Worked examples — Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere)

1.4.10 · D3 · Physics › Momentum & Collisions › Centre of mass — derivation for common shapes (rod, triangle

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Scenario matrix

Kuch bhi solve karne se pehle, chalte hain har case class ki list banate hain jo is topic mein aa sakti hai. Har worked example neeche us cell ke saath tagged hai jise wo cover karta hai.

Cell	Case class	Kya tricky hai	Example
A	Uniform 1D, positive coords	baseline sanity	E1
B	Non-uniform density $\lambda (x)$	$\lambda$ ab cancel nahi hoga	E2
C	Coordinates of both signs (origin body ke andar)	negative $x$ average ko left kheenchta hai	E3
D	2D lamina, symmetry ek axis ko khatam karti hai	symmetry use karni padegi	E4
E	Composite / "subtraction" body (hole cut out)	negative mass trick	E5
F	Degenerate / limiting input (length → 0, uniform limit)	formula finite & sensible rehni chahiye	E6
G	Real-world word problem	words ko → integral mein translate karo	E7
H	Exam twist (target COM diya hai, density nikalo)	recipe ko ulta chalao	E8

Hum 8 examples solve karte hain, ek har cell ke liye.

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E1 — Uniform rod, positive coordinates (Cell A)

Forecast: Padhne se pehle andaza lagao — tumhara gut isko kahan rakhta hai? Ek number likh lo.

1. $dm$ likho. Uniform ⇒ $\lambda =M/L$ constant hai, aur position $x$ par width $dx$ ka ek slice ka $dm=\lambda dx$ hoga. Ye step kyun? Ek rod mass ki ek line hai, isliye hum linear density $\lambda$ (mass per length) use karte hain. Constant $\lambda$ "uniform" ka fingerprint hai.
2. Slice coordinate. Slice $x$ par baitha hai; simplify karne ko kuch nahi.
3. Integrate karo. $x_{cm}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}=\frac{\lambda \frac{L^2}{2}}{\lambda L}=\frac{L}{2}=1\text{m}$ Ye step kyun? $\lambda$ upar-neeche cancel ho jaata hai — isi liye ek uniform body ka COM purely geometric hota hai (centroid).

Verify karo: Symmetry se rod apne midpoint ke baare mein left-to-right identical hai, isliye COM zaroor middle mein hoga $=1\text{m}$. Units: metres. ✓

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E2 — Non-uniform rod, $\lambda (x)={\lambda }_{0}x/L$ (Cell B)

Forecast: E1 ke comparison mein zyada mass far end ($x=L$) ki taraf hai, isliye COM midpoint $L/2$ se aage hona chahiye. Fraction guess karo.

1. $dm$ likho. $dm=\lambda (x)dx={\lambda }_0\frac{x}{L}dx$. Ye step kyun? Ab $\lambda$ position par depend karta hai, isliye wo cancel nahi hoga — yahi Cell A se saara farq hai.
2. Mass $M$. $M={\int }_0^L{\lambda }_0\frac{x}{L}dx=\frac{{\lambda }_0}{L}\cdot \frac{L^2}{2}=\frac{{\lambda }_{0}L}{2}$
3. Weighted position. ${\int }_0^{L}xdm={\int }_0^{L}x\cdot {\lambda }_0\frac{x}{L}dx=\frac{{\lambda }_0}{L}{\int }_0^L{x}^{2}dx=\frac{{\lambda }_0}{L}\cdot \frac{L^3}{3}=\frac{{\lambda }_0{L}^2}{3}$
4. Divide karo. $x_{cm}=\frac{{\lambda }_0{L}^2/3}{{\lambda }_{0}L/2}=\frac{2L}{3}$ Ye step kyun? Mass-weighted average bhaari end ki taraf khiich jaata hai, aur $\frac{2}{3}L$ par land karta hai — middle ke baad, bilkul forecast ke anusaar.

Verify karo: Limit check karo — agar $\lambda$ constant hoti, toh hume wapas $L/2$ milta. Chunki hamari density $x=L$ ki taraf jhukti hai, $\frac{2}{3}L>\frac{1}{2}L$: answer sahi direction mein gaya. Units: metres. ✓

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E3 — Do blocks, dono signs ke coordinates (Cell C)

Forecast: Bhaara block negative side par hai, isliye answer negative hona chahiye (origin ke left).

