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

1.4.10 · Physics › Momentum & Collisions

Master idea (sab kuch isi se derive karo)

YEH formula kyun hai? Discrete particles se shuru karo. COM mass-weighted average position hai: $x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2} + \dots}{m _{1} + m _{2} + \dots} = \frac{\sum m _{i} x _{i}}{\sum m _{i}}$ Ek heavy particle average ko apni taraf "kheenchta" hai. KYA hota hai continuous body ke liye? Sum ko integral se replace karo aur har $m_{i}$ ko ek infinitesimal slice $d m$ se. Bas — neeche har shape sirf ek clever slice $d m$ choose karne ki baat hai.

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

1. Uniform rod (length $L$ )

KAISE: Ek end ko origin pe rakho, rod $x$ ke along. Linear density $λ = M / L$ . Position $x$ par width $d x$ ki ek slice ka $d m = λ d x$ hai.

$x_{c m} = \frac{\int _{0}^{L} x λ d x}{\int _{0}^{L} λ d x} = \frac{λ \cdot \frac{L ^{2}}{2}}{λ L} = \frac{L}{2}$

Yeh step kyun? λ constant hai isliye cancel ho jaata hai — uniform body ke liye COM purely geometric hai (the centroid). Result: the middle of the rod.

2. Triangular lamina (uniform)

KAISE (height direction): Base $b$ , height $h$ ka triangle, base $x$ -axis par, apex height $h$ par. Similar triangles se, height $y$ par ek strip ki width $w (y) = b (1 - \frac{y}{h})$ hai. Surface density $σ$ ke saath: $d m = σ w (y) d y = σ b (1 - \frac{y}{h}) d y$

