1.2.14 · D3Rotating frames — centrifugal force, Coriolis force
Intuition Yeh page kis liye haiFoundations ne humein do fictitious-force formulas diye. Visual walkthrough ne dikhaya ki woh dikhte kaise hain. Ab hum har tarah ke case ke through grind karte hain jo ek problem tumhare saamne rakh sakti hai — cross product ke har sign ke liye, carousel ke har quadrant ke liye, degenerate "standing still" case ke liye, limiting "far from axis" case ke liye, aur ek real weather problem ke liye. Jab tum finish karo, koi bhi exam scenario naya nahi lagna chahiye.
Kuch bhi shuru karne se pehle, do tools ko fresh tarike se anchor karte hain jo hum baar baar use karte rahe, seedhe saadhe words mein:
Is topic ka har problem neeche diye gaye cells mein se ek hai (ya blend). Har worked example us cell ke saath tagged hai jo woh cover karta hai.
Cell
Case class
Covered by
C1 Centrifugal, position-only, standard magnitude
Ex 1
C2 Coriolis, velocity radially outward (z ^ × r ^
Ex 2
C3 Coriolis, velocity tangential (z ^ × θ ^
Ex 3
C4 Degenerate : frame mein object at rest (v rot = 0 Ex 4
C5 Dono forces ek saath , vectors ke roop mein add karna, aur sizes compare karnaEx 5
C6 Limiting behaviour : axis par (r ⊥ → 0 Ex 6
C7 Real-world word problem : Earth, hemisphere sign, time ke saath accumulationEx 7
C8 Exam twist : velocity axis ke along (v ∥ ω Ex 8
Hum ek convention fix karte hain jo poore note mein use hogi: spin axis z ^ ω = ω z ^ ω > 0 upar se dekha jaye toh counter-clockwise . Flat turntable par hum polar directions use karte hain: r ^ θ ^ r ^
z ^ × r ^ = θ ^ , z ^ × θ ^ = − r ^ , r ^ × θ ^ = z ^ .
Figure dekho: r ^ θ ^ z ^ z ^ r ^ θ ^
Worked example Carousel par baithnaTum r ⊥ = 2 m ω = 1.5 rad/s m = 60 kg
Forecast: bada radius ya zyada fast spin → bada fling. Aage padhne se pehle number guess karo.
Step 1 — sahi tool chuno. Seat ke relative koi velocity mention nahi hai (tum still baithe ho), toh sirf centrifugal matter karta hai. Iska magnitude position-only hai:
F cf = m ω 2 r ⊥ . Yeh step kyun? Coriolis ko v rot = 0
Step 2 — plug in karo.
F cf = 60 × ( 1.5 ) 2 × 2 = 60 × 2.25 × 2 = 270 N . Yeh step kyun? Direct substitution; units kg ⋅ ( rad/s ) 2 ⋅ m = kg⋅m/s 2 = N
Step 3 — real force interpret karo. Inertial frame mein koi outward force nahi hai. Seat tumhe 270 N andar ki taraf push karti hai — yahi centripetal force m ω 2 r ⊥ kaisi lagti hai .
Verify: 60 ⋅ 2.25 = 135 135 ⋅ 2 = 270 N
Worked example Baahir ki taraf throw karnaUsi carousel par (ω = 1.5 z ^ seedha baahir (r ^ v = 5 m/s m = 0.2 kg
Forecast: ball kis taraf curve karti dikhegi — spin ke saath ya uske against?
Step 1 — velocity ko frame directions mein likho. v rot = 5 r ^ Yeh step kyun? Coriolis sirf object ki velocity se matlab rakhta hai jaise rotating frame ke andar measure ki gayi hai , aur radial-out exactly r ^
Step 2 — cross product banao.
F Cor = − 2 m ω × v rot = − 2 ( 0.2 ) ( 1.5 ) ( z ^ × 5 r ^ ) . z ^ × r ^ = θ ^ F Cor = − 2 ( 0.2 ) ( 1.5 ) ( 5 ) θ ^ = − 3.0 θ ^ N . Yeh step kyun? Cross product velocity direction ko 90° rotate karta hai, "outward" ko "sideways" mein badal deta hai. Minus sign phir ise − θ ^ spin direction ke opposite .