1. Discrete formula use karo. $x_{cm}=\frac{\sum m_i{x}_i}{\sum m_i}=\frac{(3)(-4)+(1)(2)}{3+1}$ Ye step kyun? Point masses ke liye hum integrate nahi karte — mass-weighted average ek direct sum hai. Coordinates ke signs bina chhuye aate hain.
2. Compute karo. $x_{cm}=\frac{-12+2}{4}=\frac{-10}{4}=-2.5\text{m}$

Verify karo: COM dono blocks ($-4$ aur $+2$) ke beech hai — achha. Ye $3\text{kg}$ block ke kaafi kareeb hai, kyunki ek bhaari mass balance point ko apni taraf kheenchti hai. Sign negative hai, forecast se match karta hai. ✓

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E4 — Semicircular disc, symmetry $x$-axis ko khatam karti hai (Cell D)

Forecast: Shape left↔right mirror image hai, isliye $x_{cm}$ bilkul $0$ hona chahiye. Aur $y_{cm}$ humne parent mein $\frac{4R}{3\pi }$ derive kiya tha.

1. Pehle symmetry. Disc $y$-axis ke baare mein symmetric hai (figure mein dashed mirror line dekho). $+x$ par har slice ke liye $-x$ par ek identical slice hai; unke contributions cancel ho jaate hain. Ye step kyun? Symmetry $x_{cm}=0$ free mein deti hai — jo symmetry already answer kar de use kabhi integrate mat karo (parent note mein Error 4).
2. Vertical coordinate ke liye parent result reuse karo: $y_{cm}=\frac{4R}{3\pi }$ Ye step kyun? Hum yeh half-rings stack karke pehle hi build kar chuke hain (continuous summation); ise dobara karne ki zaroorat nahi.
3. $R=6$ plug karo. $y_{cm}=\frac{4(6)}{3\pi }=\frac{24}{3\pi }=\frac{8}{\pi }\approx 2.546\text{cm}$

Verify karo: $\frac{4R}{3\pi }\approx 0.424R=0.424\times 6=2.546\text{cm}$ ✓. Ye wire ke $\frac{2R}{\pi }\approx 3.82\text{cm}$ se neeche baitha hai, jaisa expected tha (centre ke paas filled area average ko neeche kheenchti hai). ✓

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E5 — Hole waala disc (Cell E, negative mass)

Forecast: Humne mass right side se hataya, isliye jo bacha uska COM origin se left shift hona chahiye — ek chota negative $x$.

1. Subtraction trick. "Full disc" minus "hole disc" ko do bodies mano: ek positive-mass full disc, aur ek negative-mass disc jahan hole hai. Ye step kyun? COM formula mass mein linear hai. Material hatana usi ke barabar hai jaise hole ki jagah negative mass ki body add karna — isse hum ek messy integral se bach jaate hain.
2. Masses (uniform ⇒ mass ∝ area). Full disc mass $=M$ mano. Area of full disc $\propto R^2$; area of hole $\propto (R/2{)}^2=R^2/4$. To hole ki "mass" $=M/4$. Ye step kyun? Constant $\sigma$ ke saath, mass ratios area ratios ke barabar hote hain — actual $\sigma$ ki zaroorat nahi.
3. Positions. Full disc COM $x=0$ par; hole COM $x=R/2$ par. Discrete formula apply karo hole ko $-M/4$ maanke: $x_{cm}=\frac{M(0)-\frac{M}{4}\left(\frac{R}{2}\right)}{M-\frac{M}{4}}=\frac{-\frac{MR}{8}}{\frac{3M}{4}}=-\frac{R}{6}$
4. $R=10$ plug karo. $x_{cm}=-\frac{10}{6}\approx -1.667\text{cm}$, aur $y_{cm}=0$ (dono discs $x$-axis par hain, $y$ mein symmetric hain).

Verify karo: Sign negative hai → hataye gaye material se door shift hua, forecast se match karta hai. Magnitude choti hai ($R/6$) kyunki sirf $1/4$ area hataya gaya tha. Units: cm. ✓

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E6 — Degenerate & limiting inputs (Cell F)

Forecast: (a) Ek vanishing rod ek point mein collapse hoti hai, isliye uska COM us point par hona chahiye ($x=0$). (b) Uniform rod ka COM $L/2$ hai, isliye hume wohi wapas milna chahiye.

1. (a) Limit lo. $\underset{L\to 0}{\lim }\frac{L}{2}=0$. Ye step kyun? Ek degenerate (zero-length) body origin par sirf ek point hai; formula ko blow up ya nonsense nahi dena chahiye. Ye $0$ deta hai — sensible.
2. (b) E2 generalise karo. $\lambda (x)={\lambda }_0(x/L{)}^n$ ke liye, $x_{cm}=\frac{{\int }_0^{L}x\cdot x^{n}dx}{{\int }_0^L{x}^{n}dx}=\frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)}=\frac{n+1}{n+2}L$ Ye step kyun? Ye ek formula ek saath har uniform-power rod ko contain karta hai — ek limiting-behaviour sanity net.
3. Special values check karo.
   • $n=0$ (uniform): $x_{cm}=\frac{1}{2}L$ ✓ (E1 se match karta hai).
   • $n=1$ (E2 ka linear): $x_{cm}=\frac{2}{3}L$ ✓ (E2 se match karta hai).
   • $n\to \infty$ (saara mass far end par crush): $x_{cm}=\frac{n+1}{n+2}L\to L$ ✓.

