Worked examples — Vertical circular motion — minimum speed conditions

1.2.18 · D3 · Physics › Newton's Laws & Dynamics › Vertical circular motion — minimum speed conditions

Yeh page Vertical circular motion — minimum speed conditions ka drill deck hai. Parent note ne teen master results banaye the; yahan hum unhe har us case ke against hammer karte hain jo ek problem throw kar sakta hai — circle par har position, dono constraint types (string vs rod), degenerate slow case, fast case, energy links, aur exam twists.

Shuru karne se pehle, ek reminder plain words mein. Vertical circle ka matlab sirf ek loop hai jo room mein khada hua ho, jaise ek Ferris wheel, isliye height har jagah alag hoti hai aur gravity ($mg$, mass $m$ par neeche ki taraf pull) har point par alag kaam karti hai. Letter $r$ hai radius (center se ball tak ki distance), $v$ hai speed (circle ke saath kitna fast move kar raha hai), aur $g\approx 9.8{\text{m/s}}^2$ gravity ki strength hai. Symbol $\frac{m{v}^2}{r}$ hai required inward force — koi nayi push nahi, bas woh total jitna real forces (gravity + tension + normal) milke add up karein, taaki ball straight fly karne ki jagah curving karti rahe.

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Scenario matrix

Har vertical-circle problem in cells mein se ek (ya combo) hai. Neeche har worked example us cell(s) ke saath tagged hai jis par woh fit hota hai.

Cell	Kya vary karta hai	Danger / trick jo yeh test karta hai
A — Top, string	Position = top, constraint sirf pull karta hai	$T\ge 0$; bahut slow hone par slack
B — Bottom, string	Position = bottom	tension load ko add karta hai ($6mg$)
C — Side, height $=r$	Position = 90°	gravity tangent hai, central nahi
D — Arbitrary angle $\theta$	Loop par general point	gravity ko radial + tangential mein resolve karo
E — Degenerate slow	$v_{\text{top}}<\sqrt{gr}$	ball circle chhod deti hai → projectile
F — Rigid rod	Constraint push kar sakta hai ($T<0$ allowed)	min top speed $=0$
G — Real-world word	bucket / coaster / bike	words ko ek cell mein translate karo
H — Exam twist	ek ratio ya proof poochhe	algebra, koi numbers nahi
I — Fast case	minimum se kaafi upar speed	tension badhti hai, kabhi slack nahi

Neeche ke examples har cell ko cover karne ke liye numbered hain: Ex1→A, Ex2→B, Ex3→C, Ex4→D, Ex5→E, Ex6→F, Ex7→G, Ex8→H, Ex9→I. Geometric cases (A–D) ke saath figures bhi hain.

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Example 1 — String loop ka top (Cell A)

Step 1 — Center locate karo; directions assign karo. Figure s01 mein red dot center hai aur white dot ball hai, seedha uske upar baitha hua. Toh "toward center" seedha neeche point karta hai. Blue arrow string tension $T$ hai aur yellow arrow gravity $mg$ hai — unhe aankh se trace karo: dono neeche point karte hain, center ki taraf. Green note payoff flag karta hai: kyunki dono inward hain, woh add hote hain. Yeh step kyun? Net inward force ki direction woh #1 cheez hai jo students galat karte hain — hamesha pehle center dhundo, phir arrows ko figure se padho.

Step 2 — Center ki taraf Newton's second law likho. $T+mg=\frac{m{v}_{\text{top}}^2}{r}$ Yeh step kyun? Yeh Newton's Second Law — net force form hai radius ke saath apply kiya hua: real inward forces ka sum (s01 mein dono neeche ke arrows) required $\frac{m{v}^2}{r}$ ke equal hai.

Step 3 — Physical edge $T=0$ impose karo. Ek string sirf pull kar sakti hai, isliye $T\ge 0$. Sabse slow circular motion woh hai jab tension bilkul khatam ho jaye (s01 mein blue arrow zero ho jaye): $mg=\frac{m{v}_{\text{top,min}}^2}{r}\Rightarrow v_{\text{top,min}}=\sqrt{gr}=\sqrt{9.8\times 0.80}=2.80\text{m/s}$ Yeh step kyun? $T=0$ par gravity (yellow arrow) akeli poori turning force supply karti hai — "minimum" ki defining condition yahi hai.

