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I want to figure out the COM (Center of Mass) of an infinitesimally thin semicircular ring of uniform density with mass $M$ and radius $R$.


I begin to draw the following picture:

Image of a semicircular ring

As I have understood it, the typical way to go about solving these kinds of problems is to try and think of the rigid body as it consisting of an infinite amount of point particles. Each particle will have a COM $m_{i}\bar{r_{i}}$ with $\bar{r_{i}}$ being the position vector of the COM for particle $i$. By summing up all of these infinite COMs and dividing by the total mass of the particles $M$, we obtain the COM of a system of particles.

The equations for the COM in the $y$- and $x$ directions are

$$ x_{cm} = \frac{\int x\ dm}{\int dm},\ \ \ \text{The position vector is }x = Rcos(\theta) $$

$$ y_{cm} = \frac{\int y\ dm}{\int dm},\ \ \ \text{The position vector is }y = Rsin(\theta), $$

but what is $dm$? It feels reasonable that $dm$ is the mass of the point particle whose location is given by the position vector $\bar{r_{i}}$. How do I figure out what that mass is?

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Perhaps it would be better to think of $dm$ as an infinitessimal mass element instead of a point particle. Usually you use the density to write $\text dm$ in terms of an infinitessimal volume/area/distance which you can integrate. Depending on the number of dimensions you get the following expressions for $\text dm$ \begin{align} \text dm&=\rho(x)\text dV&\text{in 3D}\\ \text dm&=\sigma(x)\text d A& \text{in 2D}\\ \text dm&=\lambda(x)\text ds&\text{in 1D} \end{align} Where $\rho,\sigma, \lambda$ are the density, area density and linear density respectively (sometimes different notation is used). Your problem is one dimensional and has uniform density so we can put $\text dm=\lambda \text ds$ where $\lambda$ is now constant. Before we can calculate $\text ds$ we first have to parametrize the domain. Define $s$ as the path length along the arc $$s(\theta)=R\theta\implies\text ds=R\text d\theta$$

We can now write the integral as \begin{align} x_\text{COM}&=\dfrac{\int \lambda(R\text d\theta)(R\cos\theta)}{\int \lambda(R\text d\theta)} \end{align}

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    $\begingroup$ The ring is meant to be infinitesimally thin, so we only want a density in one dimension, not two dimensions. $\endgroup$
    – J.G.
    Nov 10, 2021 at 19:51
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    $\begingroup$ @J.G. I totally missed that, I updated my answer $\endgroup$ Nov 10, 2021 at 20:09
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    $\begingroup$ Your explanation seems the easiest and most straight forward. We have a linear density $dm = \lambda (x) ds$ with $ds = Rd \theta$. Plugging in $\lambda(Rd \theta)$ into the integrals and evaluating yields $x_{cm} = 0$ and $y_{cm} = \frac{2R}{\pi}$ which is correct. $\endgroup$
    – NoName123
    Nov 10, 2021 at 20:22
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You're told the linear density is uniform, so it must be $\frac{M}{R\pi}$. Since the infinitesimal arc length is $Rd\theta$, $dm=\frac{M}{R\pi}Rd\theta=\frac{M}{\pi}d\theta$.

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  • $\begingroup$ So $\frac{M}{R \pi}$ is the uniform surface density call it $\sigma$. $\sigma = \frac{M}{R \pi}$ with $R \pi = A$ the surface area? I understand that the infinitesimal arc length is $Rd \theta$ but why do we multiply the arc length with the surface density and call that $dm$? If possible, would you mind visualizing $dm$ in a perhaps more detailed way than I have? $\endgroup$
    – NoName123
    Nov 10, 2021 at 20:02
  • $\begingroup$ @NoName123 It's a linear density in one dimension, so it's multiplied by a length to get a mass. If the thickness in a second dimension were some positive amount, say $\epsilon$, we'd have a surface density $\frac{M}{R\pi\epsilon}$ (well, technically $\frac{M}{\tfrac12\pi\left[R^2-(R-\epsilon)^2\right]}=\frac{2M}{\pi\epsilon(2R-\epsilon)}$). But that would be a different problem. $\endgroup$
    – J.G.
    Nov 10, 2021 at 20:07
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$\mathrm{d}m$ is an infinitesimal mass $m$, corresponding to what you call "point particles".

But they're not really point particles, they are infinitesmally small volume elements:

$$\mathrm{d}V=\mathrm{d}x\mathrm{d}y\mathrm{d}z$$

so that:

$$\mathrm{d}m=\rho \mathrm{d}V$$

Assuming $\rho$ (density) to be constant then the object's total mass $m$ is:

$$m=\rho \int_{Volume}\mathrm{d}V=\rho V_{total}$$

In thecase of your system,

$$\mathrm{d}V=\pi r \mathrm{d}r$$

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  • $\begingroup$ The ring is meant to be infinitesimally thin, so we only want a density in one dimension, not three dimensions. $\endgroup$
    – J.G.
    Nov 10, 2021 at 19:52
  • $\begingroup$ My approach was simply general. Nothing wrong with that. $\endgroup$
    – Gert
    Nov 10, 2021 at 19:57
  • $\begingroup$ It's not wrong, but it's only applicable here with some Dirac delta specifics. It's easier just to work in one dimension. $\endgroup$
    – J.G.
    Nov 10, 2021 at 19:59

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