5.6 Calculating Centers of Mass and Moments of Inertia

Big idea

Everything so far has been machinery for evaluating integrals. This section spends that machinery on a single physical question: given an object whose material is distributed unevenly, where does it balance, and how hard is it to spin?

The first question is answered by a weighted average. If a flat plate occupies a region and carries a density in mass per unit area, then the mass of a small piece is the density there times the area of the piece, and the total mass is the integral of the density over the region. The balance point is the average position, but weighted by mass rather than by area: each little piece votes on the location of the balance point in proportion to how heavy it is. Writing that average as an integral gives the center of mass, and dividing the weighted sum by the total is why every coordinate of the answer is a quotient.

The second question needs a different weighting. Resistance to rotation about an axis grows with the square of the distance from that axis, not with the distance itself, and the integral of density times squared distance is the moment of inertia. That squaring is the whole difference between the two families: mass at distance $2$ counts twice as much as mass at distance $1$ for balancing, but four times as much for spinning.

All of it transfers to three dimensions with the substitutions you would guess. Area density becomes volume density, double integrals become triple, and moments about axes become moments about coordinate planes. Nothing new is proved here; the work is setting up the right integral and then evaluating it with the techniques of the previous sections, which is why polar, cylindrical and spherical coordinates appear constantly.

Decoder

Each coordinate of the center of mass is a moment divided by the total mass, where a moment is the integral of the density times the signed distance to a reference axis or plane, and each moment of inertia is the integral of the density times the squared distance to an axis.

The word “divided” is what makes the center of mass a location rather than a total. Moments and masses both scale when you double the density everywhere, and the quotient does not, which is why a uniform plate has a balance point that does not depend on what it is made of. That special case has its own name: the centroid, the center of mass when the density is constant.

The naming of the two-dimensional moments trips people up. The moment $M_x$ is the moment about the $x$-axis, and the distance to the $x$-axis is $|y|$, so $M_x$ is an integral of $y$ times density. It then supplies $\bar{y}$, not $\bar{x}$. Reading the subscript as “which axis” rather than “which coordinate” keeps it straight.

Symmetry is the shortcut worth having. If the region is symmetric about a line and the density takes the same value at points mirrored across that line, then the center of mass lies on that line, because contributions from the two sides cancel in the perpendicular direction. That kills one integral outright. It says nothing about moments of inertia, which are built from squares and never cancel; a moment of inertia is positive whenever there is any mass at all off the axis.

Definitions and results

Density. Write $\delta(x,y)$ for mass per unit area on a plane region and $\delta(x,y,z)$ for mass per unit volume on a solid. The letter $\rho$ is reserved here for the spherical radius.

Mass. For a lamina occupying $D$, and for a solid $E$,

$$ m = \iint_D \delta(x,y)\, dA, \qquad m = \iiint_E \delta(x,y,z)\, dV $$

Moments in the plane. The moments about the coordinate axes are

$$ M_x = \iint_D y\, \delta(x,y)\, dA, \qquad M_y = \iint_D x\, \delta(x,y)\, dA $$

Center of mass in the plane. $\bar{x} = \dfrac{M_y}{m}$ and $\bar{y} = \dfrac{M_x}{m}$. When $\delta$ is constant it cancels from both quotients, and the resulting point is the centroid of the region.

Moments of inertia in the plane. About the two axes and about the origin,

$$ I_x = \iint_D y^2 \delta\, dA, \qquad I_y = \iint_D x^2 \delta\, dA, \qquad I_0 = \iint_D \big( x^2+y^2 \big)\delta\, dA $$

so $I_0 = I_x + I_y$. The quantity $I_0$ is the polar moment of inertia, the one governing rotation about the axis through the origin perpendicular to the plane.

Moments in space. Moments are taken about coordinate planes, and the distance to a plane is the coordinate perpendicular to it:

$$ M_{yz} = \iiint_E x\, \delta\, dV, \qquad M_{xz} = \iiint_E y\, \delta\, dV, \qquad M_{xy} = \iiint_E z\, \delta\, dV $$

with $\bar{x} = M_{yz}/m$, $\bar{y} = M_{xz}/m$ and $\bar{z} = M_{xy}/m$.

Moments of inertia in space. The distance from a point to the $z$-axis is $\sqrt{x^2+y^2}$, so

$$ I_z = \iiint_E \big( x^2+y^2 \big)\delta\, dV $$

and cyclically $I_x = \iiint_E (y^2+z^2)\delta\, dV$, $I_y = \iiint_E (x^2+z^2)\delta\, dV$.

Radius of gyration. The number $k$ with $I = mk^2$ is the distance from the axis at which a single point of the same total mass would have the same moment of inertia. It must land inside the object’s range of distances from the axis, which makes it a good check.

Worked examples

A triangular plate that is heavier on one side

A plate occupies the triangle $x \ge 0$, $y \ge 0$, $x + y \le 2$ with density $\delta(x,y) = x$. Find its center of mass.