Step 4 — Us minimum speed par tension. Construction se $T=0$.

Verify: $\sqrt{9.8\times 0.8}=\sqrt{7.84}=2.80$ ✓. Units: $\sqrt{({\text{m/s}}^2)(\text{m})}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✓. Tension $0\text{N}$ exactly minimum-condition answer hai, "almost zero" nahi.

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Example 2 — Usi loop ka bottom (Cell B)

Step 1 — Energy se $v_{\text{bot}}$ nikalo. Bottom aur top ke beech full Conservation of mechanical energy se shuru karo, mass $m$ visible rakhte hue. Kinetic energy $\frac{1}{2}m{v}^2$ hai aur gravitational potential energy $mgh$; bottom se top tak climb $\Delta h=2r$ hai. Tension hamesha motion ke perpendicular hoti hai, isliye woh koi kaam nahi karti: $\underset{\text{KE bottom}}{\underbrace{\frac{1}{2}m{v}_{\text{bot}}^2}}=\underset{\text{KE top}}{\underbrace{\frac{1}{2}m{v}_{\text{top}}^2}}+\underset{\text{PE gained}}{\underbrace{mg(2r)}}$ Ab har term mein same factor $m$ hai, toh poori equation ko $m$ se divide karo — mass cancel ho jaata hai aur speeds par kabhi affect nahi karta: $\frac{1}{2}{v}_{\text{bot}}^2=\frac{1}{2}{v}_{\text{top}}^2+g(2r)\Rightarrow v_{\text{bot}}^2=v_{\text{top}}^2+4gr=gr+4gr=5gr$ $v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 0.8}=\sqrt{39.2}=6.26\text{m/s}$ Yeh step kyun? Bottom ko us height ke liye "pay" karne ke liye zyada fast hona chahiye jo woh climb karega; pehle $mgh$ likhna (shortcut nahi) exactly dikhata hai ki $m$ kyun drop out hota hai, taaki koi na sooche mass kahan gaya.

Step 2 — Bottom par directions. Figure s02 mein center (red dot) ab ball ke upar hai. Center ki taraf = upar. Blue arrow (tension) upar center ki taraf point karta hai; yellow arrow (gravity) neeche point karta hai. Un dono ko figure se padhke: $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r}$ Yeh step kyun? Top se ulta geometry (s01 vs s02 mein arrows compare karo) — Cell B ka yahi poora point hai: yahan tension aur gravity ladti hain ek doosre se, add nahi hoti.

Step 3 — Solve karo. $T_{\text{bot}}=mg+\frac{m(5gr)}{r}=mg+5mg=6mg=6(0.20)(9.8)=11.76\text{N}$ Yeh step kyun? Hum jaana hua $v_{\text{bot}}^2=5gr$ radial equation mein substitute karte hain aur $T_{\text{bot}}$ isolate karte hain — yeh general force balance ko us single number mein convert karta hai jo problem ne manga tha.

Verify: $6mg=6\times 0.2\times 9.8=11.76\text{N}$ ✓. Sanity: yeh exactly weight $mg=1.96\text{N}$ ka chha guna hai ✓. Aur $T_{\text{bot}}-T_{\text{top}}=11.76-0=11.76=6mg$ ✓.

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Example 3 — Side par, height $=r$ (Cell C)

Step 1 — Energy se side par speed nikalo. Same per-unit-mass energy balance jaise Example 2, lekin bottom se climb sirf $\Delta h=r$ hai, toh $v^2$ drop $2gr$ hai: $v_{\text{side}}^2=v_{\text{bot}}^2-2gr=5gr-2gr=3gr=3\times 9.8\times 0.8=23.52$ $v_{\text{side}}=\sqrt{23.52}=4.85\text{m/s}$ Yeh step kyun? Same energy bookkeeping, bas chhoti climb.

