Physical pendulum — compound pendulum: Worked examples

1.6.7 · D3 · Physics › Oscillations & Waves › Physical pendulum — compound pendulum

Yeh page compound pendulum ke liye "sab kuch ek saath" wali page hai. Parent note ne ek master formula banaya tha $T = 2 π \frac{I}{m g d}, L_{eq} = \frac{I}{m d} .$ Yahan hum ise har tarah ki situation mein drill karenge jo yeh topic de sakta hai — normal shapes, degenerate pivots, limits, ek word problem, aur ek exam twist — taaki koi bhi case tumhe surprise na kare.

Scenario matrix

Kuch bhi work karne se pehle, yahan pura map hai un case-classes ka jo yeh topic de sakta hai. Har worked example neeche us cell/cells ko cover karta hai.

#	Case class	Kya mushkil karta hai	Kaun cover karta hai
A	Standard shape, pivot off-CM	bas recipe apply karo	Ex 1 (rod), Ex 2 (disk)
B	Pivot AT center of mass par ( $d \to 0$ )	restoring torque khatam, $T \to \infty$	Ex 3
C	Minimum-period point par pivot ( $d = k$ )	derivative $= 0$ , special length $2 k$	Ex 4
D	Reversibility / center of oscillation	do alag pivots, same $T$	Ex 5
E	Composite body (do masses milake)	$I$ add karna aur combined CM nikalna	Ex 6
F	Large-amplitude limit	small-angle formula fail ho jaata hai	Ex 7
G	Real-world word problem	words ko $m, d, I$ mein translate karo	Ex 8 (swinging sign)
H	Exam twist: unknown length ke liye solve karo	formula ulta karo	Ex 9

Neeche har numeric answer bottom mein machine-checked hai.

Example 1 — Case A: uniform rod, pivot ek end par

Worked example Rod apne end se swing karna

Ek uniform rod, length $ℓ = 1.20 m$ , mass $m$ , ek end ke baare mein swing karti hai. $T$ nikalo ( $g = 9.81 m/s^{2}$ lo).

Forecast: andaza lagao — kya $T$ , length $ℓ$ ke simple pendulum se bada hoga ya chhota? (Tip par point mass $T = 2 π ℓ / g \approx 2.20$ s dega.) Apna guess likho.

Step 1 — pivot ke baare mein moment of inertia. $I = \frac{1}{3} m ℓ^{2} .$ Yeh step kyun? Pivot end par hai, aur apne end ke baare mein rod ka standard Moment of inertia $\frac{1}{3} m ℓ^{2}$ hai — mass poori rod mein spread hai, isliye door wale bits ka zyada count hai.

Step 2 — CM tak distance. $d = \frac{ℓ}{2} = 0.60 m .$ Yeh step kyun? Ek uniform rod ka CM uske geometric middle mein hota hai; gravity wahan act karti hai.

Step 3 — plug in karo. $T = 2 π \frac{\frac{1}{3} m ℓ ^{2}}{m g ( ℓ /2 )} = 2 π \frac{2 ℓ}{3 g} = 2 π \frac{2 ( 1.20 )}{3 ( 9.81 )} \approx 1.79 s .$ Yeh step kyun? Recipe. Notice karo $m$ cancel ho jaata hai — same length ki ek bhaari rod same rate par swing karegi.

Verify: $L_{eq} = \frac{2 ℓ}{3} = 0.80 m < ℓ$ , toh rod ek chhote simple pendulum ki tarah swing karti hai → point-mass-at-tip guess se faster ( $1.79 < 2.20$ s). Mass-near-pivot-speeds-things-up rule se consistent. Units: $m / (m/s^{2}) = s$ . ✓

Example 2 — Case A: uniform disk, pivot rim par

Worked example Disk apne edge se latkana

Ek uniform disk, radius $R = 0.15 m$ , mass $m$ , apne rim par ek point se latki hai. $T$ aur $L_{eq}$ nikalo. ( $g = 9.81$ .)

Forecast: pivot CM se $R$ distance par hai. Kya tumhare hisaab se $L_{eq}$ zyada hoga $R$ se ya kam? (Yaad karo $L_{eq} = k^{2} / d + d > d$ hamesha.)

Figure mein disk (lavender), uska CM center mein marked, aur coral square rim pivot $O$ mark karta hai. Mint arrow woh shift vector hai CM se pivot tak — jiski length exactly $R$ hai, jo ki $d$ distance bhi hai AUR parallel-axis theorem ke zariye $I$ mein add hone wali amount bhi. Butter arrow CM par gravity $m g$ seedha neeche kheench raha hai.

