# Center of mass of a symmetrical rigid body

Suppose a rigid body has mirror symmetry along the $$z$$-Axis, i.e. $$\rho(x,y,z)=\rho(-x,-y,z)$$ where $$\rho$$ is the density of the body.

How can I show from this that the center of mass lies on the $$z$$-Axis and that those non-diagonal entries of the inertia tensor corresponding to $$z$$ vanish?

Both statements are very intuitive, but I would like to prove it formally.

I thought that maybe cylindrical coordinates would help, but those don't get me anywhere either.

Any hint or advice is very much appreciated!

• Divide the $x$ and $y$ integrals into two parts, one from $-\infty$ to $0$ and the other from $0$ to $\infty$. – G. Smith Dec 17 '19 at 18:27

This is a volume obtained by revolving $$z= f(\sqrt{x^2+y^2}) = f(r)$$ around z-axis. The x coordinate of the center of mass is given by

$${ \bar x } =\dfrac{\int\int \int(\rho\cdot r\sin \theta\, r\, d\theta \,dr\, dz)}{\int\int\int(\rho\, r\, d\theta \,dr\, dz)}$$ The numerator can be expressed as $${\int_0^{2 \pi} (\sin \theta \,d\theta) \cdot \int \int(\rho \,r^2 \,dr\, dz)}$$ which vanishes. Similarly for the y-coordinate the $${\bar y }$$ center of mass is zero, $$\bar z$$ is non-zero, so CM lies on z-axis.

Try doing this with summation over discrete points first to see how the logic works. Most texts do this. Consider point masses each with M, located symmetrically in the x-y plane about the z axis. For example (the simplest example), M at (x, 0, 0) and M at (-x, 0, 0). Clearly the equations for CoM and product of inertia in the x-y plane will cancel.

2Mxcm = Mx + M*(-x) = 0;

M(xy) + M(-xy) = 0

Now generalize this to integrals. As the comment suggests, divide up your integrals over x and y into two parts, -infinity to 0 and 0 to +infinity. Then you need to do some manipulation of the integrand to make the integrals along the negative axis match those along the positive axis. You will get the same expression with a relative minus sign.

I will only show the case for center of mass because the case for moment of inertia is analogous. We have for the $$x$$ component of the center of mass:

$$x_\text{com} = \frac 1M \int \rho(x,y,z) \cdot x\, \mathrm d^3 x$$

for simplicity I'm going to take $$M=1$$ in some units as the mass will not play a role in the argument.

Now do a change of coordinates $$x \to \tilde x = -x$$ and $$\tilde y = -y$$, $$\tilde z = z$$ i.e. a reflection along $$z$$ axis. Then the absolute value of the Jacobian determinant is $$1$$ and we have $$\, \mathrm d^3 x = \, \mathrm d^3 \tilde x$$.

\begin{align}\implies x_\text{com}&= \int \rho(x,y,z) \cdot x\, \mathrm d^3 x= \int \rho(\tilde x, \tilde y,\tilde z) \cdot \tilde x\, \mathrm d^3 \tilde x \\ &= \int \rho(- x, -y, z) \cdot (-x)\, \mathrm d^3 x \\ &=\int \rho(x, y, z) \cdot (-x)\, \mathrm d^3 x = - x_\text{com} \end{align}

Since $$x_\text{com} = - x_\text{com}$$ we conculude $$x_\text{com}=0$$.