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

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The scenario matrix

Before solving anything, let's list every case class this topic can throw at you. Each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What's tricky	Example
A	Uniform 1D, positive coords	baseline sanity	E1
B	Non-uniform density $\lambda (x)$	$\lambda$ no longer cancels	E2
C	Coordinates of both signs (origin inside body)	negative $x$ pulls average left	E3
D	2D lamina, symmetry kills one axis	must use symmetry	E4
E	Composite / "subtraction" body (hole cut out)	negative mass trick	E5
F	Degenerate / limiting input (length → 0, uniform limit)	formula must stay finite & sensible	E6
G	Real-world word problem	translate words → integral	E7
H	Exam twist (find density given a target COM)	run the recipe backwards	E8

We solve 8 examples, one per cell.

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E1 — Uniform rod, positive coordinates (Cell A)

Forecast: guess before reading — where does your gut put it? Write down a number.

1. Write $dm$. Uniform ⇒ $\lambda =M/L$ constant, and a slice of width $dx$ at position $x$ has $dm=\lambda dx$. Why this step? A rod is a line of mass, so we use linear density $\lambda$ (mass per length). Constant $\lambda$ is the fingerprint of "uniform."
2. Slice coordinate. The slice sits at $x$; nothing to simplify.
3. Integrate. $x_{cm}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}=\frac{\lambda \frac{L^2}{2}}{\lambda L}=\frac{L}{2}=1\text{m}$ Why this step? $\lambda$ cancels top and bottom — that's why a uniform body's COM is purely geometric (the centroid).

Verify: By symmetry the rod is identical left-to-right about its midpoint, so the COM must be at the middle $=1\text{m}$. Units: metres. ✓

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

Forecast: More mass sits toward the far end ($x=L$) than in E1, so the COM should be past the midpoint $L/2$. Guess the fraction.

1. Write $dm$. $dm=\lambda (x)dx={\lambda }_0\frac{x}{L}dx$. Why this step? Now $\lambda$ depends on position, so it will not cancel — this is the whole difference from Cell A.
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. $x_{cm}=\frac{{\lambda }_0{L}^2/3}{{\lambda }_{0}L/2}=\frac{2L}{3}$ Why this step? The mass-weighted average is dragged toward the heavy end, landing at $\frac{2}{3}L$ — past the middle, exactly as forecast.

Verify: Check the limit — if instead $\lambda$ were constant, we'd get back $L/2$. Since our density leans toward $x=L$, $\frac{2}{3}L>\frac{1}{2}L$: the answer moved the right direction. Units: metres. ✓

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E3 — Two blocks, coordinates of both signs (Cell C)

Forecast: The heavy block is on the negative side, so the answer should be negative (left of origin).

1. Use the discrete formula. $x_{cm}=\frac{\sum m_i{x}_i}{\sum m_i}=\frac{(3)(-4)+(1)(2)}{3+1}$ Why this step? For point masses we don't integrate — the mass-weighted average is a direct sum. Signs of coordinates carry through untouched.
2. Compute. $x_{cm}=\frac{-12+2}{4}=\frac{-10}{4}=-2.5\text{m}$

Verify: The COM lies between the two blocks ($-4$ and $+2$) — good. It's much closer to the $3\text{kg}$ block, as a heavy mass should pull the balance point toward it. Sign is negative, matching the forecast. ✓

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E4 — Semicircular disc, symmetry kills the $x$-axis (Cell D)

Forecast: The shape is a mirror image left↔right, so $x_{cm}$ should be exactly $0$. And $y_{cm}$ we derived in the parent as $\frac{4R}{3\pi }$.

1. Symmetry first. The disc is symmetric about the $y$-axis (see the dashed mirror line in the figure). For every slice at $+x$ there is an identical slice at $-x$; their contributions cancel. Why this step? Symmetry gives $x_{cm}=0$ for free — never integrate what symmetry already answers (Error 4 in the parent note).
2. Reuse the parent result for the vertical coordinate: $y_{cm}=\frac{4R}{3\pi }$ Why this step? We already built this by stacking half-rings (continuous summation); no need to redo it.
3. Plug in $R=6$. $y_{cm}=\frac{4(6)}{3\pi }=\frac{24}{3\pi }=\frac{8}{\pi }\approx 2.546\text{cm}$

Verify: $\frac{4R}{3\pi }\approx 0.424R=0.424\times 6=2.546\text{cm}$ ✓. It sits below the wire's $\frac{2R}{\pi }\approx 3.82\text{cm}$, as expected (filled area near the centre drags the average down). ✓

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E5 — Disc with a hole cut out (Cell E, negative mass)

Forecast: We removed mass from the right side, so the COM of what remains should shift left of the origin — a small negative $x$.

