Center of Mass

Thus far, we have been dealing only with objects that can be treated as point particles. Loosely, we’ve been considering objects with uniform distributions of mass, and using their center of mass as the point to be reckoned with.

The center of mass (or CM) is the point at which all an object’s mass can be considered to be concentrated; you can think of it as the average position of the object’s constituent particles. If we’ve got an object with a linear distribution of mass in the $ x$ dimension only, then $\displaystyle x_{\textrm{CM}}=\sum_i{\frac{m_ix_i}{m_i}}=\frac1m\sum_i{\left({m_ix_i}\right)}$ and we can develop analogous expressions for the $ y$ and $ z$ directions. It’s often helpful to consolidate the coordinates $ \left(x_i, y_i, z_i\right)$ of each chunk of mass into one expression defined as $\displaystyle \mathbf{r}_i=x_i\hat{\mathbf{i}}+y_i\hat{\mathbf{j}}+z_i\hat{\mathbf{k}}+$

Any of these techniques so far work perfectly for a system of particles, but can be applied also to larger objects with unevenly distributed mass. We simply divide the object into chunks of discrete chunks of known mass $ \Delta m_i$ and position $ \mathbf{r}_i$; we’re using the Riemann sum technique for integration.

Surely you’re guessing now (and I promise not to call you Shirley again) that we can develop an integral expression for the CM in each dimension. This is done by taking the limit as each chunk $ \Delta m_i\to0$, leaving $\displaystyle x_{\textrm{CM}}=\frac1m\lim_{\Delta m_i\to0}\sum_i{\left({m_ix_i}\right)}$ and then replacing the macroscopic chunk of mass $ \Delta m_i$ with the infinitesimal differential $ dm$ as we change the summation sign to an integral: $\displaystyle x_{\textrm{CM}}=\frac1m\int\!\! x\,dm$

The expressions for $ y_\textrm{CM}$ and $ z_\textrm{CM}$ are analogous, so we can use our previous $ \mathbf{r}_\textrm{CM}$ notation to consolidate our formulas into one: $\displaystyle \mathbf{r}_{\textrm{CM}}=\frac1m\int\!\! \mathbf{r}\,dm$

The study of center of masses is more interesting (seriously) if we concern ourselves with objects having mass distributions defined by functions. Most elementarily, if we have a rod that is $ \ell$ long and has linear density $ \lambda$ (a function of the position along the rod $ x$), how can we find its center of mass?

Since $ \lambda$ is defined as $\displaystyle \lambda = \frac{dm}{dx}$ we solve for $ dm$ in order to conveniently substitute: $ dm=\lambda\,dx$. Now we can plug these values for $ m$ and $ dm$ into our integral expression: $\displaystyle x_{\textrm{CM}}=\frac{1}{m}\int\!\! \lambda x\,dx$ and of course the integral expressions for $ y_{\textrm{CM}}$ and $ z_{\textrm{CM}}$ are similar.

We can also define an object’s area density $ \sigma(x,y)$, in which case the proof is similar to that above that $\displaystyle x_{\textrm{CM}}=\frac{1}{m}\iint_A\!\!\sigma x\,dA$ (et cetera) over a planar lamina $ A$, or with a volume density $ \rho(x,y,z)$, $\displaystyle x_{\textrm{CM}}=\frac{1}{m}\iiint_V\!\!\rho x\,dV$ again, with similar expressions for the remaining dimensions.

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