Centripetal force — what provides it in various situations: Question bank

1.2.16 · D5 · Physics › Newton's Laws & Dynamics › Centripetal force — what provides it in various situations

Definition Woh symbols jinpar hum rely karenge

$r$ = circle ka radius (center se body tak ki distance).
$v$ = speed, velocity vector $v$ ki length (jo hamesha circle ki tangent hoti hai).
$θ$ (banking / conical pendulum) = tilted surface ya string ka angle horizontal se banked road ke liye, aur vertical se conical pendulum string ke liye measure kiya jaata hai. Hum yeh convention har jagah wahan fix karenge jahan use hoti hai, kyunki yeh choice decide karti hai ki kis direction ke saath kaunsa trig function jaata hai.
$ω$ = angular velocity: body tak ka angle kitni tezi se sweep karta hai, yaani $ω = \frac{Δ θ}{Δ t}$ (radians per second). Yeh speed aur radius se $v = ω r$ ke zariye juda hua hai, kyunki time $Δ t$ mein body arc $v Δ t$ travel karti hai jo $r Δ θ$ ke barabar hai, to $Δ t$ se divide karne par $v = r ω$ milta hai.

Formula

$∣Δ v ∣ = v Δ θ$ kahan se aata hai (velocity triangle) Figure s01 dekho. Left mein circle hai: do positions jo ek chhote angle $Δ θ$ se alag hain, velocity vectors $v_{1}$ aur $v_{2}$ ke saath, har ek length $v$ ka, path ki tangent draw kiye hue.

Dono velocity arrows ko tail-to-tail copy karo (figure ke right mein). Kyunki har ek ki length $v$ hai aur tangent utne hi $Δ θ$ se rotate karti hai jitna radius ne kiya, yeh do equal sides $v$ aur apex angle $Δ θ$ wala ek isosceles triangle banate hain.
Teesri side $Δ v = v_{2} - v_{1}$ hai, velocity mein change. Chhote angle ke liye teesri side ≈ radius $v$ ke circle ka arc hai, to iska length hai (radius) × (angle) $= v Δ θ$ .
Time se divide karo: $a = \frac{∣Δ v ∣}{Δ t} = v \frac{Δ θ}{Δ t} = v ω = v \cdot \frac{v}{r} = \frac{v ^{2}}{r}$ , jo inward point karta hai (red arrow $Δ v$ center ki taraf wapas point karta hai).

Figure — Centripetal force — what provides it in various situations

Is page ko kaise use karein

True or false — justify

Uniform circular motion mein acceleration zero hota hai kyunki speed constant hai.

False — velocity ek vector hai; iska direction rotate karta hai chahe iska length fixed ho, isliye

Δ v \neq = 0

aur

a = v^{2} / r

center ki taraf point karta hai (Figure s01).

Centripetal force gravity aur electromagnetism ke saath ek alag fundamental force hai.

False — yeh sirf ek naam hai us real force ke liye jo inward pull supply karti hai; chaar fundamental forces unchanged hain.

Agar tum ek free-body diagram draw karo, to tumhe "centripetal force" label wala ek arrow add karna chahiye.

False — tum sirf real forces draw karte ho (tension, gravity, normal, friction); phir tum unke net inward component ko

m v^{2} / r

ke barabar set karte ho. Alag arrow add karna double-counting hai.

Usi curve par car ki speed double karne par inward force bhi double chahiye hoti hai.

False —

F_{c} = m v^{2} / r

v^{2}

ke saath scale karta hai, isliye double speed ke liye chaar guna inward force chahiye.

Circular orbit mein satellite par gravity kaam karti hai aur use speed up karti hai.

False — gravity center ki taraf point karti hai, velocity ke perpendicular, isliye yeh zero work karti hai; orbital speed constant rehti hai.

Frictionless banked road par exactly ek speed hoti hai jis par koi friction nahi chahiye.

True —

tan θ = v^{2} / (r g)

(jahan

θ

horizontal se hai) diye gaye angle ke liye ek single

v

fix karta hai; us speed se zyada hone par tum upar slide karte ho, kam hone par neeche, isliye us speed se alag par friction chahiye.

Magnetic field mein move karne wala charge circle kaarte waqt magnetic force se speed up hota hai.