Step 2 — Side par directions. Figure s03 mein center (red dot) ball se horizontal hai, toh blue arrow (tension, center ki taraf) horizontal hai. Gravity — yellow arrow — seedha neeche point karti hai, jo yahan circle ke tangent hai (motion ki direction mein). Green note spell out karta hai: gravity speed ko change karti hai, turning ko nahi, isliye sirf tension inward act karta hai: $T_{\text{side}}=\frac{m{v}_{\text{side}}^2}{r}$ Yeh step kyun? Yeh Cell C ka key lesson hai — gravity ka poora kaam yahan tangential hai (yellow arrow blue radial arrow ke perpendicular hai), toh woh radial equation se drop out ho jaata hai.

Step 3 — Solve karo. $T_{\text{side}}=\frac{m(3gr)}{r}=3mg=3(0.20)(9.8)=5.88\text{N}$ Yeh step kyun? $v_{\text{side}}^2=3gr$ radial equation mein plug karna use weight $mg$ ke ek clean multiple mein collapse karta hai, woh number deta hai jo problem chahta tha aur $0,3mg,6mg$ pattern expose karta hai.

Verify: $3mg=5.88\text{N}$ ✓. Pattern check: tensions padhte hain $0,3mg,6mg$ (top, side, bottom) — $1:3:5$ speed-squared ladder se match karta hai jo changing gravity component se shifted hai ✓.

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Example 4 — General angle $\theta$ (Cell D)

Step 1 — Angle figure se padho. s04 dekho: center par yellow arc angle $\theta$ hai, downward vertical se measure kiya gaya (center se bottom tak ki line) round karke dashed blue radius tak jo ball par point karta hai. Toh $\theta$ literally "bottom se kitna round" hai woh ball. Yeh anchor hai jo hamari saari signs fix karta hai. Yeh step kyun? Neeche ki har sign depend karti hai kahan se $\theta$ measure kiya gaya par; figure ise pin down karta hai taaki hum projection ulta na kar lein.

Step 2 — Gravity ko radial + tangential mein resolve karo. Gravity (s04 mein ball par yellow down-arrow) seedha neeche hai. Dashed blue radius us downward vertical ke saath angle $\theta$ banata hai (exactly woh arc jo humne abhi padha). Gravity ka woh piece jo radius ke saath lie karta hai, outward (center se door) point karte hue woh green arrow hai, size $mg\cos \theta$. Inward radial equation isliye hai: $T-mg\cos \theta =\frac{m{v}^2}{r}$ Yeh step kyun? Sirf gravity ka radial slice tension se compete karta hai; bacha hua tangential slice $mg\sin \theta$ bas ball ko speed up/slow karta hai. Hum $\cos \theta$ use karte hain kyunki $\theta$ down-direction aur radius ke beech ka angle hai, aur cosine gravity ko us radius par project karta hai (adjacent-side rule).

Step 3 — Teen landmark angles check karo.

• $\theta =0$ (bottom): $\cos 0=1\Rightarrow T-mg=\frac{m{v}^2}{r}$ ✓ (Cell B).
• $\theta =90^{\circ }$ (side): $\cos 90^{\circ }=0\Rightarrow T=\frac{m{v}^2}{r}$ ✓ (Cell C).
• $\theta =180^{\circ }$ (top): $\cos 180^{\circ }=-1\Rightarrow T+mg=\frac{m{v}^2}{r}$ ✓ (Cell A).

Yeh step kyun? Ek general formula tabhi trustworthy hai jab woh un special cases mein collapse ho jo tumne already prove kiye hain — $\cos \theta$ ki sign exactly flip hoti hai jaise s01–s03 ki geometry demand karti hai.

Step 4 — $\theta =60^{\circ }$ par numbers. Radial gravity component $=mg\cos 60^{\circ }=mg(0.5)=0.5mg$. Toh gravity yahan radius ke saath apna aadha weight supply karta hai. Yeh step kyun? Ek concrete angle evaluate karna abstract $\cos \theta$ ko ek aisa number banata hai jo feel ho sake — $60^{\circ }$ par exactly aadhi gravity tension se ladti hai, jo woh intuition hai jo Forecast ne guess karne ko bola tha.

