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

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The master idea (derive everything from this)

WHY this formula? Start from discrete particles. The COM is the mass-weighted average position: $x_{cm}=\frac{m_1x_1+m_2x_2+\cdots}{m_1+m_2+\cdots}=\frac{\sum m_i x_i}{\sum m_i}$ A heavier particle "pulls" the average toward itself. WHAT happens for a continuous body? Replace the sum by an integral and each $m_i$ by an infinitesimal slice $dm$. That's it — every shape below is just choosing a clever slice $dm$.

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1. Uniform rod (length $L$)

HOW: Put one end at origin, rod along $x$. Linear density $\lambda=M/L$. A slice of width $dx$ at position $x$ has $dm=\lambda\,dx$.

$x_{cm}=\frac{\int_0^L x\,\lambda\,dx}{\int_0^L \lambda\,dx}=\frac{\lambda\cdot \frac{L^2}{2}}{\lambda L}=\boxed{\frac{L}{2}}$

Why this step? λ is constant so it cancels — for a uniform body the COM is purely geometric (the centroid). Result: the middle of the rod.

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2. Triangular lamina (uniform)

HOW (height direction): Triangle of base $b$, height $h$, base on the $x$-axis, apex at height $h$. By similar triangles, a strip at height $y$ has width $w(y)=b\left(1-\frac{y}{h}\right)$. With surface density $\sigma$: $dm=\sigma\,w(y)\,dy=\sigma b\left(1-\frac{y}{h}\right)dy$

