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

WHAT tool: the rod result $x_{cm}=L/2$ from the parent note. WHY: uniform linear density $\lambda =M/L$ is constant, so it cancels in $\frac{\int x\lambda dx}{\int \lambda dx}$ — the COM is purely geometric, i.e. the centroid, the plain middle. $x_{cm}=\frac{L}{2}=\frac{1.2}{2}=\boxed{0.6\text{ m}}$

WHAT tool: wire $\to \frac{2R}{\pi }$, disc $\to \frac{4R}{3\pi }$. WHY different: the wire hangs all its mass on the rim; the disc piles area near the centre, which drags the average down. Same silhouette, different mass distribution → different answer. (This is exactly the trap in the parent's Error 1.) $y_{\text{wire}}=\frac{2R}{\pi }=\frac{2(7)}{\pi }\approx 4.456\text{ cm}$ $y_{\text{disc}}=\frac{4R}{3\pi }=\frac{4(7)}{3\pi }\approx 2.971\text{ cm}$

WHAT slice: a piece of width $dx$ at position $x$ carries $dm=\lambda (x)dx={\lambda }_0\frac{x}{L}dx$. WHY not just $L/2$: density is not constant, so $\lambda$ does not cancel — the heavier (right) end pulls the average toward it, so we expect $x_{cm}>L/2$. $x_{cm}=\frac{{\int }_0^{L}x{\lambda }_0\frac{x}{L}dx}{{\int }_0^L{\lambda }_0\frac{x}{L}dx}=\frac{{\int }_0^L{x}^{2}dx}{{\int }_0^{L}xdx}=\frac{L^3/3}{L^2/2}=\boxed{\frac{2L}{3}}$ As predicted, $\frac{2L}{3}>\frac{L}{2}$: the average leans toward the heavy end.

WHAT slice: a horizontal strip at height $y$, width $w(y)=b\left(1-\frac{y}{h}\right)$ by similar triangles (look at the red strip — near the apex it shrinks to zero). Then $dm=\sigma w(y)dy=\sigma b\left(1-\frac{y}{h}\right)dy$. WHY horizontal strips: we want $y_{cm}$, so we group all mass at the same height $y$ into one $dm$ — no need to vary $y$ within a slice. Step 1 — write the full integrals, carrying every factor: $y_{cm}=\frac{{\int }_0^{h}ydm}{{\int }_0^{h}dm}=\frac{{\int }_0^{h}y\sigma b\left(1-\frac{y}{h}\right)dy}{{\int }_0^h\sigma b\left(1-\frac{y}{h}\right)dy}$ Step 2 — pull the constants out of both integrals: $\sigma$ and $b$ do not depend on $y$, so each integral becomes $\sigma b\times (\text{a }y\text{-integral})$: $y_{cm}=\frac{\sigma b{\int }_0^{h}y\left(1-\frac{y}{h}\right)dy}{\sigma b{\int }_0^h\left(1-\frac{y}{h}\right)dy}$ Step 3 — WHY they vanish: the same factor $\sigma b$ sits in numerator and denominator, so it cancels exactly. This is why the answer is purely geometric — it cannot depend on how heavy ($\sigma$) or how wide ($b$) the triangle is. $y_{cm}=\frac{{\int }_0^{h}y\left(1-\frac{y}{h}\right)dy}{{\int }_0^h\left(1-\frac{y}{h}\right)dy}=\frac{h^2/2-h^2/3}{h-h/2}=\frac{h^2/6}{h/2}=\frac{h}{3}$ $y_{cm}=\frac{9}{3}=\boxed{3\text{ cm}}$ Note: this is $h/3$ regardless of $b$ — exactly because $\sigma$ and $b$ cancelled in Step 3.

WHAT tool — the subtraction (negative-mass) trick: treat the punched plate as (full disc) minus (the removed small disc). For a superposition, $x_{cm}=\frac{M_{\text{full}}{x}_{\text{full}}-M_{\text{hole}}{x}_{\text{hole}}}{M_{\text{full}}-M_{\text{hole}}}$ WHY this works: the COM formula is linear in mass, so removing material = adding negative mass. See Centre of mass — definition and system of particles.

