Visual walkthrough — Vertical circular motion — minimum speed conditions

1.2.18 · D2 · Physics › Newton's Laws & Dynamics › Vertical circular motion — minimum speed conditions

Yeh parent topic ka visual companion hai. Pehle woh padho agar tumhe words chahiye; yeh padho agar tumhe pictures chahiye.

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Step 1 — "Circle mein move karna" kya demand karta hai

KYA HAI. Figure dekho. Ek ball circle par kahin baithi hai. Lal arrow ball se seedha center ki taraf point kar raha hai.

KYU HAI. Curve karne ke liye — seedhi line mein uda jaane ki jagah curve karte rehne ke liye — ek object ko hamesha ek net force chahiye jo center ki taraf point kare. Straight-line motion ko koi sideways force nahi chahiye; curving ko chahiye. Woh inward-pointing ki zaroorat is chapter ki poori kahani hai.

PICTURE. Dhyan do ki lal inward arrow rotate karta hai jab ball move karti hai: top par woh neeche point karta hai, bottom par upar, side par sideways. Same rule, alag direction. Yeh yaad rakho — yahi ek fact hai jo top aur bottom ko itna alag banata hai.

Dekho Centripetal force and acceleration jahan se $\frac{m{v}^2}{r}$ aata hai.

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Step 2 — Top par forces (draw karo, guess mat karo)

KYA HAI. Ball bilkul top par hai. Center uske neeche hai. Toh string, ball se center tak jaati hai, neeche pull karti hai. Gravity bhi neeche pull karti hai. Dono real forces ek hi direction mein point karte hain.

PEHLE DRAW KYU KAREIN? Kyunki sabse common galti (parent ke mistake box mein dekho) yeh hai ki habit se soch lete hain ki string top par upar pull karti hai. Picture us habit ko khatam kar deti hai: center neeche hai, toh pull neeche hai.

PICTURE. Do kaale arrows lal ball se nikalte hain, dono neeche point karte hue: chhota wala $T$ hai, lamba wala $mg$. Unki combined length top par available total inward force hai.

Yeh sirf Newton's Second Law — net force form hai jismein "inward" ko positive choose kiya gaya hai.

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Step 3 — Speed ko squeeze karo aur tension ko gayab hote dekho

KYA HAI. Step 2 ko rearrange karo tension isolate karne ke liye: $T=\frac{m{v}_{\text{top}}^2}{r}-mg$

KYU HAI. Hum jaanna chahte hain ki kya hota hai jab ball top par slow hoti jaati hai. Slow matlab chhota $v_{\text{top}}$, jo $\frac{m{v}_{\text{top}}^2}{r}$ ko shrink karta hai. Kyunki $mg$ fixed hai, sirf $T$ shrink ho sakta hai match karne ke liye.

PICTURE. Figure mein, teen speeds ke liye top par ball ki teen copies: fast (lamba $T$ arrow), slower (chhota $T$), aur critical speed (koi $T$ arrow nahi — sirf gravity). Lal arrow tension hai, aur tum literally use shorten hote hue nothing tak dekh sakte ho.

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Step 4 — Kinar: tension ko exactly zero set karo

KYA HAI. $T=0$ Step 2 mein daalo: $0+mg=\frac{m{v}_{\text{top,min}}^2}{r}$

KYU HAI. Is instant par, sirf gravity hi poora inward force provide karti hai — na kuch zyada, na kuch kam.

PICTURE. Ek lal gravity arrow, aur woh exactly utna lamba hai jitna uske saath inward requirement bracket hai. Gravity aur "zaroorat" barabar hain — perfect match.

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Step 5 — Top tak pahunchna: energy bridge

KYA HAI. Bottom se top tak ball $2r$ height chadhti hai (circle ka bottom se circle ka top ek diameter hai).

ENERGY KYU, FORCES KYU NAHI? Top aur bottom par forces alag-alag directions mein point karte hain aur poore loop mein vary karte hain — unhe mid-loop track karna dardnaak hai. Energy sirf height aur speed ki parwah karti hai, direction ignore karke. "Speed height ke saath kaise badlti hai?" ke liye yeh sahi tool hai.

PICTURE. Circle ke side par ek vertical ruler: bottom height $0$ par, top height $2r$ par, side height $r$ par. Lal ball ruler chadhti hai; ek dashed line $2r$ mark karti hai jo use gain karni hai.