=\frac{\int_0^h y\left(1-\frac{y}{h}\right)dy}{\int_0^h \left(1-\frac{y}{h}\right)dy} =\frac{\frac{h^2}{2}-\frac{h^2}{3}}{h-\frac{h}{2}} =\frac{h^2/6}{h/2}=\boxed{\frac{h}{3}}$$ *Yeh step kyun?* σ aur $b$ cancel ho jaate hain (upar aur neeche dono mein hain). COM base se $h/3$ height par hai. **General result:** centroid **teen vertices ka average** hai, $\vec r_{cm}=\frac{1}{3}(\vec r_1+\vec r_2+\vec r_3)$ — yaani medians ka intersection. --- ## 3. Semicircular wire / arc (radius $R$) **KAISE:** Mass $M$, radius $R$ ka half-ring, apna flat side $x$-axis par rakhe, $+y$ mein bulge karta hua. Symmetry se $x_{cm}=0$. Ek tiny arc element ko $x$-axis se angle $\theta$ se parametrise karo: arc length $dl=R\,d\theta$, isliye $dm=\lambda\,dl=\frac{M}{\pi R}R\,d\theta=\frac{M}{\pi}d\theta$. Uski height $y=R\sin\theta$ hai. $$y_{cm}=\frac{1}{M}\int_0^\pi (R\sin\theta)\frac{M}{\pi}\,d\theta =\frac{R}{\pi}\int_0^\pi\sin\theta\,d\theta =\frac{R}{\pi}\bigl[-\cos\theta\bigr]_0^\pi=\frac{R}{\pi}(2)=\boxed{\frac{2R}{\pi}}$$ *Yeh step kyun?* Total length $\pi R$ hai isliye $\lambda=M/(\pi R)$. $\int_0^\pi\sin\theta\,d\theta=2$. Toh COM diameter se $\frac{2R}{\pi}\approx0.637R$ upar hai. --- ## 4. Semicircular disc / lamina (radius $R$) **KAISE:** Ab yeh filled hai. Thin half-rings (radius $r$, thickness $dr$) mein kaato. Radius $r$ ki ek *full* ring ki circumference $2\pi r$ hai; humari **half**-ring ki length $\pi r$, area $dA=\pi r\,dr$, isliye $dm=\sigma\,\pi r\,dr$. Hum pehle se jaante hain ki radius $r$ ki half-ring ka apna COM $\frac{2r}{\pi}$ height par hai. Toh har half-ring ko isse weight karo: $$y_{cm}=\frac{\int_0^R \frac{2r}{\pi}\,dm}{\int_0^R dm} =\frac{\int_0^R \frac{2r}{\pi}\,\sigma\pi r\,dr}{\int_0^R \sigma\pi r\,dr} =\frac{2\int_0^R r^2\,dr}{\pi\int_0^R r\,dr} =\frac{2\cdot \frac{R^3}{3}}{\pi\cdot \frac{R^2}{2}}=\boxed{\frac{4R}{3\pi}}$$ *Yeh step kyun?* Humne result #3 ko har shell ke "$y$" ki tarah reuse kiya — yahi build-up karne ki power hai. $\frac{4R}{3\pi}\approx0.424R$. Note karo yeh wire ke $\frac{2R}{\pi}$ se *kam* hai kyunki filled disc mein centre ke paas bahut saari area hai jo average ko neeche kheenchti hai. --- ## 5. Solid hemisphere (radius $R$) **KAISE:** Flat face $xy$-plane par, dome $+z$ ki taraf. $z$ ke perpendicular thin circular discs mein kaato. Height $z$ par, disc radius (sphere equation $x^2+y^2+z^2=R^2$ se) $r=\sqrt{R^2-z^2}$ hai. Slab ka volume $dV=\pi r^2\,dz=\pi(R^2-z^2)\,dz$, isliye $dm=\rho\,dV$. Har slab ka COM axis par height $z$ par hai. $$z_{cm}=\frac{\int_0^R z\,\pi(R^2-z^2)\,dz}{\int_0^R \pi(R^2-z^2)\,dz}$$ Numerator: $\int_0^R (zR^2-z^3)\,dz=\frac{R^4}{2}-\frac{R^4}{4}=\frac{R^4}{4}$. Denominator: $\int_0^R (R^2-z^2)\,dz=R^3-\frac{R^3}{3}=\frac{2R^3}{3}$. $$z_{cm}=\frac{R^4/4}{2R^3/3}=\boxed{\frac{3R}{8}}$$ *Yeh step kyun?* π cancel ho jaata hai. Flat face se $\frac{3R}{8}=0.375R$ upar. > [!example] Bonus — hemispherical **shell** (hollow) > Top se polar angle $\theta$ par surface par thin rings mein kaato. Angle $\theta$ par ek ring ka $z=R\cos\theta$, circumference $2\pi R\sin\theta$, slant width $R\,d\theta$, isliye $dm=\sigma(2\pi R\sin\theta)(R\,d\theta)$. > $$z_{cm}=\frac{\int_0^{\pi/2} R\cos\theta\,\sin\theta\,d\theta}{\int_0^{\pi/2}\sin\theta\,d\theta}=\frac{\frac12}{1}\,R=\frac{R}{2}$$ > Shell ka COM ($R/2$) solid ke COM ($3R/8$) se *zyada upar* hai — surface par mass zyada door baithti hai. --- > [!recall] Quick self-test (answers cover karo!) > - Rod COM? ::: $L/2$ > - Triangle COM ki base se height? ::: $h/3$ > - Triangle COM base ki taraf kyun hota hai? ::: Base ke paas strips wider hain → wahan zyada mass hai. > - Semicircular **wire** COM? ::: $2R/\pi$ > - Semicircular **disc** COM? ::: $4R/3\pi$ > - Solid hemisphere COM? ::: $3R/8$ > - Hemispherical **shell** COM? ::: $R/2$ --- > [!mistake] Classic errors ko steel-man karna > **Error 1: "Semicircular wire aur disc ka same COM hota hai."** > *Yeh sahi kyun lagta hai:* dono "half circles, radius $R$" hain. *Fix:* unke **mass distributions alag hain**. Wire ($2R/\pi$) ki saari mass rim par hai. Disc ($4R/3\pi$) ka area centre ke paas pile hai, average ko neeche kheenchta hai. Alag $dm$ → alag answer. > > **Error 2: Disc-slice ke liye $dm = \sigma\,dr$ use karna.