Verify: $\cos 60^{\circ }=0.5$ ✓, aur formula deta hai $T=\frac{m{v}^2}{r}+0.5mg$. $\theta =0,90,180^{\circ }$ par yeh exactly bottom/side/top laws return karta hai ✓.

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Example 5 — Top par bahut slow: circle chhod deta hai (Cell E)

Step 1 — Tension compute karo jo circle demand karega. $T=\frac{m{v}^2}{r}-mg=m\left(\frac{v^2}{r}-g\right)=m\left(\frac{2.0^2}{1.0}-9.8\right)=m(4.0-9.8)=-5.8m$ Toh equation negative tension $-5.8m$ newtons (per kg) maang rha hai. Yeh step kyun? Negative $T$ ka matlab hai string ko push karna hoga — string push nahi kar sakti (sign convention yaad karo: negative $T$ = outward push).

Step 2 — Physically interpret karo. Kyunki string demanded inward force supply nahi kar sakti, woh slack ho jaati hai ($T=0$). Ball ab gravity akeli ke under ek free projectile hai — woh circle ke andar fall karta hai aur string baad mein neeche jaate waqt re-tighten hoti hai. Usne kabhi loop complete nahi kiya circle ki tarah. Yeh step kyun? Yeh har "too slow" problem ka real-world outcome hai: chhoti circle nahi, balki constraint se departure.

Step 3 — Threshold check. Break-even exactly $v^2/r=g$ hai, yani $v=\sqrt{gr}=3.13\text{m/s}$. Isse neeche, $T<0$ (slack); is par, $T=0$; isse upar, $T>0$ (taut). Yeh step kyun? Exact break-even speed pin down karna tumhe "circular" aur "falls off" ke beech ki precise boundary batata hai, taaki tum koi bhi given top speed ek nazar mein classify kar sako instead of har baar re-solving karo.

Verify: $\frac{2.0^2}{1.0}-9.8=4.0-9.8=-5.8<0$ ✓ (negative → slack). Threshold $\sqrt{9.8}=3.13\text{m/s}$ ✓.

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Example 6 — Rigid rod vs string (Cell F)

Step 1 — $T\ge 0$ rule drop karo. Rod pull ya push kar sakta hai, toh $T$ negative ho sakta hai (sign convention yaad karo: negative $T$ = rod compression mein, outward push kar raha hai). "Top reach karne" ke liye sirf yahi requirement hai ki $v_{\text{top}}\ge 0$. Yeh step kyun? $\sqrt{gr}$ result ek string condition tha (woh entirely $T\ge 0$ demand karne se aaya tha); woh constraint remove karo aur "minimum" ki physics completely badal jaati hai.

Step 2 — (a) Min top speed aur (b) min bottom speed. $v_{\text{top,min}}=0.$ Top par zero speed ke saath pahunchne ke liye enough energy, same per-unit-mass energy balance use karke jaise Example 2: $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=0+4gr=4gr$ $v_{\text{bot,min}}=\sqrt{4gr}=2\sqrt{gr}=2\sqrt{9.8\times 0.8}=2\times 2.80=5.60\text{m/s}$ Yeh step kyun? Koi slack condition nahi hone se, rod ko sirf itna guarantee karna hai ki ball top par pahunche ($v_{\text{top}}\ge 0$); sabse sasta tarika hai $v_{\text{top}}=0$, aur energy conservation $2r$ climb ke upar phir us required bottom speed mein convert kar deta hai.

Step 3 — (c) $v_{\text{top}}=0$ par rod force top par. $v=0$ ke saath top radial equation use karo: $T+mg=\frac{m{v}^2}{r}=\frac{m(0)}{r}=0\Rightarrow T=-mg$ Negative sign ka matlab hai rod compression mein hai: woh ball ko outward (top par upward) $mg$ force se push karta hai, exactly ball ko gravity ke against up hold karta hai taaki woh wahan motionless hang kar sake. String, jo push nahi kar sakti, ball ko simply drop kar deti — yahi exactly reason hai ki rod loop complete karta hai jahan string fail hoti hai. Yeh step kyun? Yeh concretely dikhata hai, sign convention ke saath, ki "rod can push" tumhe kya deta hai: top par missing support force $mg$.