Step 1 — CM ke baare mein $I$ , phir pivot par shift karo. $I_{c m} = \frac{1}{2} m R^{2}, I = I_{c m} + m d^{2} = \frac{1}{2} m R^{2} + m R^{2} = \frac{3}{2} m R^{2} .$ Yeh step kyun? Parallel axis theorem axis ko CM se rim par shift karta hai, $m d^{2} = m R^{2}$ add karke. Figure mein mint arrow dekho: shift distance $R$ hi hai.

Step 2 — $d = R$ . Yeh step kyun? Ek uniform disk ka CM uska center hota hai; rim pivot exactly ek radius door hai.

Step 3 — plug in karo. $T = 2 π \frac{\frac{3}{2} m R ^{2}}{m g R} = 2 π \frac{3 R}{2 g} = 2 π \frac{3 ( 0.15 )}{2 ( 9.81 )} \approx 0.95 s .$ Yeh step kyun? Master formula mein $I = \frac{3}{2} m R^{2}$ aur $d = R$ substitute karo; $R$ ki ek power aur mass $m$ cancel ho jaate hain, sirf clean $3 R /2 g$ bachta hai jo sirf radius aur $g$ par depend karta hai.

Verify: $L_{eq} = I / (m d) = \frac{3}{2} R = 0.225 m > R = 0.15 m$ . ✓ $d$ se bada, bilkul forecast ke anusaar. Units pehle jaisi check hoti hain.

Example 3 — Case B: pivot AT center of mass par (degenerate)

Worked example Kya hoga agar beech mein se thok do?

Wahi rod le lo jaise Ex 1 mein, par isse exactly CM par pivot karo, phir CM se $d = 0.01 m$ dur thodi si distance par. $T$ kya hai?

Forecast: pivot CM par hone se, kya yeh fast swing karegi, slow, ya bilkul nahi?

Step 1 — pivot exactly CM par ( $d = 0$ ). $τ = - m g d sin θ = 0 for all θ .$ Yeh step kyun? Yaad karo $τ$ restoring torque hai (upar define kiya gaya). Gravity ab pivot ke through act karti hai — zero lever arm ( $d = 0$ ), isliye $τ = 0$ . Rod neutral equilibrium mein hai; kisi bhi angle par release karo, woh wahan hi reh jaayegi. Koi oscillation nahi hogi, isliye $T$ undefined hai (formally $T \to \infty$ ).

Step 2 — pivot thoda door, $d = 0.01 m$ . $T = 2 π (k^{2} + d^{2}) / (g d)$ use karo CM ke baare mein rod ke radius of gyration $k = ℓ / 12$ ke saath (Radius of gyration). $ℓ = 1.20$ ke liye: $k^{2} = ℓ^{2} /12 = 0.12 m^{2}$ . $T = 2 π \frac{0.12 + ( 0.01 ) ^{2}}{9.81 ( 0.01 )} = 2 π \frac{0.1201}{0.0981} \approx 6.95 s .$ Yeh step kyun? Jaise $d \to 0$ , numerator $\to k^{2}$ (finite) par denominator $\to 0$ , isliye $T$ bahut bada ho jaata hai — ek bahut badi, sluggish swing.

Verify: end-pivot rod se compare karo ( $T = 1.79$ s). Pivot ko CM ki taraf le jaane se period bahut zyada ho gayi (6.95 s), yeh confirm karta hai ki $T \to \infty$ jab $d \to 0$ . Yeh parent note ke counter-intuitive Forecast-then-Verify point ki baat hai. ✓

Example 4 — Case C: minimum-period pivot ( $d = k$ )

Worked example Rod sabse fast kahan swing karti hai?

$ℓ = 1.20 m$ rod ke liye, woh pivot distance $d$ nikalo jo sabse chhota period deta hai, aur woh $T_{m i n}$ .

Forecast: end ( $d = 0.6$ ) aur CM ( $d = 0$ ) ke beech kahin sabse fastest spot hai. Guess karo kahan.

Figure is rod ke liye period $T$ (vertical) ko pivot distance $d$ (horizontal) ke against plot karta hai. Yeh ek U-shaped valley hai: $T$ left mein $d \to 0$ par shoot up karta hai (pivot CM ke paas, restoring torque khatam ho jaata hai) aur right mein bhi $d$ badhne par rise karta hai. Coral dot valley ke bottom ko mark karta hai $d = k \approx 0.346$ m par, aur mint square end pivot ko mark karta hai ( $d = 0.60$ , $T \approx 1.79$ s) right wall par thoda upar. Is example ka poora point woh coral dot dhundhna hai.

