} /0.31 = 625/0.31 \approx 2016 m/s^{2}$ se karo. ✓ Dono routes agree karte hain. Woh $200 g$ se zyada hai — exactly yahi reason hai ki tyres se pathar aur mud bahar fekte hain.

Ex 10 — Cell J: exam twist, ulta kaam karo

Worked example Ek spinning disk ke rim point ko

$v = 6 m/s$ par move karte aur $a_{c} = 48 m/s^{2}$ centripetal acceleration feel karte measure kiya gaya. Radius $r$ aur angular speed $ω$ nikalo — bina kisi ko directly bataye.

Forecast: Tumhare paas do facts hain aur do unknowns hain. Kaunsa formula pair ise cleanly unlock karta hai?

$r$ paane ke liye $a_{c} = v^{2} / r$ use karo. Rearranged: $r = \frac{v ^{2}}{a _{c}} = \frac{6 ^{2}}{48} = \frac{36}{48} = 0.75 m$ . Yeh step kyun? $a_{c}$ ka yeh form sirf woh do cheezein contain karta hai jo hum jaante hain ( $v$ aur $a_{c}$ ) plus unknown $r$ — isliye hum $r$ ko immediately isolate kar sakte hain.
$ω$ paane ke liye $v = r ω$ use karo. Rearranged: $ω = \frac{v}{r} = \frac{6}{0.75} = 8 rad/s$ . Yeh step kyun? Ab $r$ pata hai, angular speed recover karne ke liye speed bridge invert karo.

Verify: Wapas $a_{c} = r ω^{2} = 0.75 \times 8^{2} = 0.75 \times 64 = 48 m/s^{2}$ mein plug karo. ✓ Given data se match karta hai, isliye $(r, ω) = (0.75 m, 8 rad/s)$ correct hai.

Ex 11 — Cell K: negative $ω$ (clockwise spin)

Worked example Ek disk

clockwise spin karti hai, jo hamare sign convention ke hisaab se $ω = - 8 rad/s$ hai. Ek rim point $r = 0.15 m$ par baitha hai. Signed tangential velocity $v$ , iska magnitude (kitna tez), aur centripetal acceleration $a_{c}$ nikalo.

Forecast: Kya minus sign point ko slower banata hai, ya yeh sirf ek direction ke roop mein bachta hai? Aur kya $a_{c}$ ab negative hai?

Signed velocity. $v = r ω = 0.15 \times (- 8) = - 1.2 m/s$ . Yeh step kyun? $ω$ ka sign rakhte hue, $v = r ω$ use inherit karta hai: minus ka matlab hai point clockwise tangential sense mein move karta hai (counter-clockwise $+$ direction ke opposite). Number negative hai lekin woh ek direction flag hai, slower speed nahi.
Magnitude (kitna tez). $∣ v ∣ = r ∣ ω ∣ = 0.15 \times 8 = 1.2 m/s$ . Yeh step kyun? "Kitna tez" sign strip kar deta hai: $∣ v ∣ = r ∣ ω ∣$ . Ex 1 ke $ω = + 8$ case jitni hi speed — sirf direction alag hai.
Centripetal accel. $a_{c} = r ω^{2} = 0.15 \times (- 8)^{2} = 0.15 \times 64 = 9.6 m/s^{2}$ . Yeh step kyun? $ω$ squared hai, isliye $(- 8)^{2} = 64$ : sign vanish ho jaata hai. Centripetal acceleration hamesha inward point karta hai aur iska magnitude care nahi karta ki tum kis taraf spin karte ho.

Verify: $a_{c}$ exactly Ex 1 se match karta hai ( $9.6 m/s^{2}$ ), confirming karta hai ki square sign khatam kar deta hai; aur $∣ v ∣$ Ex 1 ki speed se match karta hai ( $1.2 m/s$ ), confirming karta hai ki sirf $v$ ki direction flip hui. ✓

Recap: kaun se cell ko kaun sa move chahiye

Recall Matrix ke across quick self-test

$ω$ double karne par (fixed $r$ , fixed $α$ ) $v$ , $a_{t}$ , $a_{c}$ kis factor se badhte hain? ::: $v$ double ho jaata hai, $a_{t}$ unchanged rehta hai, $a_{c}$ chaar guna ho jaata hai. $r = 0$ par, $v$ , $a_{t}$ , $a_{c}$ kya hain? ::: Sab zero. Negative $α$ kaun si acceleration ko negative banata hai? ::: Sirf $a_{t}$ ko; $a_{c}$ positive (inward) rehta hai. Negative $ω$ kaun si quantities ko negative banata hai? ::: Sirf signed $v$ ko (iska direction flip ho jaata hai); $a_{c} = r ω^{2}$ positive rehta hai kyunki $ω$ squared hai. $210$ rpm ko rad/s mein convert karne ke liye tum kis se multiply karte ho? ::: $\frac{2 π}{60} = \frac{π}{30}$ .