Applications — mass, centre of mass, moments of inertia: Visual walkthrough

This page goes deeper than the parent note. It uses the same recipe but slows every line down to a drawing.

Step 1 — What is a "flat plate", and what does "density" mean here?

WHAT. Picture a thin sheet of metal lying flat on a table. We look straight down at it — it fills some region we call $R$ (just a shape on the page, like a triangle or a disc).

WHY. Before we can talk about balance, we need a way to say "there is this much stuff at this spot." A single number for the whole plate is not enough because the sheet can be thick and heavy in one corner and thin and light in another.

PICTURE. Look at the figure. The gray outline is the region $R$ . The shading shows density — dark means heavy, light means light.

Step 2 — Chop the plate into tiny tiles, each with its own tiny mass

WHAT. We cut $R$ into a grid of tiny rectangles. Pick one tile. It sits at position $(x, y)$ , it has width $d x$ and height $d y$ , so its area is a tiny number we call $d A = d x d y$ .

WHY. Over a whole plate the density changes from place to place, so we cannot use one density. But over a tile so small that density barely changes across it, we can treat density as a single number. That turns each tile into an easy point mass — the thing we already know how to handle.

PICTURE. The orange tile in the figure is our chosen piece. Its tiny mass is (density there) × (its area).

Step 3 — Add all the tiny masses: this is the total mass

WHAT. Add $d m$ over every tile in $R$ . Adding infinitely many infinitely small pieces is exactly what the symbol $\iint_{R}$ means — a double integral over the region.

WHY. Why two integral signs? Because our tiles are spread in two directions — across ( $x$ ) and up ( $y$ ). One sum sweeps a row, the second sum stacks the rows. This is the machinery from Double Integrals over General Regions.

PICTURE. The green arrows show the sweep: first fill one vertical strip (sum in $y$ ), then slide that strip across (sum in $x$ ).

Step 4 — What "balance" means: leverage, not just mass

WHAT. Set up a seesaw. Put the pivot on the vertical line $x = \overset{x}{ˉ}$ (a position we do not know yet — the $ˉ$ bar means "the balance value"). A tile at position $x$ sits a distance $(x - \overset{x}{ˉ})$ from the pivot. Its turning effect (torque) is its mass times that distance.

WHY. A small mass far from a pivot can balance a big mass close in — think of two kids on a seesaw. So position alone or mass alone cannot find the balance point; we need mass times distance-from-pivot. That product is the lever arm effect.

PICTURE. The seesaw: a light tile far left and a heavy tile near right balance about the red pivot line.

Step 5 — Solve the balance equation for $\overset{x}{ˉ}$

WHAT. Split the sum in Step 4 into two pieces, because $\overset{x}{ˉ}$ is a constant (same for every tile) and can be pulled outside the integral.

WHY. We want $\overset{x}{ˉ}$ alone on one side. Algebra on integrals is allowed: a constant factor slides out of a sum, just like $2 a + 2 b = 2 (a + b)$ .

$call this M_{y} \iint_{R} x ρ d A - \overset{x}{ˉ} = m \iint_{R} ρ d A = 0$

The first blob is a moment — mass weighted by its $x$ -position. We name it $M_{y}$ (moment about the $y$ -axis, because $x$ is the distance from the $y$ -axis).
The second blob is just the mass $m$ from Step 3, with $\overset{x}{ˉ}$ pulled out front.

PICTURE. The figure shows the two blobs literally: a stack of $x \cdot d m$ bars on the left, the total mass $m$ on the right.

Rearranging $M_{y} - \overset{x}{ˉ} m = 0$ gives:

By the identical argument balancing about a horizontal line $y = \overset{y}{ˉ}$ : $\overset{y}{ˉ} = \frac{M _{x}}{m} = \frac{\iint _{R} y ρ d A}{\iint _{R} ρ d A}, M_{x} = \iint_{R} y ρ d A .$

Step 6 — The full worked case: triangle with $ρ = x + y$

WHAT. Region $R$ : the triangle with corners $(0, 0), (1, 0), (0, 1)$ , density $ρ (x, y) = x + y$ . We run the whole recipe: find $m$ , then $M_{y}$ , then $\overset{x}{ˉ}$ .

WHY. A variable density is where the density truly matters — the balance point shifts away from the plain geometric centre. This shows the machinery earning its keep.

PICTURE. Density shading darkens toward the hypotenuse (where $x + y$ is biggest). The purple dot is the balance point we will compute.

For a fixed $x$ , the tile column runs from the bottom edge $y = 0$ up to the slanted line $y = 1 - x$ . So the inner sum is in $y$ .