Step 1 — jo cheez minimize karni hai use likho. $T$ tab sabse chhota hoga jab $L_{eq} = k^{2} / d + d$ sabse chhota hoga (kyunki $T = 2 π L_{eq} / g$ , $L_{eq}$ ke saath badhta hai). Toh hum $f (d) = \frac{k ^{2}}{d} + d$ minimize karte hain. Yeh step kyun? Poori square root se ladhne ki jagah, hum andar wali simpler quantity minimize karte hain — square root increasing hai, isliye minimum usi $d$ par hoga.

Step 2 — derivative lo aur har piece samjho. $\frac{df}{dd} = \frac{d}{dd} (\frac{k ^{2}}{d}) + \frac{d}{dd} (d) = - \frac{k ^{2}}{d ^{2}} + 1.$ Yeh step kyun? Hum valley bottom dhundhte hain, jahan slope $df / dd$ zero ho. Do derivatives: (i) $d = d^{1}$ differentiate hokar $1$ deta hai (slope $1$ ki seedhi line); (ii) $k^{2} / d = k^{2} d^{- 1}$ power rule se differentiate hota hai $d^{- 1} \to - 1 \cdot d^{- 2}$ , jo $- k^{2} / d^{2}$ deta hai. Intuition: $d$ thoda bada karne se $k^{2} / d$ term shrink hoti hai (zyada se divide ho rahi hai), isliye uska slope negative hai — isliye minus sign.

Step 3 — slope zero karo aur solve karo. $- \frac{k ^{2}}{d ^{2}} + 1 = 0 \Rightarrow d^{2} = k^{2} \Rightarrow d = k = \frac{ℓ}{12} \approx \frac{1.20}{12} = 0.3464 m .$ Yeh step kyun? Zero slope = valley ka bottom. Do competing effects (torque ka khatam hona vs. inertia ka badhna) exactly balance hote hain jab $d = k$ .

Step 4 — minimum period. $T_{m i n} = 2 π \frac{2 k}{g} = 2 π \frac{2 ( 0.3464 )}{9.81} \approx 1.67 s .$ Yeh step kyun? $d = k$ par, $L_{eq} = k^{2} / k + k = 2 k$ , isliye $T = 2 π 2 k / g$ .

Verify: $1.67 s < 1.79 s$ (end pivot) — haan, faster. Aur $d = 0.346 < 0.6$ , figure mein coral dot ki position mint square ke left mein match karta hai. ✓

Example 5 — Case D: reversibility (center of oscillation)

Worked example Do pivots, ek period

Rod jaisi pehle thi ( $ℓ = 1.20$ ), end par pivot ( $d_{1} = 0.60$ , $L_{eq} = 0.80$ ). Dikhao ki center of oscillation par pivot karne se same $T$ milta hai. Yeh Kater's pendulum ki neev hai.

Forecast: center of oscillation end pivot se $L_{eq} = 0.80$ m door hai. Woh point CM se kitna door hai?

Step 1 — center of oscillation locate karo. Yeh end pivot se $L_{eq} = 0.80 m$ door hai, yaani $0.80 - 0.60 = 0.20 m$ CM se aage (doosri taraf). Toh naya pivot distance $d_{2} = 0.20 m$ hai. Yeh step kyun? Center of oscillation by definition CM line par pivot se $L_{eq}$ distance par hota hai.

Step 2 — naye pivot $d_{2} = 0.20$ ke baare mein period. $k^{2} = 0.12$ ke saath: $T_{2} = 2 π \frac{k ^{2} + d _{2}^{2}}{g d _{2}} = 2 π \frac{0.12 + 0.04}{9.81 ( 0.20 )} = 2 π \frac{0.16}{1.962} \approx 1.79 s .$

Step 3 — Ex 1 se compare karo. Ex 1 ne end pivot se $T_{1} = 1.79 s$ diya tha. Yeh step kyun? $T_{1} = T_{2}$ reversibility theorem hai: do "conjugate" pivots roles swap karte hain aur same period rakhte hain.

Verify: naye pivot se $L_{eq}$ $= k^{2} / d_{2} + d_{2} = 0.12/0.20 + 0.20 = 0.80 m$ — same effective length, isliye same $T$ . Symmetric, jaisa promise kiya tha. ✓

Example 6 — Case E: composite body (rod + bob)

Worked example End par ek lump wali rod

Ek uniform rod, $ℓ = 1.00 m$ , mass $M = 0.50 kg$ , top end par pivot, neeche tip par ek chhota dense bob mass $m_{b} = 0.50 kg$ fix kiya hua hai. $T$ nikalo. ( $g = 9.81$ .)

Forecast: bob CM ko tip ki taraf neeche kheenchta hai. Kya $T$ bare rod se zyada hoga ya kam?

