Centripetal force — what provides it in various situations: Worked examples

1.2.16 · D3 · Physics › Newton's Laws & Dynamics › Centripetal force — what provides it in various situations

Shuru karne se pehle, ek vaada symbols ke baare mein. Hum sirf wahi letters use karenge jo humne earn kiye hain:

Definition Woh symbols jo hum baar baar use karenge

$m$ ::: moving body ki mass, kilograms (kg) mein.
$v$ ::: speed = kitna fast, direction ignore karke, metres per second (m/s) mein.
$r$ ::: circular path ki radius, metres (m) mein. String ki length, ya curve ke centre se doori socho.
$ω$ (Greek "omega") ::: angular speed = body ek second mein kitne radians angle sweep karta hai. Speed se linked hai $v = ω r$ se.
$a_{c} = \frac{v ^{2}}{r}$ ::: centripetal acceleration — velocity ki direction kitni fast ghoom rahi hai, $m/s^{2}$ mein, hamesha centre ki taraf point karti hai. Yeh woh cheez hai jo ek force produce karni chahiye (from Uniform Circular Motion).
$F_{c} = \frac{m v ^{2}}{r} = m a_{c}$ ::: required inward force. Yeh ek demand hai, ek number jitna real forces ko add up karna chahiye. Yeh koi force nahi hai jo tum draw karo.
$g = 9.8 m/s^{2}$ ::: Earth ki surface ke paas gravity ki strength.
$T$ ::: tension — ek tight string ya rope jo inward pull karti hai, newtons (N) mein. Socho string object ko tumhare haath ki taraf vapas khainch rahi hai.
$N$ ::: normal force — ek surface jo push deti hai, hamesha us surface ke perpendicular ("normal"), newtons (N) mein. Socho road seedha apne aap se upar ki taraf push kar rahi hai.
$f_{s}$ ::: static-friction force — do surfaces ke beech sideways grip jo slide nahi kar rahi, newtons (N) mein. Socho tyres road ko grip kar rahe hain.
$μ_{s}$ (Greek "mu", subscript $s$ ) ::: coefficient of static friction — ek plain number (koi units nahi) jo batata hai do surfaces kitni grippy hain. Yeh friction ki ceiling set karta hai $f_{s} \leq μ_{s} N$ se.

Agar inme se koi shaky lage, pehle parent note aur Uniform Circular Motion revisit karo.

Scenario matrix

Is topic ka har problem is table ke ek cell mein hai. Neeche ke worked examples label kiye gaye hain ki woh kaunsa cell hit karte hain — milke woh sab ko cover karte hain.

Cell	Kya special hai	$F_{c}$ ka Provider	Covered by
A Single inward force, horizontal	tension = pura kaam	Tension	Ex 1
B Force do components mein split hoti hai	vertical part weight hold karta hai, horizontal part turn karta hai	Normal / Tension	Ex 2 (banking), Ex 3 (conical pendulum)
C Ceiling / limiting value	force ka ek maximum hai → ek max speed exist karta hai	Static friction	Ex 4
D Dono forces inward milke point karti hain	gravity + contact dono add up hote hain	Gravity + Tension/Normal	Ex 5 (top of loop)
E Loop ke around sign flip hota hai	inward direction side-to-side change karti hai	Gravity vs Tension	Ex 6 (bottom of loop)
F Inverse-square provider	force $1/ r^{2}$ pe depend karti hai	Gravity (orbit)	Ex 7
G Velocity-dependent provider	force khud $v$ ke saath grow karti hai	Magnetic force	Ex 8
H Degenerate / zero input	kya hota hai jab $v \to 0$ , $r \to \infty$ , $θ \to 0$	limiting logic	Ex 9

Example 1 — Cell A: string par pathar (single inward force)

Forecast: Kya tension kuch newtons hogi, ya hundreds? Ek number guess karo, phir khud check karo.

Figure — Centripetal force — what provides it in various situations

Figure kya dikhata hai: faint circle pathar ka path hai, beech mein dot tumhara haath hai (centre). Orange arrow string ke along pathar se seedha centre tak jaati hai — woh tension $T$ hai, inward pull karne wali eklauti force. Teal arrow pathar par sideways, circle ke tangent point karti hai — woh velocity $v$ hai, aur notice karo yeh string ke right angles par hai.