Let surface density $\sigma$. Mass $\propto$ area:

• Full disc: $M_{\text{full}}=\sigma \pi R^2$, centred at $x_{\text{full}}=0$.
• Hole: radius $R/2$, so $M_{\text{hole}}=\sigma \pi (R/2{)}^2=\frac{1}{4}\sigma \pi R^2$, centred at $x_{\text{hole}}=R/2$.

$x_{cm}=\frac{\sigma \pi R^2(0)-\frac{1}{4}\sigma \pi R^2\cdot \frac{R}{2}}{\sigma \pi R^2-\frac{1}{4}\sigma \pi R^2}=\frac{-\frac{1}{8}R}{\frac{3}{4}}=-\frac{R}{6}$ $x_{cm}=-\frac{4}{6}=\boxed{-0.667\text{ cm}}$ The COM shifts away from the hole (into the solid side) — sensible, since we removed mass from the $+x$ side. By symmetry $y_{cm}=0$.

WHAT tool: treat each uniform square as a point mass at its own centre (its centroid), then take the mass-weighted average. Equal tiles → equal mass $m$; masses cancel and we average centres. WHY: each square's mass may be replaced by a point at its COM for the purpose of finding the whole body's COM — a theorem of the Centre of mass — definition and system of particles.

Centres: $A=(a/2,a/2)$, $B=(a/2,3a/2)$, $C=(3a/2,a/2)$. $x_{cm}=\frac{a/2+a/2+3a/2}{3}=\frac{5a/2}{3}=\frac{5a}{6},y_{cm}=\frac{a/2+3a/2+a/2}{3}=\frac{5a}{6}$ With $a=2$: $x_{cm}=y_{cm}=\frac{5(2)}{6}=\boxed{1.667\text{ cm}}$. The COM lies just outside the corner tile, pulled diagonally by the two outer squares.

WHAT tool: two known centroids combined by mass-weighted average. Rectangle: area $A_1=2R\cdot R=2R^2$, mass $M_1=2\sigma R^2$, centroid at its own middle $y_1=\frac{R}{2}$. Semicircle: area $A_2=\frac{\pi R^2}{2}$, mass $M_2=\frac{\sigma \pi R^2}{2}$. Its own COM is $\frac{4R}{3\pi }$ above its flat edge, and that flat edge sits at $y=R$. So its centroid height is $y_2=R+\frac{4R}{3\pi }$. WHY add $R$: the disc's $\frac{4R}{3\pi }$ is measured from its base, which we lifted to $y=R$. Always re-anchor sub-COMs to the common origin.

$y_{cm}=\frac{M_1{y}_1+M_2{y}_2}{M_1+M_2}=\frac{2\sigma R^2\cdot \frac{R}{2}+\frac{\sigma \pi R^2}{2}\left(R+\frac{4R}{3\pi }\right)}{2\sigma R^2+\frac{\sigma \pi R^2}{2}}$ Cancel $\sigma R^2$. Numerator $=R+\frac{\pi }{2}\left(R+\frac{4R}{3\pi }\right)=R+\frac{\pi R}{2}+\frac{2R}{3}=\frac{5R}{3}+\frac{\pi R}{2}$. Denominator $=2+\frac{\pi }{2}$. $y_{cm}=\frac{\frac{5}{3}+\frac{\pi }{2}}{2+\frac{\pi }{2}}R\approx \frac{3.238}{3.571}R\approx \boxed{0.907R}$ Just below the glue line ($y=R$) — the heavier rectangle keeps the balance point low, but the disc pulls it up near the seam.

WHAT tool: negative-mass subtraction with the solid-hemisphere result $z=\frac{3R}{8}$. Volume of a hemisphere $\propto R^3$; use $V=\frac{2}{3}\pi R^3$.

• Big: $M_1=\rho \cdot \frac{2}{3}\pi R^3$, $z_1=\frac{3R}{8}$.
• Removed small (radius $R/2$): $M_2=\rho \cdot \frac{2}{3}\pi (R/2{)}^3=\rho \cdot \frac{2}{3}\pi \frac{R^3}{8}=\frac{M_1}{8}$, $z_2=\frac{3(R/2)}{8}=\frac{3R}{16}$.