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Step 6 — Bottom speed solve karo

KYA HAI. Har jagah se $\frac{1}{2}m$ cancel karo (phir se mass aur half nikal jaate hain): $v_{\text{bot}}^2=v_{\text{top}}^2+4gr$ Ab Step 4 ka result $v_{\text{top}}^2=gr$ daalo: $v_{\text{bot}}^2=gr+4gr=5gr\Rightarrow \boxed{v_{\text{bot,min}}=\sqrt{5gr}}$

$4gr$ KYU? $mg(2r)$ ko cancelled $\frac{1}{2}m$ ke through wapas multiply karne par height cost $2r$ ek $v^2$ cost mein badal jaati hai $2\cdot 2gr=4gr$. Yahi chadhne ka "energy tax" hai.

PICTURE. Theme colors mein ek bar chart: bottom bar (height $5gr$) top bar ($gr$, lal) plus climbing tax ($4gr$) mein split hoti hai. Tum dekh sakte ho 5 kahan se aata hai: $1+4$.

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Step 7 — Bottom tension, aur famous $6mg$

KYA HAI. Bottom par center ball ke upar hota hai, toh tension upar point karti hai jabki gravity neeche. Net upward force = requirement: $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r}$

MINUS SIGN KYU? Kyunki yahan dono real forces oppose karte hain: upar minus neeche. Compare karo Step 2 (top) se jahan woh add hote the. Same law, opposite geometry — bilkul Step 1 ka "center kahan hai?" wala rule.

PICTURE. Bottom par ball: ek lamba lal $T$ arrow upar, ek chhota $mg$ arrow neeche; net up arrow inward requirement ke barabar hai.

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Step 8 — Degenerate case: ek rigid rod

KYA HAI. Rod ke saath, top par sirf yeh requirement bachti hai ki ball wahan pahunche: $v_{\text{top}}\ge 0$. Toh minimum top speed $0$ hai.

YEH KYUN MATTER KARTA HAI. Parent ka mnemonic ($5$–$3$–$1$, $6mg$) ek string/track result hai. Constraint badlo, answer badal jaata hai. Kabhi $\sqrt{gr}$ ko prakriti ka niyam mat samjho — yeh "no pushing" ka consequence hai.

PICTURE. Do panels. Left: top par ek string $v_{\text{top}}=\sqrt{gr}$ ke saath, ek taut lal line. Right: top par rod $v_{\text{top}}\to 0$ ke saath, rod upar push kar raha hai (lal arrow reverse) ball ko circle par rakhne ke liye.

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Ek-picture summary

KYA HAI. Ek figure poora loop gather karta hai: circle ball ke saath bottom, side, top par; squared-speed labels $5gr$, $3gr$, $gr$; tensions $6mg$ aur $0$; har point par inward arrows; aur height ruler $0$, $r$, $2r$. Yeh poori derivation compressed hai.

Recall Feynman retelling — plain words mein poora walkthrough bolo

Curve karne ke liye, tumhe hamesha beech ki taraf ek pull chahiye. Ek loop ke top par, beech tumhare neeche hota hai, toh gravity pehle se tumhe inward pull karne mein help kar rahi hai. Agar tum dheere jaate ho, circle ko sirf ek gentle inward pull chahiye — gravity se bhi gentle — toh string ko tumhe wapas hold karne ke liye push karna padega, aur strings push nahi kar saktin. Sabse dheere tum ja sakte ho woh tab hai jab gravity ka pull exactly wahi hai jo circle ko chahiye, aur string kuch nahi karti: woh speed $\sqrt{gr}$ hai. Us speed ke saath top par pahunchne ke liye, tumhe bottom par zyada fast start karna hoga, kyunki do radii ki height chadhna speed churaata hai — energy bookkeeping us chadhne ko ek extra $4gr$ mein badal deti hai, toh bottom speed squared $1+4=5$ times $gr$ hoti hai. Bottom par string ko tumhara weight bhi uthana hota hai aur bada inward pull bhi provide karna hota hai, toh woh $6mg$ par strain karta hai jabki top par kuch nahi. String ki jagah ek stiff rod lagao aur "no pushing" rule gayab ho jaata hai, toh tum zero speed se top par crawl kar sakte ho — magic $\sqrt{gr}$ kabhi koi law nahi tha, bas ek string ki honest limit thi.

Recall Paanch numbers, memory se

Squared speeds $5gr,3gr,1gr$ bottom / side / top par. Tensions $6mg$ bottom par, $0$ top par. Rod: $2\sqrt{gr}$ bottom par.

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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