Steps.

Real forces identify karo jo inward point kar rahi hain. Yeh step kyun? Figure mein orange arrow bilkul string ke along hai, centre ki taraf aimed hai: sirf tension $T$ pathar ko us taraf khinchti hai. Koi aur arrow us inward line ke along point nahi karta, isliye tension akele kaam karti hai.
Net inward force = demand set karo: $T = \frac{m v ^{2}}{r} .$ Yeh step kyun? Inward direction mein Newton's 2nd law (Newton's Second Law) kehta hai net inward force $m a_{c} = m v^{2} / r$ ke barabar hai.
Numbers daalo: $T = \frac{0.20 \times 4. 0 ^{2}}{0.50} = \frac{0.20 \times 16}{0.50} = 6.4 N .$

Verify: Units: $\frac{kg \cdot ( m/s ) ^{2}}{m} = \frac{kg \cdot m}{s ^{2}} = N$ . ✓ Sanity: speed double karo → tension chaar guna ho jaati hai (kyunki $v^{2}$ ), jo "faster whirl feel much harder" se match karta hai. ✓

Example 2 — Cell B: frictionless banked road par car

Forecast: Gentle 20° tilt — kya tum walking pace, city speed, ya highway speed expect karte ho?

Figure kya dikhata hai: thick dark line tilted road surface hai, plum square car hai uske upar. Seedha neeche point karta dark arrow gravity $m g$ hai. Road ke right angles par car se nikalti orange arrow normal force $N$ hai. Do teal dashed lines $N$ ko uske shadow mein vertical wall aur horizontal floor par drop karti hain: vertical dashed piece $N cos θ$ hai (upar point karta hai), aur horizontal dashed piece $N sin θ$ hai (curve ke centre ki taraf inward point karta hai).

Steps.

Do real forces draw karo: gravity $m g$ seedha neeche, aur normal force $N$ tilted road ke perpendicular. Yeh step kyun? Frictionless bank par sirf yahi do exist karti hain — friction deliberately switch off hai.
$N$ ko pieces mein split karo (do teal dashed lines follow karo). Vertical piece $N cos θ$ weight hold karti hai; horizontal piece $N sin θ$ centre ki taraf point karti hai. $N cos θ = m g, N sin θ = \frac{m v ^{2}}{r} .$ Yeh step kyun? Car sink ya fly nahi karti, isliye vertical forces cancel ho jaate hain; bacha hua horizontal slice hi eklauta cheez hai jo car ko turn kar sakti hai — isliye woh demand ke barabar hona chahiye.
$N$ aur $m$ ko khatam karne ke liye dono equations divide karo: $tan θ = \frac{v ^{2}}{r g} \Rightarrow v = r g tan θ .$ Yeh step kyun? Divide karna woh trick hai jo unknown $N$ remove kar deta hai — humein sirf components ka ratio chahiye, jo $tan θ$ hai (dekho Banking of Roads).
Numbers: $v = 80 \times 9.8 \times tan 2 0^{\circ} = 80 \times 9.8 \times 0.3640 = 285.3 \approx 16.9 m/s .$

Verify: $16.9 m/s \approx 61 km/h$ — ek sensible highway-curve speed. ✓ Note karo $m$ cancel ho gaya: design speed truck aur scooter dono ke liye same hai. ✓

Example 3 — Cell B: conical pendulum (tension splits)

Forecast: Banking jaisi hi $tan θ = v^{2} / (r g)$ shape — lekin yahan $r$ $L$ ke andar hidden hai. Kya tumhe iska zaroorat padegi?

Figure kya dikhata hai: dark slanted line string hai jo support se neeche plum bob tak jaati hai. Dotted vertical line support se seedha neeche drop hoti hai, aur orange arc uske aur string ke beech angle $θ$ mark karta hai. Teal dashed horizontal line us vertical se bob tak circle ka radius $r$ hai — yeh string ka horizontal shadow hai, isliye $r = L sin θ$ hai, $L$ nahi. String ke upar orange arrow tension $T$ hai; seedha neeche dark arrow weight $m g$ hai.

Steps.