** > *Yeh sahi kyun lagta hai:* tumne "1D" pattern copy kiya. *Fix:* $dm=\sigma\,dA$, aur half-ring shell ke liye $dA=\pi r\,dr$ (length × thickness) hai. Tumhe sirf thickness nahi, slice ke **area** ki geometry include karni hogi. > > **Error 3: Triangle COM ko $h/2$ par rakhna.** > *Yeh sahi kyun lagta hai:* rod $L/2$ par tha, isliye tum "middle" assume karte ho. *Fix:* rod ki uniform width hai; triangle apex ki taraf **narrow** hota hai, isliye mass bottom-heavy hai → $h/3$, $h/2$ nahi. > > **Error 4: Symmetry use karna bhool jaana.** > *Yeh sahi kyun lagta hai:* tum dutifully dono $x$ aur $y$ integrate karte ho. *Fix:* ek symmetry axis COM ko ussi par force karta hai — off-axis coordinate ko turant $0$ set karo aur aadha kaam bachao. --- > [!mnemonic] Answers yaad karo > **"Rod Two, Tri Three, Wire Two-over-pi, Disc Four-over-three-pi, Solid hemi Three-eighths, Shell a Half."** > Radius shapes ka pattern (fraction of $R$): wire $\frac{2}{\pi}\!\approx\!0.64$ → disc $\frac{4}{3\pi}\!\approx\!0.42$ → solid hemi $\frac{3}{8}\!=\!0.375$ → shell $\frac{1}{2}\!=\!0.50$. > *Filled cheezein apne wire/shell versions se neeche baithti hain* (shell vs solid ko chhodkar, jahan surface-mass zyada upar hoti hai). --- > [!recall]- Feynman: ek 12-saal ke bachche ko explain karo > Ek see-saw imagine karo. Ise balance karane ke liye, pivot jahan rakho woh spot dono sides ka poora weight "feel" kare — heavy bacche balance point ko apni taraf kheenchte hain. Centre of mass bilkul wahi perfect balance point hai. > Ek stick ke liye, yeh beech mein hota hai. Triangle pizza slice ke liye, wide bottom heavy hai, isliye balance point neeche ke paas hota hai (ek-tehai upar). Ek dome (hemisphere) ke liye, zyada material neeche ke paas hai, isliye balance point middle se neeche hai (teen-aathwa raaste tak). Ise find karne ke liye, shape ko tiny pieces mein kaato, har tiny piece ke weight ko uski door se multiply karo, sab add karo, aur total weight se divide karo. Yahi poora raaz hai. --- ## #flashcards/physics COM master formula for a continuous body? ::: $\vec r_{cm}=\frac{1}{M}\int \vec r\,dm$ Line / area / volume ke liye $dm$ kya hai? ::: $\lambda\,dx$ / $\sigma\,dA$ / $\rho\,dV$ Uniform rod COM? ::: $L/2$ (centre) Triangle COM ki base se height, aur general rule? ::: $h/3$; also = average of 3 vertices Triangle COM $h/3$ par kyun hai $h/2$ nahi? ::: Strips base ki taraf wide hoti hain → mass bottom-heavy hai Semicircular wire COM? ::: $\frac{2R}{\pi}\approx0.64R$ Semicircular disc COM? ::: $\frac{4R}{3\pi}\approx0.42R$ Solid hemisphere COM flat face ke upar? ::: $\frac{3R}{8}=0.375R$ Hemispherical shell COM? ::: $\frac{R}{2}$ Semicircular disc ke liye, radius $r$ par half-ring ka $dm$ kya hai? ::: $\sigma\pi r\,dr$ Solid hemisphere mein, height $z$ par slab radius? ::: $\sqrt{R^2-z^2}$, area $\pi(R^2-z^2)$ Uniform-body COM = centroid kyun? ::: Constant density cancel ho jaati hai; sirf geometry bachti hai --- ## Connections - [[Centre of mass — definition and system of particles]] - [[Centroid vs centre of mass vs centre of gravity]] - [[Moment of inertia — derivation for common shapes]] (same slicing technique!) - [[Momentum conservation]] (external force acts at COM) - [[Integration as continuous summation]] - [[Symmetry arguments in physics]] ## 🖼️ Concept Map ```mermaid flowchart TD COM[Centre of mass] -->|"mass-weighted avg position"| DISC["Discrete: sum m_i x_i over sum m_i"] DISC -->|"sum to integral"| MASTER["r_cm = 1/M integral r dm"] MASTER -->|"3-step recipe"| RECIPE[Pick slice, write coord, integrate] RECIPE -->|"line dm=lambda dx"| ROD[Uniform rod] RECIPE -->|"area dm=sigma dA"| TRI[Triangular lamina] RECIPE -->|"area dm=sigma dA"| SEMI[Semicircle] RECIPE -->|"volume dm=rho dV"| HEMI[Hemisphere] ROD -->|"gives"| RRES["x_cm = L/2"] TRI -->|"strips parallel to base"| TRES["y_cm = h/3"] MASTER -->|"uniform body"| CENTROID[COM = geometric centroid] CENTROID -->|"exploit"| SYMM[Symmetry kills one axis] SYMM -->|"simplifies"| RECIPE ```

Master idea (sab kuch isi se derive karo)

1. Uniform rod (length L)

2. Triangular lamina (uniform)

1. Uniform rod (length $L$ )