=\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}}$$ *Why this step?* σ and $b$ cancel (top and bottom share them). The COM is at height $h/3$ above the base. **General result:** the centroid is the **average of the three vertices**, $\vec r_{cm}=\frac{1}{3}(\vec r_1+\vec r_2+\vec r_3)$ — i.e. the intersection of the medians. --- ## 3. Semicircular wire / arc (radius $R$) **HOW:** Half-ring of mass $M$, radius $R$, lying with its flat side on the $x$-axis, bulging up in $+y$. By symmetry $x_{cm}=0$. Parametrise a tiny arc element by angle $\theta$ from the $x$-axis: arc length $dl=R\,d\theta$, so $dm=\lambda\,dl=\frac{M}{\pi R}R\,d\theta=\frac{M}{\pi}d\theta$. Its height $y=R\sin\theta$. $$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}}$$ *Why this step?* Total length is $\pi R$ so $\lambda=M/(\pi R)$. $\int_0^\pi\sin\theta\,d\theta=2$. So COM is at $\frac{2R}{\pi}\approx0.637R$ above the diameter. --- ## 4. Semicircular disc / lamina (radius $R$) **HOW:** Now it's filled. Slice into thin half-rings (radius $r$, thickness $dr$). A *full* ring of radius $r$ has circumference $2\pi r$; our **half**-ring has length $\pi r$, area $dA=\pi r\,dr$, so $dm=\sigma\,\pi r\,dr$. We already know a half-ring of radius $r$ has its own COM at height $\frac{2r}{\pi}$. So weight each half-ring by that: $$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}}$$ *Why this step?* We reused result #3 as the "$y$" of each shell — that's the power of building up. $\frac{4R}{3\pi}\approx0.424R$. Note it's *less* than the wire's $\frac{2R}{\pi}$ because the filled disc has lots of area near the centre pulling the average down. --- ## 5. Solid hemisphere (radius $R$) **HOW:** Flat face on $xy$-plane, dome along $+z$. Slice into thin circular discs perpendicular to $z$. At height $z$, the disc radius (from sphere equation $x^2+y^2+z^2=R^2$) is $r=\sqrt{R^2-z^2}$. Volume of slab $dV=\pi r^2\,dz=\pi(R^2-z^2)\,dz$, so $dm=\rho\,dV$. Each slab's COM is on the axis at height $z$. $$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}}$$ *Why this step?* The π cancels. $\frac{3R}{8}=0.375R$ above the flat face. > [!example] Bonus — hemispherical **shell** (hollow) > Slice into thin rings on the surface at polar angle $\theta$ from the top. A ring at angle $\theta$ has $z=R\cos\theta$, circumference $2\pi R\sin\theta$, slant width $R\,d\theta$, so $dm=\sigma(2\pi R\sin\theta)(R\,d\theta)$. > $$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}$$ > The shell's COM ($R/2$) is *higher* than the solid's ($3R/8$) — mass on the surface sits farther out. --- > [!recall] Quick self-test (cover the answers!) > - Rod COM? ::: $L/2$ > - Triangle COM height above base? ::: $h/3$ > - Why does the triangle COM sit toward the base? ::: Strips near the base are wider → more mass there. > - Semicircular **wire** COM? ::: $2R/\pi$ > - Semicircular **disc** COM? ::: $4R/3\pi$ > - Solid hemisphere COM? ::: $3R/8$ > - Hemispherical **shell** COM? ::: $R/2$ --- > [!mistake] Steel-manning the classic errors > **Error 1: "Semicircular wire and disc have the same COM."** > *Why it feels right:* both are "half circles, radius $R$." *The fix:* they have **different mass distributions**. The wire ($2R/\pi$) has all mass on the rim. The disc ($4R/3\pi$) has area piled near the centre, dragging the average down. Different $dm$ → different answer. > > **Error 2: Using $dm = \sigma\,dr$ for a disc-slice.** > *Why it feels right:* you copied the "1D" pattern. *The fix:* $dm=\sigma\,dA$, and for a half-ring shell $dA=\pi r\,dr$ (length × thickness). You must include the geometry of the slice's **area**, not just its thickness. > > **Error 3: Putting the triangle COM at $h/2$.** > *Why it feels right:* the rod was at $L/2$, so you assume "middle." *The fix:* the rod has uniform width; the triangle **narrows** toward the apex, so mass is bottom-heavy → $h/3$, not $h/2$. > > **Error 4: Forgetting to use symmetry.** > *Why it feels right:* you dutifully integrate both $x$ and $y$. *The fix:* a symmetry axis forces the COM onto it — set the off-axis coordinate to $0$ instantly and save half the work. --- > [!mnemonic] Remember the answers > **"Rod Two, Tri Three, Wire Two-over-pi, Disc Four-over-three-pi, Solid hemi Three-eighths, Shell a Half."** > Pattern for radius shapes (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 things sit lower than their wire/shell versions* (except the shell vs solid, where surface-mass sits higher). --- > [!recall]- Feynman: explain it to a 12-year-old > Imagine a see-saw. To make it balance, the spot you put the pivot has to "feel" all the weight equally on both sides — heavy kids pull the balance point toward them. The centre of mass is exactly that perfect balance point. > For a stick, it's the middle. For a triangle pizza slice, the wide bottom is heavier, so the balance point sits closer to the bottom (one-third up). For a dome (hemisphere), most of the material is near the bottom, so the balance point is below the middle (three-eighths of the way up). To find it, we slice the shape into tiny pieces, multiply each tiny piece's weight by how far it is, add them all up, and divide by the total weight. That's the whole secret. --- ## #flashcards/physics COM master formula for a continuous body? ::: $\vec r_{cm}=\frac{1}{M}\int \vec r\,dm$ What is $dm$ for a line / area / volume? ::: $\lambda\,dx$ / $\sigma\,dA$ / $\rho\,dV$ Uniform rod COM? ::: $L/2$ (centre) Triangle COM height above base, and general rule? ::: $h/3$; also = average of 3 vertices Why is triangle COM at $h/3$ not $h/2$? ::: Strips widen toward base → mass bottom-heavy Semicircular wire COM? ::: $\frac{2R}{\pi}\approx0.64R$ Semicircular disc COM? ::: $\frac{4R}{3\pi}\approx0.42R$ Solid hemisphere COM above flat face? ::: $\frac{3R}{8}=0.375R$ Hemispherical shell COM? ::: $\frac{R}{2}$ For the semicircular disc, what is $dm$ of a half-ring at radius $r$? ::: $\sigma\pi r\,dr$ For solid hemisphere, slab radius at height $z$? ::: $\sqrt{R^2-z^2}$, area $\pi(R^2-z^2)$ Why does uniform-body COM = centroid? ::: Constant density cancels; only geometry remains --- ## 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Centre of mass ka matlab hai woh ek single point jahan agar tum poore body ka maas concentrate kar do, toh translation ke liye body bilkul ek point particle ki tarah behave karegi. Sabse important formula sirf ek hai: body ko chhote chhote tukdo $dm$ me kaato, har tukde ki position ko uske maas se multiply karo, sab ko add (integrate) karo, aur total maas se divide kar do. Bas yahi sab shapes me lagana hai — $x_{cm}=\frac{1}{M}\int x\,dm$. > > Har shape me sirf ek cheez change hoti hai: tum slice kaise choose karte ho aur uska $dm$ kaise likhte ho. Line ke liye $dm=\lambda\,dx$, area ke liye $dm=\sigma\,dA$, volume ke liye $dm=\rho\,dV$. Rod uniform hai toh COM bilkul beech me, $L/2$. Triangle me niche ki strips chaudi hoti hain (zyada maas), isliye COM base se $h/3$ upar hota hai, $h/2$ nahi — yeh ek common trap hai. Symmetry ka full faayda lo: agar koi symmetry axis hai toh COM usi par hoga, doosra coordinate seedha $0$. > > Semicircle me do alag answer yaad rakho: wire (sirf rim par maas) ka $2R/\pi$, aur filled disc ka $4R/3\pi$. Disc ka chhota isliye hai kyunki centre ke paas bahut saara area hota hai jo average ko niche khinch leta hai. Solid hemisphere ke liye discs me kaato, slab ka radius $\sqrt{R^2-z^2}$, integrate karke aata hai $3R/8$. Hollow shell ka $R/2$ — kyunki surface ka maas bahar ki taraf zyada hota hai. > > Exam tip (80/20): formula ratta mat maaro, **process** yaad rakho — slice, $dm$, position, integrate, divide. Phir agar answer bhool bhi jao, derive kar loge. Aur moment of inertia me bhi yahi slicing technique kaam aayegi, toh yeh ek baar acche se samajh lo. ![[audio/1.4.10-Centre-of-mass-—-derivation-for-common-shapes-(rod,-triangle,-semicircle,-hemispher.mp3]]