Circular motion — centripetal acceleration derivation: Question bank

1.2.15 · D5 · Physics › Newton's Laws & Dynamics › Circular motion — centripetal acceleration derivation

Shuru karne se pehle, is page par sirf yahi symbols use hote hain:

$v$ = speed (kitna tez, ek plain number, metres per second).
$v$ = velocity (speed with direction — ek arrow).
$r$ = circle ka radius (metres).
$m$ = moving object ki mass (kilograms).
$ω$ = angular speed (radians turned per second — dekho Angular velocity and period).
$a_{c} = \frac{v ^{2}}{r} = ω^{2} r = v ω$ = inward (centre-seeking) acceleration ki magnitude.
$F_{c} = m a_{c}$ = woh net inward force jo turn ko jaari rakhne ke liye chahiye.

Do choti pictures neeche ke kai traps ke liye geometry set up karti hain. Inhe saamne rakhna.

$a_{c} = v^{2} / r$ ke peeche ke do triangles. Left mein, points A aur B tak ke do radius arrows (length $r$ ) swept angle $Δ θ$ banate hain; chord AB unki short side hai. Right mein, do velocity arrows (length $v$ ), har ek apne radius se $9 0^{\circ}$ muda hua, wahi $Δ θ$ banate hain — to do triangles similar hain, aur $∣Δ v ∣/ v = ∣ AB ∣/ r$ .

Figure — Circular motion — centripetal acceleration derivation

Chord vs. arc. Straight chord AB hamesha thodi choti hoti hai us curved arc se jo woh katti hai. Sirf jab $Δ θ$ zero ki taraf shrink hota hai tab dono lengths milti hain — yehi wajah hai ki derivation ko limit lena padta hai, yeh claim nahi karna ki woh equal hain.

True or false — justify karo

Har jawab mein wajah deni hai, sirf verdict nahi.

Constant speed par circle par, acceleration zero hai.

False. Speed constant hai lekin

v

ki direction har instant mein ghoomti hai, isliye

Δ v \neq = 0

; acceleration centre ki taraf point karta hai magnitude

v^{2} / r

ke saath.

Centripetal acceleration moving object par koi work nahi karta.

True. Yeh hamesha

v

ke perpendicular hota hai, aur work

= F \cdot d

perpendicular vectors ke liye zero ho jaata hai, isliye yeh sirf direction badalta hai, kabhi speed nahi.

Agar speed constant hai, toh

a

ki magnitude bhi constant hai.

Fixed radius ke circle ke liye True:

a_{c} = v^{2} / r

mein teeno quantities fixed hain, isliye

∣ a ∣

constant hai, chahe

a

ki direction andar ki taraf rotate karti rahe.

Centripetal force ek extra force hai jo tension ya gravity ke upar add karni padti hai.

False. Yeh net inward force ka naam hai, aur woh net force tension, gravity, friction ya normal force se supply hoti hai — dono draw karna double-counting hai (dekho Newton's Second Law).

Radius double karna (same speed) centripetal acceleration ko half kar deta hai.

True.

a_{c} = v^{2} / r

, aur

r

denominator mein first power par hai, isliye

r \to 2 r

a_{c} \to a_{c} /2

ho jaata hai.

Speed double karna (same radius) centripetal acceleration ko double kar deta hai.

False.

a_{c} \propto v^{2}

, isliye

v

double karne se

a_{c}

chaar guna ho jaata hai — yehi wajah hai ki fast corners itne unforgiving hote hain.

Uniform circular motion mein velocity aur acceleration vectors same direction mein point karte hain.

False.

v

tangent hai (motion ke saath);

a

radial-inward hai. Yeh har instant par exactly

9 0^{\circ}

apart hain.

Circular orbit mein ek satellite Earth ki taraf "gir" raha hai.

