Applications — mass, centre of mass, moments of inertia: Worked examples

Before we start, one symbol we lean on the whole page:

The scenario matrix

Every problem this topic can throw is one of these cells. Each example below is tagged with the cell it covers.

Cell	What makes it different	Example
C1 Constant density, simple shape	$ρ$ cancels → pure geometry (centroid)	Ex 1
C2 Variable density	$ρ (x, y)$ must stay inside the integral	Ex 2
C3 Region touching all four quadrants / sign of $x, y$	lever arms go negative in both $x$ and $y$ — do they cancel?	Ex 3
C4 Symmetry shortcut	a moment is zero by symmetry — spot it, skip work	Ex 4
C5 Polar coordinates (curved region)	the mandatory Jacobian $r$	Ex 5
C6 Degenerate / limiting shape	thin rod, or $a \to 0$ — check formula survives	Ex 6
C7 Real-world word problem	translate words → region + density	Ex 7
C8 Perpendicular-axis / radius of gyration	combine $I_{x}, I_{y}$ ; find $R_{g}$	Ex 8
C9 Triple integral (3-D solid)	$d m = ρ d V$ , one more dimension	Ex 9
C10 Exam twist — density blows up / non-uniform axis	a trap that punishes autopilot	Ex 10

We now fill every cell.

Example 1 — C1: constant density, the centroid of a triangle

Worked example Statement

A uniform plate ( $ρ =$ constant) is the triangle with corners $(0, 0)$ , $(2, 0)$ , $(0, 3)$ . Find its centre of mass.

Forecast: where do you think the balance point sits? Guess before reading. (Hint: the medians of a triangle meet at a point one-third of the way from each side.)

Step 1 — Describe the region. The slanted edge runs from $(2, 0)$ to $(0, 3)$ . Its line is $\frac{x}{2} + \frac{y}{3} = 1$ , i.e. $y = 3 - \frac{3}{2} x$ . So for each $x$ between $0$ and $2$ , $y$ runs from $0$ up to $3 - \frac{3}{2} x$ . Why this step? Before integrating we must know the exact limits — the picture (look at the vertical dashed strip) tells us $y$ starts at the base and stops at the slanted red line.

Step 2 — Mass. Since $ρ$ is constant, $m = ρ \cdot (area) = ρ \cdot \frac{1}{2} (2) (3) = 3 ρ$ . Why this step? For a uniform plate mass is just density × area — no integral needed, though $\iint ρ d A$ gives the same.

Step 3 — Moment $M_{y} = \iint x ρ d A$ (defined in the box above — lever arm $x$ ). $M_{y} = ρ \int_{0}^{2} \int_{0}^{3 - \frac{3}{2} x} x d y d x = ρ \int_{0}^{2} x (3 - \frac{3}{2} x) d x .$ Inner integral: $\int_{0}^{3 - \frac{3}{2} x} d y = 3 - \frac{3}{2} x$ (the height of the strip). Outer: $ρ \int_{0}^{2} (3 x - \frac{3}{2} x^{2}) d x = ρ [\frac{3}{2} x^{2} - \frac{1}{2} x^{3}]_{0}^{2} = ρ (6 - 4) = 2 ρ$ . Why this step? $M_{y}$ weights each strip by its $x$ -position — the lever arm to the $y$ -axis.

Step 4 — $\overset{x}{ˉ} = M_{y} / m = 2 ρ /3 ρ = \frac{2}{3}$ . By the same computation with the other moment $M_{x} = \iint y ρ d A$ we get $M_{x} = 3 ρ$ and $\overset{y}{ˉ} = M_{x} / m = 1$ . Why this step? $M_{x}$ (lever arm $y$ , defined in the box above) gives the vertical balance; dividing each moment by the total mass converts "total leverage" into "the single position that reproduces it."

Verify: the centroid of a triangle is the average of its vertices: $\overset{x}{ˉ} = \frac{0 + 2 + 0}{3} = \frac{2}{3}$ , $\overset{y}{ˉ} = \frac{0 + 0 + 3}{3} = 1$ . ✓ Exactly matches. Notice $ρ$ cancelled — that is what makes this a centroid (see Centroids and the Pappus Theorems).

Example 2 — C2: variable density strip

Worked example Statement

A plate occupies $0 \leq x \leq 1$ , $0 \leq y \leq 1$ (unit square) with density $ρ (x, y) = x$ (denser on the right, zero on the left edge). Find $m$ and $\overset{x}{ˉ}$ .

Forecast: more mass sits on the right, so should $\overset{x}{ˉ}$ be greater or less than $0.5$ ?