Verify: $2\sqrt{9.8\times 0.8}=2\sqrt{7.84}=2\times 2.8=5.60\text{m/s}$ ✓, jo string ke $\sqrt{5gr}=6.26\text{m/s}$ se chhota hai ✓ (rod easier hai). Rod force $=-mg$ top par (compression) ✓.

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Example 7 — Bucket of water (real-world word problem, Cell G)

Step 1 — (a) Words ko ek cell mein translate karo. Yeh modelling step hai jo har word problem ko chahiye. Paani tabhi "girega" jab woh bucket bottom se door fall karne lagta hai — yaani agar bucket bottom (woh surface jo paani ko push karti hai) ko paani ko inward pull karna pade taaki woh circle mein rahe. Bottom ek normal force $N$ exert karta hai, aur ek surface sirf push kar sakti hai, toh $T\ge 0$ — exactly wahi maths jaise Cell A mein string. "Ball" paani hai, "string" bucket bottom hai. Toh minimum: $v_{\text{top,min}}=\sqrt{gr}=\sqrt{9.8\times 1.2}=\sqrt{11.76}=3.43\text{m/s}$ Yeh step kyun? Har real-world problem ka aadha kaam yeh spot karna hai ki woh kaun sa cell hai; constraint naam karna ($N\ge 0$) bucket ko Cell A reduce kar deta hai.

Step 2 — (b) $v=4.0\text{m/s}$ par effective weight. Top par paani ka free body banao: gravity $mg$ neeche, normal force $N$ neeche (bucket bottom paani ko center ki taraf push karta hai, jo neeche hai). Mass $m$ ke paani ke liye top radial equation padhti hai: $N+mg=\frac{m{v}^2}{r}\Rightarrow N=m\left(\frac{v^2}{r}-g\right)=m\left(\frac{16.0}{1.2}-9.8\right)=m(13.33-9.8)=3.53m$ Toh paani bucket par $N=3.53m$ newtons per kg se press karta hai — yeh top par uska "effective weight" hai. Yeh step kyun? Newton's third law se bucket jo force paani par push karta hai ($N$) woh us force ke equal hai jo paani bucket par push karta hai, toh radial equation ko $N$ ke liye solve karna hi "effective weight" hai jo question ne manga — aur iska sign batata hai ki paani andar rahega ($N>0$) ya girega ($N<0$).

Verify: (a) $\sqrt{9.8\times 1.2}=\sqrt{11.76}=3.43\text{m/s}$ ✓. (b) $\frac{16}{1.2}-9.8=13.333-9.8=3.533>0$ ✓ (positive → paani ruka rahega, free-fall se zyada press karta hai). Kyunki $4.0>3.43$, koi spill nahi ✓.

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Example 8 — Exam twist: prove karo $T_{\text{bot}}-T_{\text{top}}=6mg$ KISI BHI speed par (Cell H)

Step 1 — Dono radial equations likho. $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r},T_{\text{top}}+mg=\frac{m{v}_{\text{top}}^2}{r}$ Yeh step kyun? Directions differ karte hain (top par center neeche, bottom par center upar), toh $mg$ signs flip hoti hain — bottom $mg$ subtract karta hai, top add karta hai. Yeh sirf Cells B aur A hain side by side likhe numbers ki jagah symbols ke saath.

Step 2 — Top equation ko bottom se subtract karo. $T_{\text{bot}}-T_{\text{top}}=\frac{m}{r}\left(v_{\text{bot}}^2-v_{\text{top}}^2\right)+2mg$ Yeh step kyun? Dono $mg$ terms combine hokar $+2mg$ bante hain (top ka $+mg$ subtract karne se left par $-(+mg)$ milta hai, yani $+mg$ over move hota hai, bottom ke $-(-mg)$ mein add hota hai); speeds ek bracket mein gather ho jaate hain, energy substitution ke liye ready.