Verify karo: Har limit ek independently known answer reproduce karta hai, aur extreme $n\to \infty$ COM ko far tip par le jaata hai jahan saara mass concentrate hota hai. ✓

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E7 — Real-world word problem (Cell G)

Forecast: Insaan bilkul front par extra mass add karta hai, isliye combined COM canoe ke apne middle ($2.0\text{m}$) se aage hona chahiye par front tak nahi.

1. Do point masses ki tarah model karo. Canoe ka apna COM (uniform rod) uske midpoint par: $x_{\text{canoe}}=2.0\text{m}$. Insaan front tip par: $x_{\text{person}}=4.0\text{m}$. Ye step kyun? Hum extended canoe ko uske centroid par ek point mein collapse karte hain — COM exist karne ka poora reason yahi hai (dekho Centre of mass — definition and system of particles).
2. Weighted average. $x_{cm}=\frac{(60)(2.0)+(40)(4.0)}{60+40}=\frac{120+160}{100}=\frac{280}{100}=2.8\text{m}$

Verify karo: $2.8\text{m}$ canoe centre ($2.0$) aur insaan ($4.0$) ke beech hai, canoe ki taraf lean karta hai kyunki canoe bhaari hai ($60>40$). Units: metres. ✓

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E8 — Exam twist: recipe ko ulta chalao (Cell H)

Forecast: Positive $b$ rod ko far end ki taraf bhaari banata hai, $x_{cm}$ ko $L/2$ se aage push karta hai. Chunki $\frac{3}{5}L>\frac{1}{2}L$, hum positive $b/a$ expect karte hain.

1. Dono integrals set up karo. $M={\int }_0^L(a+bx)dx=aL+\frac{b{L}^2}{2}$ ${\int }_0^{L}x(a+bx)dx=\frac{a{L}^2}{2}+\frac{b{L}^3}{3}$ Ye step kyun? Hume abhi $b/a$ nahi pata, isliye hum dono letters carry karte hain aur end mein target impose karte hain — yahi recipe ka "backwards" chalana hai.
2. COM condition likho. $x_{cm}=\frac{\frac{a{L}^2}{2}+\frac{b{L}^3}{3}}{aL+\frac{b{L}^2}{2}}=\frac{3}{5}L$
3. Solve karo. Numerator aur denominator ko $L$ se divide karo, $k=b/a$ lo: $\frac{\frac{L}{2}+\frac{k{L}^2}{3}}{1+\frac{kL}{2}}=\frac{3L}{5}$ Cross-multiply karo aur $L$ cancel karo: $5\left(\frac{1}{2}+\frac{kL}{3}\right)=3\left(1+\frac{kL}{2}\right)\Rightarrow \frac{5}{2}+\frac{5kL}{3}=3+\frac{3kL}{2}$ $\frac{5kL}{3}-\frac{3kL}{2}=\frac{1}{2}\Rightarrow kL\left(\frac{10-9}{6}\right)=\frac{1}{2}\Rightarrow \frac{kL}{6}=\frac{1}{2}\Rightarrow kL=3$ To $\frac{b}{a}=\frac{3}{L}$.

Verify karo: Step 2 mein $k=3/L$ wapas plug karo: numerator $=\frac{a{L}^2}{2}+\frac{a}{L}\cdot \frac{L^3}{3}\cdot 3=\frac{a{L}^2}{2}+a{L}^2=\frac{3a{L}^2}{2}$; denominator $=aL+\frac{a}{L}\cdot \frac{L^2}{2}\cdot 3=aL+\frac{3aL}{2}=\frac{5aL}{2}$; ratio $=\frac{3a{L}^2/2}{5aL/2}=\frac{3L}{5}$ ✓. Aur $b/a>0$, forecast se match karta hai. ✓

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Recall Which cell did each example hit?

E1 (A: uniform 1D) ::: rod midpoint $L/2$ E2 (B: non-uniform density) ::: $2L/3$ E3 (C: both-sign coordinates) ::: $-2.5\text{m}$ E4 (D: symmetry kills an axis) ::: $x=0,y=4R/3\pi$ E5 (E: subtraction / negative mass) ::: $-R/6$ E6 (F: degenerate & limiting) ::: general $\frac{n+1}{n+2}L$ E7 (G: word problem) ::: $2.8\text{m}$ back se E8 (H: reverse-engineer density) ::: $b/a=3/L$