Step 1 — Mass. $m = \int_{0}^{1} \int_{0}^{1} x d y d x = \int_{0}^{1} x d x = \frac{1}{2} .$ Why this step? The inner $y$ -integral of $x$ (constant in $y$ ) just multiplies by the height $1$ .

Step 2 — $M_{y} = \iint x \cdot ρ d A = \iint x \cdot x d A$ . $M_{y} = \int_{0}^{1} \int_{0}^{1} x^{2} d y d x = \int_{0}^{1} x^{2} d x = \frac{1}{3} .$ Why this step? Here the "distance" $x$ and the density $x$ both appear, giving $x^{2}$ — do not forget the density factor (this is the classic mistake in the parent note).

Step 3 — $\overset{x}{ˉ} = M_{y} / m = \frac{1/3}{1/2} = \frac{2}{3}$ . Why this step? Dividing the moment by total mass gives the single $x$ where all the mass could sit and balance the same.

Verify: $\overset{x}{ˉ} = \frac{2}{3} > 0.5$ ✓ — the balance point leans right toward the heavy side, exactly as forecast. If we had (wrongly) ignored density, we'd get the geometric centre $\overset{x}{ˉ} = 0.5$ ; the density pulled it out to $\frac{2}{3}$ .

Example 3 — C3: a region spanning ALL FOUR quadrants (signs of both lever arms)

Worked example Statement

Plate: the rectangle $- 1 \leq x \leq 3$ , $- 2 \leq y \leq 2$ , uniform density $ρ$ . It straddles both axes, so tiles live in all four quadrants. Find $\overset{x}{ˉ}$ and $\overset{y}{ˉ}$ . Do the negative $x$ and negative $y$ values matter?

Forecast: the plate reaches further right ( $x$ to $3$ ) than left ( $x$ to $- 1$ ), so $\overset{x}{ˉ} > 0$ . But it is symmetric top–bottom ( $y$ from $- 2$ to $2$ ), so what should $\overset{y}{ˉ}$ be?

Step 1 — Mass. Area $= (3 - (- 1)) \times (2 - (- 2)) = 4 \times 4 = 16$ , so $m = 16 ρ$ . Why this step? We will divide both moments by this total mass to reach the balance point; the mass is the "weight" that converts leverage into position. For a uniform plate it is density × area.

Step 2 — $M_{y} = ρ \iint x d A$ (horizontal balance). The lever arm $x$ is negative in the two left quadrants (violet in the figure) and positive in the two right quadrants (orange). They partly cancel. $M_{y} = ρ \int_{- 2}^{2} \int_{- 1}^{3} x d x d y = ρ \int_{- 2}^{2} [\frac{x ^{2}}{2}]_{- 1}^{3} d y = ρ \int_{- 2}^{2} \frac{9 - 1}{2} d y = ρ \cdot 4 \cdot 4 = 16 ρ .$ Why this step? $\int_{- 1}^{3} x d x = \frac{9}{2} - \frac{1}{2} = 4$ — the negative contribution from $x \in [- 1, 0]$ is real and subtracts, but the bigger positive side wins.

Step 3 — $M_{x} = ρ \iint y d A$ (vertical balance). Now the lever arm $y$ is negative below the $x$ -axis and positive above, over the symmetric range $[- 2, 2]$ : $M_{x} = ρ \int_{- 1}^{3} \int_{- 2}^{2} y d y d x = ρ \int_{- 1}^{3} [\frac{y ^{2}}{2}]_{- 2}^{2} d x = ρ \int_{- 1}^{3} \frac{4 - 4}{2} d x = 0.$ Why this step? Because $y$ ranges symmetrically about $0$ , every tile at height $+ y$ is cancelled by its mirror at $- y$ — a zero moment, so no integral toil is truly needed once you spot it.

Step 4 — Centre of mass. $\overset{x}{ˉ} = \frac{M _{y}}{m} = \frac{16 ρ}{16 ρ} = 1$ , and $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{0}{16 ρ} = 0$ . Why this step? Dividing each total moment (leverage) by the total mass yields the one point that reproduces the same leverage about every axis — the balance point.

Verify: the geometric centre of the rectangle is $(\frac{- 1 + 3}{2}, \frac{- 2 + 2}{2}) = (1, 0)$ ✓. So $\overset{x}{ˉ} = 1 > 0$ (leans right, as forecast) and $\overset{y}{ˉ} = 0$ (top–bottom symmetry, as forecast). Lesson: signed lever arms in both directions cancel automatically — you never split the integral by quadrant; the algebra handles every sign.