Step 3 — Energy conservation se speeds ko khatam karo. Bottom se top tak height $2r$ par, per-unit-mass energy balance ek fixed speed-squared drop deta hai, independent of kitna fast tum swing karo: $v_{\text{bot}}^2-v_{\text{top}}^2=4gr$ Substitute karo: $T_{\text{bot}}-T_{\text{top}}=\frac{m}{r}(4gr)+2mg=4mg+2mg=6mg$ Yeh step kyun? Speed cancel ho jaata hai kyunki $v^2$ mein difference geometry se set hota hai (woh $2r$ climb) alone, launch speed se nahi — exactly isliye identity universal hai, jo problem ne prove karne ko bola tha.

Verify: $\frac{m}{r}(4gr)+2mg=4mg+2mg=6mg$, aur result mein koi $v$ nahi ✓ — yeh har taut speed ke liye hold karta hai.

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Example 9 — Fast case: tension badhti jaati hai (Cell I)

Step 1 — Top par tension. $T_{\text{top}}=\frac{m{v}_{\text{top}}^2}{r}-mg=\frac{0.20\times 6.0^2}{0.80}-0.20\times 9.8=9.0-1.96=7.04\text{N}$ Yeh step kyun? Same top radial equation jaise Cell A, lekin ab $v_{\text{top}}$ bada hai, toh $\frac{m{v}^2}{r}$ $mg$ se kaafi zyada hai aur string firmly taut hai ($T>0$) — Example 5 ke slack case ka ulta.

Step 2 — Energy se bottom speed. $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=36.0+4(9.8)(0.80)=36.0+31.36=67.36{\text{m}}^2/{\text{s}}^2$ Yeh step kyun? Same per-unit-mass energy link jaise Example 2 — fixed $4gr$ boost — fast top speed ko bottom tak le jaata hai.

Step 3 — Bottom par tension. $T_{\text{bot}}=\frac{m{v}_{\text{bot}}^2}{r}+mg=\frac{0.20\times 67.36}{0.80}+1.96=16.84+1.96=18.80\text{N}$ Yeh step kyun? Cell B ka bottom radial equation; bada $v_{\text{bot}}^2$ substitute karne se dikhata hai ki tension speed ke saath scale karta hai aur kabhi slack nahi hota.

Step 4 — Universal identity check karo. $T_{\text{bot}}-T_{\text{top}}=18.80-7.04=11.76\text{N}=6mg$ Yeh step kyun? Example 8 confirm karta hai: minimum se bahut door bhi, difference $6mg$ par locked hai — sirf individual tensions badhte hain.

Verify: (a) $\frac{0.2\times 36}{0.8}-0.2\times 9.8=9.0-1.96=7.04\text{N}$ ✓. (b) $\frac{0.2\times 67.36}{0.8}+1.96=16.84+1.96=18.80\text{N}$ ✓. Difference $18.80-7.04=11.76=6mg$ ✓ — identity hold karti hai, dono tensions positive hain (kabhi slack nahi) ✓.

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Recall

Recall Kaun sa cell "gravity tangent hai, central nahi" test karta hai?

Cell C — side point (height $r$), jahan sirf tension inward hai aur minimum loop ke liye $T=3mg$ hai.

Recall Bottom se

$\theta$ par, radial equation kya hai? $T-mg\cos \theta =\frac{m{v}^2}{r}$; yeh $\theta =0/90/180^{\circ }$ par bottom/side/top laws deta hai.

Recall Bahut slow ball chhoti circle kyun nahi banati?

String push nahi kar sakti, toh $T=0$ (slack) aur ball loop ke andar free-falling projectile ban jaati hai.

Recall Rod ki minimum bottom speed string ke comparison mein badi hai ya chhoti?

Chhoti: $2\sqrt{gr}$ vs $\sqrt{5gr}$, kyunki rod ko sirf top reach karna hai, taut nahi rehna.

Recall Fast case mein (minimum se kaafi upar), kya constant rehta hai aur kya badhta hai?

Dono individual tensions speed ke saath badhte hain, lekin difference $T_{\text{bot}}-T_{\text{top}}$ $6mg$ par locked rehta hai.

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Connections

• Centripetal force and acceleration
• Newton's Second Law — net force form
• Conservation of mechanical energy
• Tension and constraint forces
• Normal force in circular tracks
• Banked curves and horizontal circular motion
• Free-body diagrams