1. The subtraction trick. Treat "full disc" minus "hole disc" as two bodies: a positive-mass full disc, and a negative-mass disc where the hole is. Why this step? The COM formula is linear in mass. Removing material is the same as adding a body of negative mass at the hole's location — this saves us a messy integral.
2. Masses (uniform ⇒ mass ∝ area). Let full disc mass $=M$. Area of full disc $\propto R^2$; area of hole $\propto (R/2{)}^2=R^2/4$. So hole "mass" $=M/4$. Why this step? With constant $\sigma$, mass ratios equal area ratios — no need for the actual $\sigma$.
3. Positions. Full disc COM at $x=0$; hole COM at $x=R/2$. Apply the discrete formula with the hole as $-M/4$: $x_{cm}=\frac{M(0)-\frac{M}{4}\left(\frac{R}{2}\right)}{M-\frac{M}{4}}=\frac{-\frac{MR}{8}}{\frac{3M}{4}}=-\frac{R}{6}$
4. Plug in $R=10$. $x_{cm}=-\frac{10}{6}\approx -1.667\text{cm}$, and $y_{cm}=0$ (both discs sit on the $x$-axis, symmetric in $y$).

Verify: Sign is negative → shifted away from the removed material, as forecast. Magnitude is small ($R/6$) because only $1/4$ of the area was removed. Units: cm. ✓

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E6 — Degenerate & limiting inputs (Cell F)

Forecast: (a) A vanishing rod collapses to a point, so its COM should sit at that point ($x=0$). (b) A uniform rod's COM is $L/2$, so we should recover that.

1. (a) Take the limit. $\underset{L\to 0}{\lim }\frac{L}{2}=0$. Why this step? A degenerate (zero-length) body is just a point at the origin; the formula must not blow up or give nonsense. It gives $0$ — sensible.
2. (b) Generalise E2. For $\lambda (x)={\lambda }_0(x/L{)}^n$, $x_{cm}=\frac{{\int }_0^{L}x\cdot x^{n}dx}{{\int }_0^L{x}^{n}dx}=\frac{L^{n+2}/(n+2)}{L^{n+1}/(n+1)}=\frac{n+1}{n+2}L$ Why this step? This one formula contains every uniform-power rod at once — a limiting-behaviour sanity net.
3. Check the special values.
   • $n=0$ (uniform): $x_{cm}=\frac{1}{2}L$ ✓ (matches E1).
   • $n=1$ (E2's linear): $x_{cm}=\frac{2}{3}L$ ✓ (matches E2).
   • $n\to \infty$ (all mass crushed at the far end): $x_{cm}=\frac{n+1}{n+2}L\to L$ ✓.

Verify: Each limit reproduces an independently known answer, and the extreme $n\to \infty$ pushes the COM onto the far tip exactly where all the mass concentrates. ✓

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E7 — Real-world word problem (Cell G)

Forecast: The person adds mass at the very front, so the combined COM should sit past the canoe's own middle ($2.0\text{m}$) but not all the way to the front.

1. Model as two point masses. Canoe's own COM (uniform rod) is at its midpoint: $x_{\text{canoe}}=2.0\text{m}$. Person at the front tip: $x_{\text{person}}=4.0\text{m}$. Why this step? We collapse the extended canoe to a point at its centroid — the whole reason COM exists (see 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: $2.8\text{m}$ lies between the canoe centre ($2.0$) and the person ($4.0$), leaning toward the canoe because the canoe is heavier ($60>40$). Units: metres. ✓

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E8 — Exam twist: run the recipe backwards (Cell H)

Forecast: Positive $b$ makes the rod heavier toward the far end, pushing $x_{cm}$ past $L/2$. Since $\frac{3}{5}L>\frac{1}{2}L$, we expect a positive $b/a$.

1. Set up both integrals. $M={\int }_0^L(a+bx)dx=aL+\frac{b{L}^2}{2}$ ${\int }_0^{L}x(a+bx)dx=\frac{a{L}^2}{2}+\frac{b{L}^3}{3}$ Why this step? We don't yet know $b/a$, so we carry both letters and impose the target at the end — that's "backwards" running of the recipe.
2. Write the COM condition. $x_{cm}=\frac{\frac{a{L}^2}{2}+\frac{b{L}^3}{3}}{aL+\frac{b{L}^2}{2}}=\frac{3}{5}L$
3. Solve. Divide numerator and denominator by $L$, let $k=b/a$: $\frac{\frac{L}{2}+\frac{k{L}^2}{3}}{1+\frac{kL}{2}}=\frac{3L}{5}$ Cross-multiply and cancel $L$: $5\left(\frac{1}{2}+\frac{kL}{3}\right)=3\left(1+\frac{kL}{2}\right)\Rightarrow \frac{5}{2}+\frac{5kL}{3}=3+\frac{3kL}{2}$ $\frac{5kL}{3}-\frac{3kL}{2}=\frac{1}{2}\Rightarrow kL\left(\frac{10-9}{6}\right)=\frac{1}{2}\Rightarrow \frac{kL}{6}=\frac{1}{2}\Rightarrow kL=3$ So $\frac{b}{a}=\frac{3}{L}$.

Verify: Plug $k=3/L$ back into step 2: 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 $=\frac{3a{L}^2/2}{5aL/2}=\frac{3L}{5}$ ✓. And $b/a>0$, matching the forecast. ✓

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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) ::: $2.8\text{m}$ from back E8 (H: reverse-engineer density) ::: $b/a=3/L$