Radius $L$ nahi hai! Figure mein teal dashed line padhke, horizontal reach $r = L sin θ$ hai. Yeh step kyun? String slanted hypotenuse hai; circle ka radius uska horizontal shadow hai. $r = 1.2 sin 3 0^{\circ} = 1.2 \times 0.5 = 0.60 m .$
Tension split karo: vertical part weight carry karta hai, horizontal part bob ko turn karta hai. $T cos θ = m g, T sin θ = \frac{m v ^{2}}{r} .$ Yeh step kyun? Banking jaisi identical logic — sirf tilted force ab string hai, road nahi.
Unknown $T$ (aur mass $m$ ) cancel karne ke liye dono equations divide karo, phir $v$ ke liye solve karo: $tan θ = \frac{v ^{2}}{r g} \Rightarrow v = r g tan θ .$ Yeh step kyun? Humein $T$ ya $m$ kabhi bataya nahi gaya, isliye humein unhe hataana hoga. Horizontal equation ko vertical se divide karne par sirf ratio $sin θ / cos θ = tan θ$ bachta hai — ek equation known quantities mein, jise hum phir $v$ isolate karne ke liye rearrange karte hain. $v = 0.60 \times 9.8 \times tan 3 0^{\circ} = 0.60 \times 9.8 \times 0.5774 = 3.395 \approx 1.84 m/s .$

Verify: Root ke neeche units: $m \cdot m/s^{2} = m^{2} / s^{2}$ , root m/s deta hai. ✓ Agar $θ \to 0$ (string almost vertical), $tan θ \to 0$ isliye $v \to 0$ — barely-swinging bob, sahi. ✓

Example 4 — Cell C: flat curve, friction ki ceiling hai

Forecast: Banking se zyada friction — isliye Example 2 ki 16.9 m/s se fast ya slow?

Figure kya dikhata hai: flat curve ka bird's-eye (top-down) view. Faint arc road ka edge hai; plum square car hai. Orange arrow curve ke centre ki taraf inward point karta hai — woh static friction $f_{s}$ hai, eklauta sideways grip jo car ko turn karta hai. Teal arrow car ki velocity $v$ hai, arc ke tangent. Ek taraf dashed orange line ceiling height $μ_{s} m g$ mark karti hai: friction us line tak grow kar sakti hai aur aage nahi.

Steps.

Flat road par, eklauta inward force static friction $f_{s}$ hai. Yeh step kyun? Gravity vertical hai, normal vertical hai; friction hi akela horizontal (inward) player hai (Friction) — exactly figure mein orange arrow.
Friction apni ceiling cross nahi kar sakti: $f_{s} \leq μ_{s} N = μ_{s} m g .$ Yeh step kyun? Static friction exactly utna supply karti hai jitna chahiye maximum $μ_{s} m g$ tak (dashed ceiling line). Isse aage push karo aur tyres slip kar jaate hain.
Maximum speed par, demand = ceiling: $\frac{m v _{m a x}^{2}}{r} = μ_{s} m g \Rightarrow v_{m a x} = μ_{s} g r .$
Numbers: $v_{m a x} = 0.60 \times 9.8 \times 45 = 264.6 \approx 16.3 m/s .$

Verify: $m$ phir cancel ho gaya — ek loaded truck same speed par skid karta hai jitna ek light car (surprising but true). ✓ Ex 2 se compare karo: yeh flat curve 16.3 m/s par top out hoti hai; road bank karne se tum zyada speed par ja sakte ho kam friction wear ke saath. ✓

Example 5 — Cell D: vertical loop ka top (dono forces inward)

Forecast: Bilkul top par, gravity neeche point karti hai, jo centre ki taraf hai. Kya yeh string ki help karta hai ya hurt?

Figure kya dikhata hai: faint circle vertical loop hai, beech mein dot uska centre hai, aur plum dot bilkul top par baitha hai. Ball se nikalte dono arrows seedha neeche point karte hain: orange arrow tension $T$ hai aur dark arrow weight $m g$ hai. Key visual yeh hai ki top par centre ball ke neeche hai, isliye "inward" ka matlab "downward" hai — aur dono forces usi direction mein line up hoti hain.

Steps.