False — magnetic force

q v B

hamesha

v

ke perpendicular hoti hai, isliye yeh koi work nahi karti; yeh sirf direction change karti hai, speed constant rakhti hai.

Vertical loop ke top par string tension upar ki taraf point karti hai.

False — top par string center ki taraf jaati hai, jo bob se neeche hai, isliye tension neeche point karti hai, gravity mein add hoti hai:

T + m g = m v^{2} / r

(Figure s02, right).

Centrifugal force hi woh hai jo tumhe sharp turn mein car door ke against throw karti hai.

False — ground (inertial) frame mein koi outward force nahi hoti; tumhari body ki inertia use straight jaate rehne deti hai jabki door curve karke tumhare mein aati hai. Centrifugal force sirf rotating frame mein ek pseudo-force ke roop mein exist karti hai.

Curve ka radius badhana kisi bhi given speed par hamesha use safer banata hai.

True — required

F_{c} = m v^{2} / r

r

badhne par kam hoti hai, isliye wider curve ko usi speed ke liye kam friction/banking chahiye, skidding se pehle zyada margin hoti hai.

Error dhundho

Ek student flat curve par car ke liye likhta hai: "

f_{s} = μ_{s} m g

hamesha, isliye

v = μ_{s} g r

har speed par."

Error:

f_{s}

sirf skid limit par

μ_{s} m g

ke barabar hoti hai. Us se neeche, friction exactly

m v^{2} / r

supply karta hai, jo maximum se kam hai;

μ_{s} g r

maximum speed hai, actual wali nahi.

Ek student conical pendulum ke liye likhta hai: "

T cos θ

centripetal force hai." (Yahan

θ

string ka vertical se angle hai, jaise Figure s03 mein.)

Error: jab

θ

vertical se measure kiya jaata hai, to horizontal component

T sin θ

hota hai — woh inward point karta hai aur centripetal force hai; vertical component

T cos θ

gravity balance karta hai (

T cos θ = m g

Ek student loop ke top par likhta hai: "

T = m g + m v^{2} / r

Error: sign flip.

T

aur

m g

dono inward (neeche) point karte hain, aur saath mein required inward force ke barabar hain:

T + m g = m v^{2} / r

, isliye

T = m v^{2} / r - m g

Ek student orbit ko "

m g = m v^{2} / r

isliye

v = g r

" set up karta hai Earth se door satellite ke liye.

Error: wahan gravity

GM m / r^{2}

hai,

m g

nahi (

g = 9.8

sirf surface value hai). Correct:

GM m / r^{2} = m v^{2} / r \Rightarrow v = GM / r

Ek student kehta hai: "Banked road par friction kabhi involve nahi hoti, isliye

tan θ = v^{2} / (r g)

har speed par hold karta hai."

Error:

tan θ = v^{2} / (r g)

(jahan

θ

horizontal se hai) ek specific speed ke liye frictionless design equation hai. Doosri speeds par friction contribute karta hai, aur real safe range us ideal speed ke aas-paas ek band hoti hai.

Ek student field mein charge ke liye likhta hai: "

r = m v / (q B)

, isliye tez charges chhhote loops mein circle karte hain."

Error:

r

v

ke proportional hai (yeh numerator mein hai), isliye tez charges bade loops mein circle karte hain, chhote mein nahi.

Ek student, banked road ke liye, "

θ

horizontal se measure karte hue

N cos θ

centripetal force" likhta hai.

Error:

θ

horizontal se hone par,

N

ka horizontal (inward) component

N sin θ

hota hai; wahi centripetal force hai, jabki

N cos θ

vertical hai aur

m g

balance karta hai (Figure s03, left).

Why questions

Ball string release karte hi straight line mein kyun fly off karti hai, spiral out karne ki bajay?

Kyunki tension khatam hone par koi inward force nahi hoti; Newton's first law se ball apni current velocity maintain karti hai, jo circle ki tangent hai — ek straight line.

Required force ko "centripetal" (center-seeking) kyun kaha jaata hai, "centrifugal" kyun nahi?

Kyunki net real force jo path ko bend karti hai, motion ko inward curve karne ke liye center ki taraf point karni chahiye; outward net force object ko circle se door push kar degi.

Flat curve par static friction unlimited centripetal force kyun supply nahi kar sakti?