Step 3 — picture karo. Rider ki seat se ball spin ke against peeche ki taraf mur jaati hai. Baahir se, ball seedha udti hai; rider (aur turntable) uske neeche rotate karta hai.
Verify: 2 ⋅ 0.2 ⋅ 1.5 ⋅ 5 = 3.0 N − θ ^ r ^ F Cor ⋅ v rot = 0
Worked example Spin ke saath sideways move karnaWahi carousel. Ab ek mouse tangentially, spin direction mein (+ θ ^ v = 4 m/s m = 0.1 kg
Forecast: pichhli baar radial motion ne sideways push kiya. Tangential motion… kahan push karegi?
Step 1 — frame mein velocity. v rot = 4 θ ^ Yeh step kyun? Spin ke along daudna matlab velocity θ ^
Step 2 — doosri handedness ke saath cross product. z ^ × θ ^ = − r ^ F Cor = − 2 ( 0.1 ) ( 1.5 ) ( z ^ × 4 θ ^ ) = − 2 ( 0.1 ) ( 1.5 ) ( 4 ) ( − r ^ ) = + 1.2 r ^ N . Yeh step kyun? Yeh matrix mein doosra sign case hai. Tangential motion ek radial Coriolis push produce karta hai — yahan outward (+ r ^ saath move kar raha hai, effectively apni circling speed up kar raha hai aur frame jo provide karta hai usse zyada inward force ki zaroorat hai.
Step 3 — direction reverse karo, sign flip karo (C3 ki sanity). Agar mouse spin ke against daudta (− θ ^ − 1.2 r ^ v
Verify: 2 ⋅ 0.1 ⋅ 1.5 ⋅ 4 = 1.2 N r ^ θ ^
Worked example Khade rehnaEk 70 kg r ⊥ = 3 m khada rehta hai , ω = 1.5 rad/s
Forecast: kya koi Coriolis hota hai jab tum move nahi karte?
Step 1 — Coriolis. v rot = 0 F Cor = − 2 m ω × 0 = 0. Yeh step kyun? Coriolis velocity frame ke andar ke proportional hai. Zero velocity ⇒ zero Coriolis. Yeh sabse common student error hai jo bachni chahiye.
Step 2 — Centrifugal abhi bhi act karta hai.
F cf = m ω 2 r ⊥ = 70 × ( 1.5 ) 2 × 3 = 70 × 2.25 × 3 = 472.5 N , outward . Yeh step kyun? Centrifugal sirf position par depend karta hai, velocity par nahi — jab tak tum spinning platform par baithe ho tab tak yeh kabhi off nahi hota.
Verify: 70 ⋅ 2.25 = 157.5 157.5 ⋅ 3 = 472.5 N = 0 = 0
Worked example Kaun dominate karta hai?Ek 0.5 kg radially outward v = 6 m/s r ⊥ = 1 m ω = 2 rad/s
Forecast: 1 2
Step 1 — centrifugal (radial, outward, + r ^
F cf = m ω 2 r ⊥ = 0.5 × ( 2 ) 2 × 1 = 2.0 N ( + r ^ ) . Yeh step kyun? Position-based, hamesha r ^
Step 2 — Coriolis (radial-out velocity → tangential force). v rot = 6 r ^ z ^ × r ^ = θ ^ F Cor = − 2 ( 0.5 ) ( 2 ) ( 6 ) θ ^ = − 12 θ ^ N . Yeh step kyun? Ex 2 jaisi C2 machinery; yahan number bahut bada hai kyunki ω v
Step 3 — perpendicular vectors ki tarah add karo. + 2.0 r ^ − 12 θ ^ ∣ F total ∣ = 2. 0 2 + 1 2 2 = 4 + 144 = 148 ≈ 12.17 N . Yeh step kyun? r ^ ⊥ θ ^
Verify: 2 ⋅ 0.5 ⋅ 2 ⋅ 6 = 12 4 + 144 = 148 ≈ 12.166
Worked example Formula ki edgesEk bug ek turntable par ω = 3 rad/s radially outward v = 1 m/s karta hai, mass m = 0.01 kg axis par r ⊥ = 0 r ⊥ = 10 m
Forecast: centre par kaun si force disappear hoti hai, aur kaun si dono jagah same rehti hai?