Top par, dono gravity $m g$ aur tension $T$ downward = inward point karte hain. Yeh step kyun? Jaise figure dikhata hai, loop ka centre ball ke neeche hai, isliye "inward" yahan "down" hai. Dono real forces conveniently line up hoti hain. $T + m g = \frac{m v ^{2}}{r} .$
Minimum speed tab hai jab string slack hone ki verge par hoti hai, $T = 0$ : $m g = \frac{m v _{m i n}^{2}}{r} \Rightarrow v_{m i n} = g r = 9.8 \times 1.0 \approx 3.13 m/s .$ Yeh step kyun? Agar gravity akele hi demand se zyada exceed kare, string push nahi kar sakti (strings sirf pull karti hain), isliye woh slack ho jaati hai aur ball circle se girti hai.
Ab $v = 5.0$ par tension: $T = \frac{m v ^{2}}{r} - m g = \frac{0.30 \times 5. 0 ^{2}}{1.0} - 0.30 \times 9.8 = 7.5 - 2.94 = 4.56 N .$

Verify: $T > 0$ ✓ (string taut hai, $v = 5.0 > v_{m i n} = 3.13$ se consistent). Agar hum $v = v_{m i n}$ plug karte, humein exactly $T = 0$ milta. ✓

Example 6 — Cell E: loop ka bottom (sign flip hota hai)

Forecast: Bottom par, kya string gravity ke saath kaam karta hai ya uske against? Top se badi ya choti tension?

Figure kya dikhata hai: wahi faint loop aur centre dot, lekin ab plum ball bilkul bottom par hai. Centre ball ke upar hai, isliye "inward" ab "upward" matlab hai. Orange arrow upar point karta hai (tension $T$ , inward) jabki dark arrow neeche point karta hai (weight $m g$ , outward) — dono forces ab ek doosre ko oppose karti hain, Example 5 ke bilkul opposite.

Steps.

Bottom par, centre ball ke upar hai, isliye inward = up. Jaise figure dikhata hai, tension (orange) upar point karti hai (inward, $+$ ); gravity (dark) neeche point karti hai (outward, $-$ ). Yeh step kyun? Inward direction sides flip ho gayi — yahi Cell E ka poora point hai. $T - m g = \frac{m v ^{2}}{r} .$
Solve karo: $T = \frac{m v ^{2}}{r} + m g = 7.5 + 2.94 = 10.44 N .$ Yeh step kyun? Ab string ko poori demand supply karni hai aur weight bhi hold karna hai, isliye tension yahan sabse badi hai.

Verify: Units: har term $kg \cdot m/s^{2} = N$ hai. ✓ Sanity: $10.44 N$ (bottom) $> 4.56 N$ (top, Ex 5) — string bottom par snap karne ke sabse zyada chances hain, jo swings aur buckets ke real experience se match karta hai. ✓ Difference $10.44 - 4.56 = 5.88 N = 2 m g$ , exactly gravity ka sign flip ( $2 \times 0.30 \times 9.8$ ). ✓

Example 7 — Cell F: satellite (inverse-square provider)

Forecast: ISS roughly isi height par orbit karta hai. Kya tum kuch sau m/s expect karte ho, ya kai kilometres per second?

Steps.

Provider gravity hai, jiska strength $1/ r^{2}$ se girti hai (Gravitation and Orbits): $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} .$ Yeh step kyun? Orbit mein gravity hi eklauti inward force hai — use demand ke barabar set karo.
Ek $r$ aur mass $m$ cancel karo: $v = \frac{GM}{r} .$ Yeh step kyun? Satellite ki apni mass drop out ho jaati hai — orbit speed sirf isi pe depend karti hai ki tum kahan ho, satellite kitna heavy hai iska nahi.
Numbers: $v = \frac{3.986 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} = 5.694 \times 1 0^{7} \approx 7.55 \times 1 0^{3} m/s .$

Verify: Almost $7.5 km/s$ — well-known low-Earth-orbit speed. ✓ Bada $r$ → chota $v$ : dur orbits slower hain, jaise Moon vs ISS observe hota hai. ✓

Example 8 — Cell G: magnetic field mein charge (velocity-dependent provider)

Forecast: Microscopic mass, decent speed — kya radius centimetres hogi, ya kilometres?

Steps.