Example 4 — C4: symmetry makes a moment vanish

Worked example Statement

A uniform semicircular plate of radius $a$ sits with its flat edge on the $x$ -axis: $x^{2} + y^{2} \leq a^{2}$ , $y \geq 0$ . Find $\overset{x}{ˉ}$ without integrating, then find $\overset{y}{ˉ}$ .

Forecast: the shape is a mirror image across the $y$ -axis. What does that force $\overset{x}{ˉ}$ to be?

Step 1 — Spot the symmetry. For every tile at $(x, y)$ there is a mirror tile at $(- x, y)$ with the same density (uniform). Their $x$ -moments $x ρ d A$ and $(- x) ρ d A$ cancel (see the mirrored violet/orange tiles in the figure). Why this step? Recognising symmetry saves an entire integral — $M_{y} = 0$ , hence $\overset{x}{ˉ} = 0$ .

Step 2 — $\overset{y}{ˉ}$ needs work; use polar. $x^{2} + y^{2} \leq a^{2}$ , $y \geq 0$ means $0 \leq r \leq a$ , $0 \leq θ \leq π$ , with $y = r sin θ$ and $d A = r d r d θ$ . $m = ρ \cdot \frac{1}{2} π a^{2}, M_{x} = ρ \int_{0}^{π} \int_{0}^{a} (r sin θ) r d r d θ .$ Why polar? A circular boundary is trivial in polar ( $r = a$ ) but ugly in rectangular. See Polar Coordinates and the Jacobian.

Step 3 — Evaluate $M_{x}$ . $M_{x} = ρ \int_{0}^{π} sin θ d θ \int_{0}^{a} r^{2} d r = ρ \cdot [- cos θ]_{0}^{π} \cdot \frac{a ^{3}}{3} = ρ \cdot 2 \cdot \frac{a ^{3}}{3} = \frac{2 ρ a ^{3}}{3} .$

Step 4 — $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{2 ρ a ^{3} /3}{ρ π a ^{2} /2} = \frac{4 a}{3 π} .$

Verify: the standard textbook value for a semicircle's centroid is $\overset{y}{ˉ} = \frac{4 a}{3 π} \approx 0.4244 a$ , which sits below halfway ( $0.5 a$ ) — correct, since more area bunches near the flat base — and above $0$ and below the top $a$ . And $\overset{x}{ˉ} = 0$ ✓ by symmetry, exactly as forecast.

Example 5 — C5: why polar tiles wear an $r$ (moment of inertia of a disc)

Worked example Statement

A uniform disc of radius $a$ , density $ρ$ , centred at the origin. Find $I_{0}$ , the moment of inertia about the axis through the centre (the $z$ -axis).

Forecast: from the parent's " $1, r, r^{2}$ " idea, $I_{0} = \iint r^{2} d m$ , and areas scale as (length) $^{2}$ . So $I_{0}$ should grow like $a^{4}$ . Guess the exact constant.

Step 1 — Justify $d A = r d r d θ$ . Look at the shaded polar tile in the figure: it spans a small radial width $d r$ and a small angle $d θ$ . Its two "long" sides are arcs of length $r d θ$ (arc length = radius × angle). So the tile is nearly a rectangle of sides $d r$ and $r d θ$ , area $= r d r d θ$ . Why this step? This is the Jacobian factor. Forgetting the $r$ is the single most common blunder — it makes tiles near the rim count as small as tiles near the centre, which is geometrically false.

Step 2 — Set up $I_{0}$ . Distance from the axis is $r$ , so lever arm $^{2} = r^{2}$ : $I_{0} = ρ \int_{0}^{2 π} \int_{0}^{a} r^{2} \cdot d A r d r d θ = ρ \int_{0}^{2 π} d θ \int_{0}^{a} r^{3} d r .$ Why $r^{2} \cdot r = r^{3}$ ? One $r^{2}$ is the physics (distance²); the extra $r$ is the geometry (Jacobian). They are different and both required.

Step 3 — Integrate. $I_{0} = ρ \cdot 2 π \cdot \frac{a ^{4}}{4} = \frac{π ρ a ^{4}}{2} .$

Step 4 — Per-mass form. Mass $m = ρ π a^{2}$ , so $I_{0} = \frac{π ρ a ^{4}}{2} = \frac{( ρ π a ^{2} ) a ^{2}}{2} = \frac{m a ^{2}}{2}$ .

Verify: $I_{0} \propto a^{4}$ ✓ as forecast, constant $\frac{π ρ}{2}$ ; the per-mass form $\frac{1}{2} m a^{2}$ is the famous "solid disc" result from Rotational Kinetic Energy and Angular Momentum.