True. Gravity uski centripetal acceleration supply karti hai, isliye woh continuously Earth ke centre ki taraf accelerate karta hai — woh bas itni tez sideways jaata hai ki miss karta rehta hai (dekho Gravitation — orbital motion).

Agar ghoomte hue patthar ki string kaat do, toh patthar radially bahar ud jaata hai.

False. Koi inward force nahi toh inertia le leta hai aur woh tangent ke saath ud jaata hai (straight line), radius ke saath bahar nahi.

Centrifugal force ground frame mein object par act karne wali real force hai.

False. Inertial (ground) frame mein koi outward force nahi hoti; centrifugal force ek bookkeeping term hai jo sirf rotating frame mein appear hoti hai (dekho Centrifugal force (non-inertial frames)).

Formulas

v^{2} / r

aur

ω^{2} r

same motion ke liye alag numbers de sakte hain.

False. Yeh

v = ω r

ke zariye algebraically identical hain; agar consistent values plug karo toh zaroor match hona chahiye — mismatch ka matlab arithmetic slip hai.

Error dhundho

Har statement mein flaw dhundho, phir batao sahi kya hai.

a_{c} = v^{2} r

, isliye bada radius matlab bada acceleration."

Radius denominator mein jaata hai:

a_{c} = v^{2} / r

. Bada radius (same speed) matlab gentler turn, isliye chota acceleration.

"Kyunki

a ⊥ v

hai, acceleration velocity ko bilkul nahi badal sakta."

Yeh magnitude (speed) nahi badalta, lekin

v

ki direction badalta hai — woh directional turning hi puri acceleration ki point hai.

a_{c} = ω^{2} / r

— angular speed squared, radius se divide."

r

ki jagah galat hai: yeh

ω^{2} r

hai (radius multiply hota hai). Dimension check:

(rad/s)^{2} \cdot m = m/s^{2}

✓, jabki

ω^{2} / r

deta hai

1/ (s^{2} m)

, jo acceleration nahi hai.

"Similar-triangles derivation mein, velocity vectors ek alag angle banate hain radius vectors se."

Yeh same angle

Δ θ

banate hain — upar figure mein do triangles dekho. Har

v

apne radius ke perpendicular hai, aur dono radii ko

9 0^{\circ}

rotate karne par unke beech ka angle preserve hota hai.

"Chord AB arc ke barabar hai, isliye

∣ AB ∣ = r Δ θ

exactly."

Yeh sirf approximately equal hai, aur sirf chhote

Δ θ

ke liye (upar chord-vs-arc figure dekho); derivation limit

Δ t \to 0

lene se bach jaati hai jahan chord aur arc ek ho jaate hain.

"Circle par constant speed par ek body Newton's first law follow karti hai — koi net force nahi."

Curved path straight nahi hai, isliye ek net force hai (centripetal wali). First law sirf tab straight-line motion guarantee karta hai jab net force zero ho.

Why questions

Mechanism explain karo, sirf label nahi.

Centripetal acceleration centre ki taraf kyun point karta hai, motion ke saath kyun nahi?

Kyunki velocity mein change,

Δ v = v_{B} - v_{A}

, chhote step ki limit mein centre ki taraf inward point karta hai — wahi ek direction hai jo path ko speed badlaye bina bend karta hai.

Speed constant rehne par bhi acceleration kyun hai?

Speedometer sirf

∣ v ∣

read karta hai; acceleration kisi bhi change par respond karta hai

v

mein, aur direction continuously change ho rahi hai.

Required centripetal force speed ke square ke saath kyun badhta hai?

Do effects mili kar demand ko

v

ke factor se multiply karte hain: (1) ek tez object same chhote angle

Δ θ

ko kam time mein cross karta hai,

Δ t = Δ θ / ω

jahan

ω = v / r

, isliye

Δ t \propto 1/ v

; (2) velocity triangle se velocity change ki size

∣Δ v ∣ = v Δ θ

hai, jo khud

v

ke saath badhti hai. Ek

∣Δ v ∣

v

ke scale karta hai us

Δ t

se divide karo jo

1/ v

scale karta hai, milta hai

a_{c} \propto v^{2}

— neeche teesra figure dono scalings visible karta hai.