Provider magnetic force $q v B$ hai, hamesha velocity ke perpendicular (Magnetic Force on Moving Charges), isliye yeh path curve karta hai bina speed change kiye — ek perfect centripetal force. $q v B = \frac{m v ^{2}}{r} .$ Yeh step kyun? Ek force jo hamesha motion ke right angles par ho woh koi work nahi karta (speed change nahi) lekin continuously direction turn karta hai — exactly wahi jo ek circle ko chahiye.
Ek $v$ cancel karo aur solve karo: $r = \frac{m v}{q B} .$ Yeh step kyun? Notice karo force khud $v$ ke saath grow karti hai, isliye $v$ ka ek factor cancel ho jaata hai — radius sirf linearly speed ke saath badhti hai, $v^{2}$ ki tarah nahi.
Numbers: $r = \frac{1.67 \times 1 0 ^{- 27} \times 2.0 \times 1 0 ^{6}}{1.60 \times 1 0 ^{- 19} \times 0.50} = \frac{3.34 \times 1 0 ^{- 21}}{8.0 \times 1 0 ^{- 20}} \approx 0.0418 m .$

Verify: Almost $4.2 cm$ — ek plausible cyclotron-scale radius. ✓ Units: $\frac{kg \cdot m/s}{C \cdot T} = m$ ✓ (kyunki $T = kg / (C \cdot s)$ ).

Example 9 — Cell H: degenerate aur limiting cases

Forecast: Yeh "kya agar basically circle hi nahi hai?" checks hain. Padhne se pehle har ek predict karo.

Steps.

$v \to 0$ : demand $F_{c} = m v^{2} / r \to 0$ . Yeh step kyun? Barely move karne wale body ko almost koi inward force nahi chahiye — ek parked car ko "turn" karne ke liye zero friction chahiye. Formula agree karta hai: koi motion nahi, koi requirement nahi.
$r \to \infty$ : $F_{c} = m v^{2} / r \to 0$ . Yeh step kyun? Infinitely bada circle ek straight line hai, aur straight line ko bilkul koi inward force nahi chahiye (Newton's 1st law). Formula smoothly "no force" answer return karta hai.
$θ \to 0$ (flat bank): frictionless design speed $v = r g tan θ$ hai. Jaise $θ \to 0$ , $tan θ \to 0$ , isliye $v \to 0$ . Yeh step kyun? Ek bilkul flat road zero horizontal normal component contribute karta hai ( $N sin 0 = 0$ ), isliye friction switch off ke saath "safe" frictionless speed sirf khadi rehna hai. Isliye real flat curves tilt pe akele rely nahi kar sakti aur friction borrow karni padti hai (jo Cell C hai, Example 4). Banking picture aur friction picture $θ = 0$ par cleanly milte hain.

Verify: Teeno limits "koi inward force needed / available nahi" par collapse hoti hain, har ek straight-line motion se consistent. Koi formula blow up nahi hota ya nonsense negative nahi deta, aur case (c) kaam friction ko wapas de deta hai — models ke beech koi gap nahi. ✓

Recall Quick self-test

Inme se har ek kaun sa cell hai? (i) swing ke top par paani ki bucket. (ii) spinning turntable par coin jo slide nahi kar raha. (iii) Earth ke around Moon. (i) ::: Cell D — gravity aur tension/normal dono inward (Ex 5 jaisa). (ii) ::: Cell C — static friction $F_{c}$ provide karta hai ceiling ke saath (Ex 4 jaisa). (iii) ::: Cell F — gravity, inverse-square provider (Ex 7 jaisa).

Connections

Newton's Second Law — inward projection $\sum F_{in} = m a_{c}$ har example mein use hoti hai.
Uniform Circular Motion — jahan se $v = ω r$ aur $a_{c} = v^{2} / r$ aate hain.
Banking of Roads — Ex 2 ki geometry detail mein.
Friction — Ex 4 ke peeche $μ_{s} N$ ceiling.
Gravitation and Orbits — Ex 7 ka provider.
Magnetic Force on Moving Charges — Ex 8 ka provider.
Pseudo-forces and Non-inertial Frames — kyun "centrifugal" inhe ground-frame solutions mein kabhi appear nahi karta.
← Parent topic par wapas