Example 6 — C6: degenerate limit (thin rod as a shrinking rectangle)

Worked example Statement

Take the rectangle $0 \leq x \leq L$ , $0 \leq y \leq b$ , uniform, with $I_{x} = \frac{m b ^{2}}{3}$ and $I_{y} = \frac{m L ^{2}}{3}$ (parent Example B). What happens to a thin rod — let the height $b \to 0$ ? Find $I_{y}$ of the resulting rod about one end.

Forecast: as the plate becomes a needle along the $x$ -axis, what happens to $I_{x}$ (spin about the long axis) and $I_{y}$ (spin about the short end)?

Step 1 — Take $b \to 0$ in $I_{x}$ . $I_{x} = \frac{m b ^{2}}{3} \to 0$ . Why? When the rod is infinitely thin, no mass is any distance from the $x$ -axis, so it offers zero resistance to spinning about its own length. Degenerate but sensible.

Step 2 — $I_{y}$ survives. $I_{y} = \frac{m L ^{2}}{3}$ does not involve $b$ , so it stays finite as $b \to 0$ . Why this step? $I_{y}$ measures spin about the short end; the mass is spread along $x \in [0, L]$ , still a real distance from the $y$ -axis.

Verify: a rod of length $L$ , mass $m$ , rotated about one end has the classic moment of inertia $\frac{1}{3} m L^{2}$ ✓ (textbook value). Our 2-D rectangle formula degenerated exactly into the 1-D rod formula — a good consistency check that the machinery is right in the limit.

Example 7 — C7: a real-world word problem

Worked example Statement

A rectangular metal sheet, $0 \leq x \leq 4$ m, $0 \leq y \leq 2$ m, was heated so that its areal density (kg/m²) increases linearly with height: $ρ (x, y) = 2 + y$ (kg/m²). Find its total mass and the height $\overset{y}{ˉ}$ of its balance point.

Forecast: density is bigger at the top ( $y = 2$ gives $ρ = 4$ ) than bottom ( $y = 0$ gives $ρ = 2$ ), so $\overset{y}{ˉ}$ should be pushed above the geometric mid-height $1$ m.

Step 1 — Translate words to a region + density. Region: rectangle $[0, 4] \times [0, 2]$ . Density: $ρ = 2 + y$ , shown as a colour gradient (light at the bottom, dark at the top) in the figure. No $x$ -dependence, so all $x$ -integrals just multiply by the width $4$ . Why this step? Word problems are just region + density in disguise; naming them turns prose into an integral.

Step 2 — Mass. $m = \int_{0}^{4} \int_{0}^{2} (2 + y) d y d x = 4 \int_{0}^{2} (2 + y) d y = 4 [2 y + \frac{y ^{2}}{2}]_{0}^{2} = 4 (4 + 2) = 24 kg .$

Step 3 — $M_{x} = \iint y ρ d A$ . $M_{x} = 4 \int_{0}^{2} y (2 + y) d y = 4 \int_{0}^{2} (2 y + y^{2}) d y = 4 [y^{2} + \frac{y ^{3}}{3}]_{0}^{2} = 4 (4 + \frac{8}{3}) = \frac{80}{3} kg\cdotm .$ Why this step? $\overset{y}{ˉ}$ balances about the $x$ -axis, whose lever arm is $y$ ; so we use $M_{x}$ , not $M_{y}$ .

Step 4 — $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{80/3}{24} = \frac{80}{72} = \frac{10}{9} \approx 1.11$ m.

Verify: units: $M_{x}$ is kg·m, $m$ is kg, so $\overset{y}{ˉ}$ is metres ✓. Value $1.11$ m $> 1$ m ✓ — leans toward the denser top, exactly as forecast.

Example 8 — C8: perpendicular-axis theorem & radius of gyration

Worked example Statement

For the uniform disc of Example 5 ( $I_{0} = \frac{1}{2} m a^{2}$ ), find $I_{x}$ and $I_{y}$ separately, then the radius of gyration about the centre.

Forecast: the disc looks identical whether you spin it about $x$ or $y$ (it has circular symmetry), so what must be true of $I_{x}$ vs $I_{y}$ ?

Step 1 — Symmetry gives $I_{x} = I_{y}$ . Rotating the disc $9 0^{\circ}$ maps the $x$ -axis onto the $y$ -axis and leaves the disc unchanged (see the two axes drawn through the same disc in the figure), so the two moments of inertia are equal. Why this step? Symmetry again lets us avoid a second integral.