Static friction

μ_{s} N = μ_{s} m g

par cap hoti hai; jab demand

m v^{2} / r

is ceiling se zyada ho jaati hai, friction saturate ho jaati hai, wheel grip break ho jaata hai, aur car outward skid karti hai.

Banking a road car ko friction par rely kiye bina tez turn kyun karne deta hai?

Road ko tilt karne se normal force tilt hoti hai taaki use ek inward horizontal component

N sin θ

mile (jahan

θ

horizontal se hai), jo centripetal kaam akele karta hai, isliye turn karne ke liye grip ki zaroorat nahi.

Vertical loop ke top par minimum speed exist kyun karta hai lekin bottom par nahi?

Top par gravity already inward point karti hai; agar

v

bahut chhota ho to required

m v^{2} / r

m g

se kam hota hai, isliye string push nahi kar sakti (woh slack ho jaati hai) aur object fall kar jaata hai. Bottom par gravity outward point karti hai, isliye tension hamesha koi bhi

F_{c}

supply karne ke liye badhayi ja sakti hai.

Magnetic force charge ke path ko kyun curve karti hai lekin uski speed kabhi nahi badhalti?

Kyunki

F = q v \times B

hamesha

v

ke perpendicular hoti hai, isliye

F \cdot v = 0

: zero work, isliye constant kinetic energy aur constant speed, sirf turning.

Derivation

a = v ω

position triangle ki bajay velocity triangle kyun use karta hai?

Kyunki acceleration velocity ka rate of change hai; hume

∣Δ v ∣

chahiye, isliye hum do velocity vectors (har ek length

v

) se isosceles triangle banate hain jo

Δ θ

se turn karta hai (Figure s01).

Edge cases

Centripetal acceleration kya hota hai jab

v = 0

ho (object momentarily at rest, jaise pendulum apne highest swing point par)?

v = 0

ke saath object instantaneously bilkul bhi circle mein move nahi kar raha, isliye strictly

r

(ek circular-path radius) motion se define nahi hoti; limit lete hue, circular motion ke liye required inward acceleration

v^{2} / r \to 0

hai — us instant koi centripetal force nahi chahiye kyunki path abhi curve nahi kar raha.

Minimum top-of-loop speed

g r

ka kya hota hai jab radius zero ki taraf shrink hota hai?

v_{m i n} = g r \to 0

: ek tiny loop ko top par almost koi speed nahi chahiye, kyunki required

m v^{2} / r

chhote

v

ke liye bhi

m g

se match kar sakta hai.

Flat road par car ke liye, maximum turning speed kya hai agar road perfectly frictionless ho (

μ_{s} = 0

v_{m a x} = μ_{s} g r = 0

— koi friction aur koi banking nahi hone par

F_{c}

supply karne ke liye kuch nahi hai, isliye car bilkul turn nahi kar sakti; woh straight jaati hai.

Banking formula

tan θ = v^{2} / (r g)

mein, straight road (

r \to \infty

) ke liye kitna banking angle chahiye?

tan θ = v^{2} / (r g) \to 0

, isliye

θ \to 0

(horizontal se measure kiya): straight road ko koi banking nahi chahiye, jo consistent hai koi inward force required nahi hone ke saath.

Agar satellite ka orbital radius

r \to \infty

ho, to uski required orbital speed ka kya hoga?

v = GM / r \to 0

: far-out orbits slower hoti hain, kyunki weakening gravity kam centripetal force supply karti hai aur chhota

v^{2} / r

demand karti hai.

Straight line mein constant velocity se move karne wali body par kaunsi net inward force act karti hai?

Zero — straight path ka infinite radius hota hai, isliye

m v^{2} / r \to 0

; koi centripetal force nahi chahiye, jo Newton's first law se match karta hai.

Recall Har trap ki one-line summary

Centripetal force ek requirement hai ( $m v^{2} / r$ , jahan $ω = v / r$ ) jo real forces meet karti hain; yeh hamesha inward point karti hai, motion ke perpendicular hone par koi work nahi karti, friction se supply hone par cap hoti hai, loop ke top vs bottom par kaunsi forces help karti hain yeh flip hota hai, aur har degenerate limit mein vanish ho jaati hai ( $v = 0$ , $r \to \infty$ , $μ_{s} = 0$ ).