Step 1 — combined center of mass (pivot se). $d = \frac{M ( ℓ /2 ) + m _{b} ℓ}{M + m _{b}} = \frac{0.50 ( 0.50 ) + 0.50 ( 1.00 )}{1.00} = 0.75 m .$ Yeh step kyun? Composite ka CM parts ki positions ka mass-weighted average hota hai. Bhaari tip CM ko $0.50$ se $0.75$ m neeche kheenchti hai.

Step 2 — pivot ke baare mein total moment of inertia. $I = rod, end \frac{1}{3} M ℓ^{2} + point bob at tip m_{b} ℓ^{2} = \frac{1}{3} (0.50) (1.00)^{2} + (0.50) (1.00)^{2} = 0.1667 + 0.50 = 0.6667 kg\cdotm^{2} .$ Yeh step kyun? Moments of inertia ek common axis ke baare mein add hote hain. Distance $ℓ$ par ek point mass $m_{b} ℓ^{2}$ contribute karta hai.

Step 3 — total mass ke saath plug in karo. $T = 2 π \frac{I}{( M + m _{b} ) g d} = 2 π \frac{0.6667}{( 1.00 ) ( 9.81 ) ( 0.75 )} \approx 1.89 s .$ Yeh step kyun? Master formula mein " $m$ " total swinging mass $M + m_{b}$ hai, kyunki gravity poori body par act karti hai distance $d$ par combined CM ke through. Woh combined $I$ aur combined $d$ use karo jo humne abhi nikale, parts ke alag-alag values nahi.

Verify: bare rod (Ex 1 style, $ℓ = 1.0$ ) deta hai $T = 2 π 2 (1.0) / (3 \cdot 9.81) \approx 1.64$ s. Bob ne $T$ ko $1.89$ s tak badha diya — bhaari far end ka matlab bada $I$ jo $d$ se zyada tezi se badhta hai, isliye slower. Physically reasonable. ✓

Example 7 — Case F: large-amplitude limit (formula fail ho jaata hai)

Worked example Jab small-angle formula jhooth bolta hai

End-pivoted rod (Ex 1, $T_{0} = 1.79 s$ tiny swings ke liye) $θ_{0} = 9 0^{\circ}$ se release kiya gaya. True period estimate karo aur dekho simple formula kitna off hai.

Forecast: bada swing — kya real period $1.79$ s se upar jaayegi ya neeche?

Step 1 — simple formula kyun fail hota hai. Humne $sin θ \approx θ$ use kiya, jo sirf chhote $θ$ ke liye valid hai. $9 0^{\circ}$ par woh approximation turning points ke paas slow-down ko understate karti hai, jahan restoring torque $m g d sin θ$ linear $m g d θ$ ke suggest se zyada weak hota hai. Yeh step kyun? Validity domain ke baare mein honesty — parent note ki teesri galti.

Step 2 — angle ko radians mein convert karo, phir correction apply karo. Amplitude-correction series ke liye amplitude $θ_{0}$ radians mein chahiye, kyunki yeh $sin θ$ ke Taylor expansion se aata hai jahan $θ$ radians mein hona chahiye. Pehle convert karo: $θ_{0} = 9 0^{\circ} = 90 \times \frac{π}{180} rad = \frac{π}{2} \approx 1.5708 rad .$ Leading amplitude correction hai $T \approx T_{0} (1 + \frac{1}{16} θ_{0}^{2} + \dots), θ_{0} in radians,$ $T \approx 1.79 (1 + \frac{( 1.5708 ) ^{2}}{16}) = 1.79 (1 + 0.1542) \approx 2.07 s .$ Yeh step kyun? $\frac{1}{16} θ_{0}^{2}$ term exact elliptic-integral period expand karne ka pehla term hai; isme radians use karne padte hain, isliye $9 0^{\circ}$ convert karna bhool jaana ( $1.570 8^{2}$ ki jagah $9 0^{2}$ use karna) ek nonsense answer dega. Yeh classic unit trap hai.

Verify: $2.07 > 1.79 s$ — bade swings slow hote hain, approximately $15%$ zyada. Toh yahan $1.79$ s quote karna noticeable margin se galat hoga. Correction term positive aur small-ish hai, jaisa ek first-order term hona chahiye. ✓

Example 8 — Case G: real-world word problem

Worked example Swinging shop sign

Ek rectangular shop sign, height $h = 0.80 m$ (vertical), ek horizontal rod se apni top edge se latka hai, toh woh us top edge ke baare mein swing karta hai. Ise ek uniform flat plate, mass $m$ , maano. Ek poora swing kitne time mein hoga? ( $g = 9.81$ .)