Step 2 — Perpendicular-axis theorem. For a flat lamina, $I_{0} = I_{x} + I_{y}$ . With $I_{x} = I_{y}$ : $I_{0} = 2 I_{x} \Rightarrow I_{x} = I_{y} = \frac{1}{2} I_{0} = \frac{1}{2} \cdot \frac{1}{2} m a^{2} = \frac{1}{4} m a^{2} .$ Why this theorem? It links the spin about the perpendicular ( $z$ ) axis to the two in-plane axes — but only for a flat plate (the parent's mistake box warns it fails for 3-D solids).

Step 3 — Radius of gyration about centre. $R_{g} = I_{0} / m = \frac{1}{2} m a^{2} / m = \frac{a}{2}$ . Why this quantity? $R_{g}$ is the single radius at which you could put all the mass (as a thin ring) to get the same $I_{0}$ .

Verify: $I_{x} + I_{y} = \frac{1}{4} m a^{2} + \frac{1}{4} m a^{2} = \frac{1}{2} m a^{2} = I_{0}$ ✓ — the perpendicular-axis check closes. And $R_{g} = a / 2 \approx 0.707 a$ lies between $0$ and $a$ ✓ — a plausible "average radius."

Example 9 — C9: a triple integral (3-D solid)

Worked example Statement

A solid cube $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , $0 \leq z \leq 1$ has volume density $ρ = 1$ (mass per unit volume). Find its mass and $\overset{z}{ˉ}$ . (Now $d m = ρ d V$ , one extra dimension.)

Forecast: uniform cube → balance point at the dead centre $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ . So $\overset{z}{ˉ} = ?$

Step 1 — Mass by triple integral. $m = ∭_{V} ρ d V = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 1 d z d y d x = 1.$ Why a triple integral? A solid needs a volume element $d V = d x d y d z$ ; the recipe is identical to 2-D but with three nested integrals. See Triple Integrals.

Step 2 — Moment about the $x y$ -plane, $M_{x y} = ∭ z ρ d V$ . $M_{x y} = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} z d z d y d x = \int_{0}^{1} \int_{0}^{1} [\frac{z ^{2}}{2}]_{0}^{1} d y d x = \frac{1}{2} .$ Why this step? In 3-D the lever arm to the horizontal $x y$ -plane is the height $z$ ; $\overset{z}{ˉ} = M_{x y} / m$ .

Step 3 — $\overset{z}{ˉ} = M_{x y} / m = \frac{1/2}{1} = \frac{1}{2}$ .

Verify: $\overset{z}{ˉ} = \frac{1}{2}$ ✓, the geometric centre of a uniform cube, exactly as forecast. By identical symmetry $\overset{x}{ˉ} = \overset{y}{ˉ} = \frac{1}{2}$ too.

Example 10 — C10: exam twist (density that blows up at the origin)

Worked example Statement

A plate is the quarter disc $r \leq 1$ , first quadrant, with density $ρ = \frac{1}{r}$ (heavier near the centre, singular at $r = 0$ ). Does the mass stay finite? Find it.

Forecast: $ρ \to \infty$ as $r \to 0$ . Autopilot screams "infinite mass!" But the Jacobian $r$ might save us. Guess: finite or infinite?

Step 1 — Set up in polar, keep the Jacobian. $m = \int_{0}^{π /2} \int_{0}^{1} \frac{1}{r} \cdot d A r d r d θ = \int_{0}^{π /2} \int_{0}^{1} 1 d r d θ .$ Why this is the trap: the $\frac{1}{r}$ from density cancels the $r$ from the Jacobian! What looked singular becomes the integral of $1$ — perfectly finite. Whoever forgets the Jacobian $r$ gets $\int_{0}^{1} \frac{1}{r} d r = \infty$ and the wrong answer.

Step 2 — Integrate. $m = \int_{0}^{π /2} [r]_{0}^{1} d θ = \int_{0}^{π /2} 1 d θ = \frac{π}{2} .$

Verify: $m = \frac{π}{2} \approx 1.571$ , finite ✓. The singular density was tamed by the polar area element — the exam's whole point. This is why "polar tiles wear an $r$ " is not a cosmetic rule but the thing that decides finiteness.

Recall Self-test: match each example to its matrix cell

Ex 1 :::: C1 constant density / centroid Ex 2 :::: C2 variable density Ex 3 :::: C3 signed lever arms across all four quadrants Ex 4 :::: C4 symmetry ⇒ moment $= 0$ Ex 5 :::: C5 polar Jacobian Ex 6 :::: C6 degenerate/limiting shape Ex 7 :::: C7 word problem Ex 8 :::: C8 perpendicular-axis + radius of gyration Ex 9 :::: C9 triple integral (3-D) Ex 10 :::: C10 exam twist (singular density)