The mass comes first:

$$ m = \int_0^2 \int_0^{2-x} x\, dy\, dx = \int_0^2 x(2-x)\, dx = 4 - \frac{8}{3} = \frac{4}{3} $$

Then the two moments:

$$ M_y = \int_0^2 \int_0^{2-x} x^2\, dy\, dx = \int_0^2 x^2(2-x)\, dx = \frac{16}{3} - 4 = \frac{4}{3} $$

$$ M_x = \int_0^2 \int_0^{2-x} xy\, dy\, dx = \int_0^2 x\,\frac{(2-x)^2}{2}\, dx = \frac{1}{2}\left( 8 - \frac{32}{3} + 4 \right) = \frac{2}{3} $$

So $\bar{x} = \tfrac{4/3}{4/3} = 1$ and $\bar{y} = \tfrac{2/3}{4/3} = \tfrac{1}{2}$, giving the point $\left(1, \tfrac{1}{2}\right)$.

Check against the uniform case. With constant density the centroid is the average of the vertices, $\left(\tfrac{2}{3}, \tfrac{2}{3}\right)$. Weighting by $x$ should pull the balance point toward larger $x$, and $1 > \tfrac{2}{3}$. It should also pull $\bar{y}$ down, since the heavy end near $x = 2$ is where the triangle is thin, and $\tfrac{1}{2} < \tfrac{2}{3}$.

A half disk, using symmetry

Find the centroid of the upper half of the disk of radius $a$, with constant density. The region is symmetric about the $y$-axis and the density is constant, so $\bar{x} = 0$ without computing anything.

For $\bar{y}$, take $\delta = 1$, which is legitimate because the constant cancels. The mass is the area $\tfrac{\pi a^2}{2}$, and polar coordinates handle the moment:

$$ M_x = \int_0^{\pi} \int_0^a \big( r\sin\theta \big) r\, dr\, d\theta = \frac{a^3}{3} \Big[ -\cos\theta \Big]_0^{\pi} = \frac{2a^3}{3} $$

$$ \bar{y} = \frac{2a^3/3}{\pi a^2/2} = \frac{4a}{3\pi} $$

That is about $0.42a$, comfortably inside the half disk and below the midheight $0.5a$, which is right because the region is widest near the bottom.

Spinning a uniform disk

Find the moments of inertia of a disk of radius $a$ and constant density $\delta$ about its center. In polar coordinates the squared distance to the center is $r^2$:

$$ I_0 = \int_0^{2\pi} \int_0^a r^2 \cdot \delta\, r\, dr\, d\theta = 2\pi\delta \cdot \frac{a^4}{4} = \frac{\pi \delta a^4}{2} $$

The mass of the disk is $m = \delta \pi a^2$, so $I_0 = \tfrac{1}{2} m a^2$, the standard result for a disk spun about its axis. The radius of gyration is $a/\sqrt{2} \approx 0.71a$, inside the disk as it must be.

By symmetry $I_x = I_y$, and since they add to $I_0$ each equals $\tfrac{1}{4}m a^2$ without a further integral.

A solid bowl in three dimensions

Let $E$ be the solid between the paraboloid $z = x^2+y^2$ and the plane $z = 4$, with constant density $\delta = 1$. Its volume, computed earlier by projecting onto the disk of radius $2$, is $8\pi$, and that is also its mass.

Symmetry about the $z$-axis gives $\bar{x} = \bar{y} = 0$. For $\bar{z}$, work in cylindrical coordinates:

$$ M_{xy} = \int_0^{2\pi} \! \int_0^2 \! \int_{r^2}^{4} z\, r\, dz\, dr\, d\theta = \pi \int_0^2 \big( 16r - r^5 \big) dr = \pi \left( 32 - \frac{32}{3} \right) = \frac{64\pi}{3} $$

$$ \bar{z} = \frac{64\pi/3}{8\pi} = \frac{8}{3} $$

About $2.67$, which is between $0$ and $4$ and above the halfway mark, as it should be: the solid is widest near the top.

For the moment of inertia about the axis of symmetry the squared distance is $r^2$:

$$ I_z = \int_0^{2\pi} \! \int_0^2 \! \int_{r^2}^{4} r^2 \cdot r\, dz\, dr\, d\theta = 2\pi \int_0^2 r^3 \big( 4 - r^2 \big) dr = 2\pi\left( 16 - \frac{32}{3} \right) = \frac{32\pi}{3} $$

Check with the radius of gyration: $k^2 = I_z/m = \tfrac{32\pi/3}{8\pi} = \tfrac{4}{3}$, so $k \approx 1.15$. Every point of the solid is within $2$ of the axis, so $k$ had to come out below $2$, and it did.

Practice

Begin with laminas. Compute the mass first, then the moments, then divide, and use symmetry to skip whichever coordinate it hands you. Watch the subscripts on $M_x$ and $M_y$.

Then solids. The pattern is identical with one more integral, but the geometry is usually easier in cylindrical or spherical coordinates, so convert before setting up.

Quiz

Five items on mass, moments, centers of mass and centroids, and moments of inertia for both laminas and solids.