Mass. $m = \int_{0}^{1} \int_{0}^{1 - x} (x + y) d y d x .$ Inner: $\int_{0}^{1 - x} (x + y) d y = x (1 - x) + \frac{( 1 - x ) ^{2}}{2} = \frac{1 - x ^{2}}{2} .$ Outer: $\int_{0}^{1} \frac{1 - x ^{2}}{2} d x = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{3} .$ So $m = \frac{1}{3}$ .

Moment $M_{y} = \iint x ρ d A$ . Inner: $\int_{0}^{1 - x} x (x + y) d y = x \cdot \frac{1 - x ^{2}}{2} .$ Outer: $\int_{0}^{1} \frac{x - x ^{3}}{2} d x = \frac{1}{2} (\frac{1}{2} - \frac{1}{4}) = \frac{1}{8} .$

Balance point. $\overset{x}{ˉ} = \frac{M _{y}}{m} = \frac{1/8}{1/3} = \frac{3}{8} .$ Since $ρ = x + y$ is symmetric in $x$ and $y$ , the same computation gives $\overset{y}{ˉ} = \frac{3}{8}$ . The balance point is $(\frac{3}{8}, \frac{3}{8})$ — pushed toward the heavy hypotenuse, not at the plain centroid $(\frac{1}{3}, \frac{1}{3})$ .

Step 7 — Edge and degenerate cases (never hit a surprise)

WHAT / WHY / PICTURE — three failure-proofing scenarios.

Uniform density. If $ρ$ is a constant $c$ , it factors out of top and bottom and cancels: $\overset{x}{ˉ} = \frac{c \iint x d A}{c \iint d A} = \frac{\iint x d A}{area} .$ Now the balance point is pure geometry — the centroid. See Centroids and the Pappus Theorems.
A shape with a hole / concave region. The balance point can land outside the material (e.g. a crescent). That is fine — $\overset{x}{ˉ}, \overset{y}{ˉ}$ are still the correct averages; nothing says the pivot must sit on the plate.
Zero density everywhere ( $ρ \equiv 0$ ). Then $m = 0$ and $\overset{x}{ˉ} = 0/0$ is undefined — correctly so: an empty plate has no balance point. The formula warns you by dividing by zero.
A single dominant heavy point (density spike). As one tiny region gets extremely heavy, $\overset{x}{ˉ}, \overset{y}{ˉ}$ slide toward that point — in the limit the balance point is that point. The mass-weighted average degrades gracefully to the point-mass answer.

Step 8 — Polar version: the same recipe, one extra factor

WHAT. For round regions we switch to polar coordinates: $x = r cos θ$ , $y = r sin θ$ . A polar tile is a curved wedge, not a rectangle.

WHY. The wedge's two sides are a radial length $d r$ and an arc length $r d θ$ (arc = radius × angle). So its area is $d A = radial d r \times arc r d θ = r d r d θ,$ not $d r d θ$ . That extra $r$ is the Jacobian factor from Polar Coordinates and the Jacobian.

PICTURE. The wedge with its short inner arc and long outer arc — clearly wider far from the centre.

The same $r^{2}$ idea powers moments of inertia and Rotational Kinetic Energy and Angular Momentum; the whole apparatus lifts to solids via Triple Integrals and Change of Variables and Jacobians.

The one-picture summary

The whole derivation on one canvas: chop $R$ into tiles → each tile is a point mass $d m = ρ d A$ → add them for $m$ → weight by position and add for $M_{y}, M_{x}$ → divide to get the balance point $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .

Recall Feynman retelling — the whole walkthrough in plain words

Imagine an unevenly cheesy pizza lying on the table. First I cut it into hundreds of tiny squares. Each little square is small enough that its cheese is basically even, so I can say "this square weighs this much" — density times its little area. Add up all the squares and I know the total weight of the pizza; that is the mass. Now I want the fingertip spot where it balances. I imagine a pivot line and ask each square to push down with a turning force equal to its weight times how far it sits from the pivot. When the left pushes exactly cancel the right pushes, that pivot line is the balance line. Doing the algebra, the balance $x$ turns out to be the weighted average of $x$ — every square's position counted according to how heavy it is. Do the same left-right and up-down and I get the balance point. If the cheese is perfectly even the weights don't matter and I get the plain geometric centre. If the pizza has a bite taken out, the balance point can even float over empty air — and if there's no pizza at all, there's simply no balance point, which is exactly what dividing by zero is telling me. For round pizzas I use pie-slice tiles, and because the outer arc is longer than the inner arc, each slice-tile secretly carries an extra factor of $r$ . Same recipe every time: chop, weight, add, divide.