Step 1 — centrifugal at r ⊥ = 0
F cf = m ω 2 ⋅ 0 = 0. Yeh step kyun? Centrifugal ∝ r ⊥ nahi hai, toh yeh vanish ho jaata hai. Yeh "on the axis" limiting cell hai.
Step 2 — centrifugal at r ⊥ = 10 m
F cf = 0.01 × 9 × 10 = 0.9 N . Yeh step kyun? Centrifugal distance ke saath linearly grow karta hai — bahut door par yeh chhote mass ke liye bhi bada ho sakta hai.
Step 3 — dono jagah Coriolis.
F Cor = 2 mω v = 2 ( 0.01 ) ( 3 ) ( 1 ) = 0.06 N ( same at r ⊥ = 0 aur 10 m ) . Yeh step kyun? Coriolis mein koi r ⊥ — yeh velocity par depend karta hai, position par nahi. Toh jab bug baahir move karta hai tab bhi yeh unchanged rehta hai. Yahi contrast (centrifugal grow karta hai, Coriolis constant) is cell ka poora point hai.
Verify: 0.01 ⋅ 9 ⋅ 10 = 0.9 N 2 ⋅ 0.01 ⋅ 3 ⋅ 1 = 0.06 N = 0
Worked example Northern Hemisphere ke storms counter-clockwise kyun spin karte hainHawa ek low-pressure centre ki taraf v = 10 m/s ω ⊕ = 7.3 × 1 0 − 5 rad/s ω local vertical component lo taaki effective ω zameen se upar point kare. Coriolis acceleration nikalo aur rotation ka sense explain karo. (Dekho Coriolis Effect in Weather .)
Forecast: aata hua air apni left mein mudta hai ya right mein?
Step 1 — Coriolis acceleration magnitude.
a Cor = 2 ω ⊕ v = 2 ( 7.3 × 1 0 − 5 ) ( 10 ) = 1.46 × 1 0 − 3 m/s 2 . Yeh step kyun? Hum per-mass effect (acceleration) chahte hain, toh force formula ko m a Cor = 2 ω v
Step 2 — direction. ω − 2 ω × v right ki taraf deflect karta hai. Right-deflected inflow incoming streams ko centre ke around ek counter-clockwise swirl mein curl kar deta hai.
Yeh step kyun? z ^ × v
Step 3 — "tiny" kyun phir bhi win karta hai. L = 100 km = 1 0 5 m t = L / v = 1 0 5 /10 = 1 0 4 s Δ v = a Cor t = 1.46 × 1 0 − 3 × 1 0 4 = 14.6 m/s , Yeh step kyun? Coriolis locally weak hai lekin large distances aur long times par accumulate hota hai — yahi Coriolis Effect in Weather aur Foucault Pendulum ke peeche ka 80/20 idea hai.
Verify: 2 ⋅ 7.3 × 1 0 − 5 ⋅ 10 = 1.46 × 1 0 − 3 1.46 × 1 0 − 3 ⋅ 1 0 4 = 14.6 m/s
Worked example Trap questionEk rocket seedha rotation axis ke along upar move karta hai, v rot = v z ^ v = 200 m/s ω = 0.8 z ^ m = 50 kg r ⊥ = 0
Forecast: yeh fast move kar raha hai — surely Coriolis bahut bada hai?
Step 1 — parallel vectors ka cross product.
F Cor = − 2 m ω × v rot = − 2 ( 50 ) ( 0.8 ) ( z ^ × 200 z ^ ) . z ^ × z ^ = 0 F Cor = 0. Yeh step kyun? Cross product magnitude mein sin ( angle ) parallel to ω = 0 sin 0 = 0 koi Coriolis nahi , chahe kitni bhi fast speed ho. Sirf v perpendicular hai Coriolis produce karta hai.
Step 2 — yahan centrifugal. Axis par r ⊥ = 0 F cf = m ω 2 r ⊥ = 0 Yeh step kyun? Is configuration mein dono fictitious forces vanish ho jaate hain — "gotcha" yeh hai ki badi velocity number tumhe bada answer dene ki taraf attract karti hai.
Verify: z ^ × z ^ = 0 ⇒ F Cor = 0 r ⊥ = 0 ⇒ F cf = 0
Recall Matrix ke across quick self-testCoriolis zero hota hai jab object frame mein rest par ho ::: True (Ex 4) — v rot = 0 ω sin 0 = 0 r ⊥ = 0 r ⊥ v