$z_{cm}=\frac{M_1{z}_1-M_2{z}_2}{M_1-M_2}=\frac{M_1\frac{3R}{8}-\frac{M_1}{8}\frac{3R}{16}}{M_1-\frac{M_1}{8}}$ Cancel $M_1$: numerator $=\frac{3R}{8}-\frac{3R}{128}=\frac{48R-3R}{128}=\frac{45R}{128}$. Denominator $=\frac{7}{8}$. $z_{cm}=\frac{45R/128}{7/8}=\frac{45R}{128}\cdot \frac{8}{7}=\frac{45R}{112}\approx \boxed{0.402R}$ Higher than the plain $0.375R$: we scooped mass from low down (the small hemisphere hugs the flat face), so the average rises.

WHAT parts: the arc and the diameter, both uniform with linear density $\lambda$.

• Arc: length $\pi R$, mass $M_a=\lambda \pi R$, its COM at $y_a=\frac{2R}{\pi }$ (parent result #3).
• Diameter: length $2R$, mass $M_d=2\lambda R$, lying on the axis so $y_d=0$.

$y_{cm}=\frac{M_a{y}_a+M_d{y}_d}{M_a+M_d}=\frac{\lambda \pi R\cdot \frac{2R}{\pi }+0}{\lambda \pi R+2\lambda R}=\frac{2\lambda R^2}{\lambda R(\pi +2)}=\frac{2R}{\pi +2}$ $y_{cm}=\frac{2R}{\pi +2}\approx \boxed{0.389R}$ Limiting discussion: the diameter sits at $y=0$, so adding it can only pull the COM down. Hence $0<\underset{\approx 0.389R}{\underbrace{\frac{2R}{\pi +2}}}<\underset{\approx 0.637R}{\underbrace{\frac{2R}{\pi }}}.$ The pure arc ($\frac{2R}{\pi }$) is the upper bound (no dead weight on the axis); the lower bound $0$ would need infinite mass on the diameter. The "D" lands between — exactly as a mass-weighted average must lie between its extremes. This is the Symmetry arguments in physics principle plus the Centre of mass — definition and system of particles averaging law together.

The argument: the solid hemisphere contains material at every depth from the flat face up to the pole, including a lot of mass near the flat face (the big base discs). The shell puts all its mass on the curved surface, which on average sits farther from the flat plane. A mass-weighted average of farther material lands higher. Hence solid $<$ shell. No numbers needed — just "where does the mass live?"

Quick verification (shell) — WHAT slice: a ring at polar angle $\theta$ from the top has height $z=R\cos \theta$ and mass $dm=\sigma (2\pi R\sin \theta )(Rd\theta )$ (circumference × slant width × density). Step 1 — write $z_{cm}$ carrying every factor: $z_{cm}=\frac{{\int }_0^{\pi /2}(R\cos \theta )\sigma (2\pi R\sin \theta )(Rd\theta )}{{\int }_0^{\pi /2}\sigma (2\pi R\sin \theta )(Rd\theta )}$ Step 2 — pull constants out: every factor except $\cos \theta ,\sin \theta$ is a constant in $\theta$. Numerator carries $\sigma \cdot 2\pi R^2\cdot R=2\pi \sigma R^3$; denominator carries $\sigma \cdot 2\pi R^2=2\pi \sigma R^2$: $z_{cm}=\frac{2\pi \sigma R^3{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta }{2\pi \sigma R^2{\int }_0^{\pi /2}\sin \theta d\theta }$ Step 3 — WHY the constants collapse to a single $R$: the shared factor $2\pi \sigma R^2$ cancels top and bottom; only the extra $R$ from the numerator's $z=R\cos \theta$ survives. So $z_{cm}=R\times (\text{ratio of the two }\theta \text{-integrals})$ — the answer must be a pure number times $R$, as any length should be. $z_{cm}=R\frac{{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta }{{\int }_0^{\pi /2}\sin \theta d\theta }=R\cdot \frac{1/2}{1}=\frac{R}{2}$ Indeed $0.5R>0.375R$. ✓