Forecast: yeh ek plate hai jo apne top par pivot hai. Guess karo $L_{eq}$ zyada hoga ya kam $h$ se.

Step 1 — words ko $m, d, I$ mein translate karo. Swing top edge ke baare mein hai; sirf vertical extent $h$ matter karta hai (horizontal width pivot axis ke along hai aur swinging moment ko affect nahi karta). Ise ek thin lamina ki tarah model karo jo apni top edge ke baare mein swing kar rahi hai = is motion ke liye length $h$ wali rod apne end ke baare mein swing karne jaisa hi. Yeh step kyun? Top edge ke baare mein rotation sirf vertical direction mein distributed mass ko "feel" karta hai; width axis ke parallel hai.

Step 2 — numbers. $I = \frac{1}{3} m h^{2}, d = \frac{h}{2} = 0.40 m .$ Yeh step kyun? Step 1 mein plate ko "apne end par pivot length $h$ wali rod" mein reduce karne ke baad, hum rod-at-end results reuse karte hain: $I = \frac{1}{3} m h^{2}$ aur CM mid-height $h /2$ par hai.

Step 3 — period. $T = 2 π \frac{\frac{1}{3} m h ^{2}}{m g ( h /2 )} = 2 π \frac{2 h}{3 g} = 2 π \frac{2 ( 0.80 )}{3 ( 9.81 )} \approx 1.46 s .$ Yeh step kyun? Master formula mein $I$ aur $d$ substitute karo; $m$ aur $h$ ki ek power cancel ho jaati hai, wahi $2 h /3 g$ shape bachti hai jaise rod mein — jaisi honi chahiye, kyunki humne ise rod ki tarah model kiya tha.

Verify: $L_{eq} = \frac{2 h}{3} = 0.533 m > d = 0.40$ m (hamesha $L_{eq} > d$ ). Ek real sign $0.40$ m string-pendulum se noticeably slow swing karta hai. Units seconds mein. ✓

Example 9 — Case H: exam twist (unknown ke liye invert karo)

Worked example Chhupa hua bob dhundho

Ek light (massless) rod length $ℓ = 1.00 m$ top pivot se $x$ distance par ek point bob mass $m$ carry karta hai. Measured period $T = 1.60 s$ . Bob kahan hai? ( $g = 9.81$ .)

Forecast: ek light rod par ek point mass simply length $x$ ka simple pendulum hai. Toh $x$ kya aana chahiye?

Step 1 — $I$ aur $d$ set up karo. $I = m x^{2}, d = x .$ Yeh step kyun? Saari mass point bob hai distance $x$ par; rod massless hai toh kuch contribute nahi karta.

Step 2 — period simple pendulum mein reduce ho jaata hai. $T = 2 π \frac{m x ^{2}}{m g x} = 2 π \frac{x}{g} .$ Yeh step kyun? $I / (m d) = x^{2} / x = x$ , isliye $L_{eq} = x$ — yeh special case wapas Simple pendulum mein degenerate ho jaata hai. Achha sanity anchor hai.

Step 3 — $x$ ke liye solve karo. $x = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.81 ( 1.60 ) ^{2}}{4 π ^{2}} = \frac{25.11}{39.48} \approx 0.636 m .$ Yeh step kyun? $T = 2 π x / g$ ke dono sides square karo, $T^{2} = 4 π^{2} x / g$ milta hai, phir $x$ isolate karo. Yeh "formula ulta karna" wala move hai jo exam test kar raha hai.

Verify: $x = 0.636 m < ℓ = 1.00$ m, toh bob rod par hai — physically valid. Wapas plug karo: $2 π 0.636/9.81 \approx 1.60$ s ✓, diye gaye period se match karta hai. Aur yeh confirm karta hai forecast ki ek point mass $L_{eq} = x$ deta hai. Units: metres. ✓

Case-coverage recap

Recall Kaun sa example kaun sa case hai?

Ex 1 rod / Ex 2 disk ::: Case A (standard off-CM pivot) Ex 3 pivot at CM ::: Case B (degenerate, $T \to \infty$ ) Ex 4 fastest pivot ::: Case C ( $d = k$ , minimum period) Ex 5 do pivots same $T$ ::: Case D (reversibility, Kater's pendulum) Ex 6 rod plus bob ::: Case E (composite: $I$ add karo, weighted CM) Ex 7 $9 0^{\circ}$ release ::: Case F (large-amplitude, formula fails) Ex 8 shop sign ::: Case G (real-world translation) Ex 9 hidden bob ::: Case H (length ke liye solve karne ke liye invert karo)