Gravity akele kisi planet ko circular orbit mein kisi rope ya track ke bina kyun rakh sakti hai?

Gravity central body ki taraf inward pull provide karti hai, aur jab gravitational force ki magnitude

m v^{2} / r

ke barabar ho jaati hai toh planet bilkul sahi rate par ghoom ke circle trace karta hai (dekho Gravitation — orbital motion).

Fast curve par road ko bank (tilt) kyun karte hain friction par rely karne ki jagah?

Tilting se normal force ek inward horizontal component gain karta hai jo centripetal force supply karta hai, isliye turn thodi ya zero friction mein bhi kaam karta hai (dekho Banking of roads).

Position vector ko twice differentiate karna

a_{c}

nikalne ka valid doosra tarika kyun hai?

Acceleration by definition position ka second time-derivative hai; ise

r = r (cos ω t, sin ω t)

par apply karne se

- ω^{2} r

milta hai, jiski magnitude

ω^{2} r = v^{2} / r

hai — jo geometric result se match karta hai.

Calculus method mein

a = - ω^{2} r

mein minus sign kyun aata hai?

Minus signal karta hai ki

a

outward radius vector

r

ke opposite point karta hai, yaani seedha centre ki taraf — wahi "centre-seeking" conclusion jo geometry deti hai (dekho Vectors — derivative of a unit vector).

Edge cases

Boundary aur degenerate scenarios jinhe formulas ko phir bhi handle karna hai.

a_{c}

kya hai agar speed exactly zero ho?

Zero:

a_{c} = v^{2} / r = 0

. Momentarily stationary object ka koi centripetal acceleration nahi — koi direction of motion hi nahi hai jo bend ho.

a_{c}

ka kya hota hai jab radius bina bound ke badhta jaata hai (path seedha ho jaata hai)?

a_{c} = v^{2} / r \to 0

jab

r \to \infty

. "Infinite-radius circle" ek straight line hai, jise koi inward acceleration nahi chahiye — Newton's First Law (Inertia) ke saath consistent.

Kya centripetal acceleration defined hai jab radius zero ho?

Nahi:

a_{c} = v^{2} / r

blow up ho jaata hai (zero se division). Ek point par move karne ke liye koi circle nahi, isliye situation physically meaningless hai.

Agar speed khud change ho rahi hai (turn karte hue speed up), toh kya

a_{c} = v^{2} / r

phir bhi poora acceleration hai?

Nahi. Ab ek additional tangential acceleration hai

v

ke saath;

v^{2} / r

sirf inward (centripetal) part deta hai, aur total dono ka vector sum hai.

Vertical loop ke exact top par, centripetal acceleration kis direction mein point karta hai?

Seedha neeche, loop ke centre ki taraf — isliye gravity (bhi neeche) usse supply karne mein help kar sakti hai, yehi wajah hai ki contact maintain karne ke liye minimum speed exist karti hai.

String par lage patthar ko vertical circle ke bottom par same reason se minimum speed chahiye jaise top par?

Nahi — bottom par, tension upar (inward) point karta hai jabki gravity neeche (outward), isliye tension ko gravity overcome bhi karni hoti hai aur

m v^{2} / r

supply bhi karni hoti hai; tension yahan sabse zyada hoti hai, speed constraint nahi.

Recall Ek-line self-test

Agar koi dost kahe "fast turn mein door mujhe bahar ki taraf push karta hai," toh ek sentence mein correction kya hai? Aapka inertia seedha jaana chahta hai; door actually aapko andar ki taraf push karta hai taaki aap curve ke saath ghoom sako — "baahri" sensation ek outward force ki absence hai, uski presence nahi.