KAHIBARO
Table of Contents
From Particles to Continuous Matter
For a small number of particles, the center of mass is found by averaging positions with mass as the weight. For a continuous object, such as a rod, plate, sphere, or block, the same idea remains true, but the mass is spread smoothly through space instead of being concentrated at separate points.
To handle this, we imagine cutting the object into many very small pieces. Each tiny piece has a small mass $dm$ and a position. The center of mass is then the mass weighted average of all these tiny positions.
If the total mass is $M$, then the position vector of the center of mass is
$$
\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm
$$
This is the continuous version of the center of mass formula for particles.
For a continuous object, the center of mass is found from
$$
\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm
$$
where $dm$ is a tiny mass element and $M = \int dm$ is the total mass.
Coordinate Form
In Cartesian coordinates, if the object occupies space with coordinates $(x,y,z)$, then the center of mass coordinates are
$$
x_{\text{cm}} = \frac{1}{M}\int x\,dm
$$
$$
y_{\text{cm}} = \frac{1}{M}\int y\,dm
$$
$$
z_{\text{cm}} = \frac{1}{M}\int z\,dm
$$
These equations are often the most practical way to calculate the center of mass.
Mass Density
To use the integral formulas, we need to express $dm$ in terms of geometry. This is done with mass density. The form of density depends on the kind of object.
For a thin wire or rod, mass is distributed along a line. We use linear density,
$$
\lambda = \frac{dm}{dl}
$$
so that
$$
dm = \lambda\,dl
$$
For a thin sheet or plate, mass is spread over an area. We use surface density,
$$
\sigma = \frac{dm}{dA}
$$
so that
$$
dm = \sigma\,dA
$$
For a three dimensional object, mass fills a volume. We use volume density,
$$
\rho = \frac{dm}{dV}
$$
so that
$$
dm = \rho\,dV
$$
| Type of object | Density symbol | Definition | Mass element |
|---|---|---|---|
| Thin rod or wire | $\lambda$ | $\lambda = dm/dl$ | $dm = \lambda\,dl$ |
| Thin plate | $\sigma$ | $\sigma = dm/dA$ | $dm = \sigma\,dA$ |
| Solid object | $\rho$ | $\rho = dm/dV$ | $dm = \rho\,dV$ |
If the density is constant, the object is uniform. Then calculations become much simpler, because the density can be taken outside the integral.
Choose the correct mass element before integrating:
$$
dm = \lambda\,dl, \qquad dm = \sigma\,dA, \qquad dm = \rho\,dV
$$
A correct center of mass calculation begins with a correct choice of $dm$.
Symmetry
Before doing any integration, it is smart to look for symmetry. Symmetry often tells us the location of the center of mass immediately.
If a uniform object is symmetric about a plane, then its center of mass lies on that plane. If it is symmetric about an axis, then its center of mass lies on that axis. If it is symmetric in all directions around a point, then the center of mass is at that point.
For example, the center of mass of a uniform sphere is at its geometric center. The same is true for a uniform circular disk and a uniform rectangular block.
Symmetry can determine some or all coordinates of the center of mass without calculation.
Uniform Rod
Consider a thin uniform rod of length $L$ placed along the $x$ axis from $x=0$ to $x=L$. Because the rod is uniform, the linear density $\lambda$ is constant:
$$
\lambda = \frac{M}{L}
$$
The center of mass coordinate is
$$
x_{\text{cm}} = \frac{1}{M}\int x\,dm
$$
Using $dm = \lambda\,dx$,
$$
x_{\text{cm}} = \frac{1}{M}\int_0^L x\lambda\,dx
$$
Since $\lambda = M/L$,
$$
x_{\text{cm}} = \frac{1}{M}\cdot \frac{M}{L}\int_0^L x\,dx
= \frac{1}{L}\left[\frac{x^2}{2}\right]_0^L
= \frac{L}{2}
$$
So the center of mass is at the midpoint.
Uniform Rectangular Plate
Now consider a thin uniform rectangular plate of width $a$ and height $b$, with one corner at the origin and sides along the axes. Because the plate is uniform, the surface density is constant:
$$
\sigma = \frac{M}{ab}
$$
The center of mass coordinates are
$$
x_{\text{cm}} = \frac{1}{M}\int x\,dm, \qquad
y_{\text{cm}} = \frac{1}{M}\int y\,dm
$$
Using $dm = \sigma\,dA = \sigma\,dx\,dy$,
$$
x_{\text{cm}} = \frac{1}{M}\int_0^a \int_0^b x\sigma\,dy\,dx
$$
Since $x$ does not depend on $y$,
$$
x_{\text{cm}} = \frac{\sigma}{M}\int_0^a \left(\int_0^b dy\right)x\,dx
= \frac{\sigma b}{M}\int_0^a x\,dx
= \frac{\sigma b}{M}\cdot \frac{a^2}{2}
$$
Since $M = \sigma ab$,
$$
x_{\text{cm}} = \frac{a}{2}
$$
Similarly,
$$
y_{\text{cm}} = \frac{b}{2}
$$
So the center of mass is at the geometric center of the rectangle.
Nonuniform Objects
If the density changes from place to place, the center of mass shifts toward the heavier region. In that case, density cannot be treated as constant.
For example, suppose a rod lies along the $x$ axis from $x=0$ to $x=L$, but its linear density varies with position:
$$
\lambda(x) = kx
$$
where $k$ is a constant. Then
$$
dm = \lambda(x)\,dx = kx\,dx
$$
First find the total mass:
$$
M = \int_0^L dm = \int_0^L kx\,dx = \frac{kL^2}{2}
$$
Now compute the center of mass:
$$
x_{\text{cm}} = \frac{1}{M}\int_0^L x\,dm
= \frac{1}{M}\int_0^L x(kx)\,dx
= \frac{k}{M}\int_0^L x^2\,dx
= \frac{k}{M}\cdot \frac{L^3}{3}
$$
Substituting $M = kL^2/2$ gives
$$
x_{\text{cm}} = \frac{2L}{3}
$$
This is farther to the right than $L/2$, which makes sense because the rod is denser at larger $x$.
For nonuniform objects, always compute the total mass first:
$$
M = \int dm
$$
Then use
$$
x_{\text{cm}} = \frac{1}{M}\int x\,dm
$$
and similarly for $y_{\text{cm}}$ and $z_{\text{cm}}$.
Common Results for Uniform Shapes
Many standard uniform objects have centers of mass at simple geometric locations.
| Object | Center of mass location |
|---|---|
| Uniform rod of length $L$ | Midpoint, $L/2$ from either end |
| Uniform rectangular plate | Geometric center |
| Uniform circular disk | Center of the disk |
| Uniform solid sphere | Center of the sphere |
| Uniform rectangular block | Geometric center |
These results come from symmetry, though they can also be verified by integration.
Strategy for Solving Problems
A center of mass problem for a continuous object usually follows a clear pattern. First, choose coordinates that match the shape. Second, write the correct mass element $dm$. Third, find the total mass if needed. Fourth, evaluate the coordinate integrals. Finally, check whether the answer agrees with symmetry and common sense.
If the object has strong symmetry, the work may be very short. If the density varies, the integral becomes more important.
Physical Meaning
The center of mass is the balance point of the object's mass distribution. If you support an object exactly at its center of mass, it can balance without rotating, provided no other torques act. For continuous objects, this balance point depends both on shape and on how the mass is distributed inside the shape.
This is why a uniform object often has its center of mass at the geometric center, while a nonuniform object does not.