Finding the Centroid of an arc, and then of a sector, with heuristic arguments

[etotheipi]

Homework Statement

A uniform circular arc of radius $a$ subtends an angle $2\alpha$. The distance from the centre of the arc to the centre of mass of the arc is a function of $d = f(\alpha)$. By cutting the arc into two similar arcs, show that $f(\alpha) = f(\frac{\alpha}{2})\cos{\frac{\alpha}{2}}$.

Then assume that $f(\alpha) = \frac{A\sin{\alpha}}{\alpha}$ to be the correct form for $f(\alpha)$ and derive that $d$ for a lamina in the shape of a sector of a circle is $\frac{2a\sin{\alpha}}{3\alpha}$.

Relevant Equations

N/A

The first part is not a problem, I let one radius lie along the $x$ axis and then we can write down

$S_x = \frac{M}{2}f(\frac{\alpha}{2})\cos{\frac{3\alpha}{2}} + \frac{M}{2}f(\frac{\alpha}{2})\cos{\frac{\alpha}{2}} = Mf(\alpha)\cos{\alpha}$

from which we can then get the following after using the identity $\cos{\frac{3\alpha}{2}} = 4\cos^{3}{\frac{\alpha}{2}}-3\cos{\frac{\alpha}{2}}$,

$f(\frac{\alpha}{2})\left[\frac{\cos{\frac{3\alpha}{2}}+\cos{\frac{\alpha}{2}}}{\cos{\alpha}}\right] = 2f(\alpha) \implies f(\alpha) = f(\frac{\alpha}{2}) \cos{\frac{\alpha}{2}}$

It's the second part that's tripping me up. We're now considering a sector which I set up such that the radius again lies along the $x$ axis. The fact that it is a sector does not change the general form of $f(\alpha)$ since we could just replace linear density with area density in the above working.

I consider an arbitrarily small sector of width $\delta \theta$ at an angle $\theta$ from the $x$ axis such that the centre of mass is $\frac{a}{2}$ from the centre of the sector. I can write the expression for $\delta S_x$ (using areas for simplicity) and then integrate from $0$ to $2\alpha$,

$\delta S_x = \frac{a}{2}\cos{\theta} \times \frac{\delta \theta}{2} a^2$
$S_x = \frac{a^3}{4}\sin{2\alpha} = \alpha a^2 \bar{x}$

and then once I have $\bar{x}$ I can just divide by $\cos{\alpha}$ to get $d$:

$d = \frac{\bar{x}}{\cos{\alpha}} = \frac{a\sin{\alpha}}{2\alpha}$

It's of the right form, but the constant out the front is all wrong. I'm not sure where I've gone wrong; any help would be appreciated. Thanks!

 

[haruspex]

such that the centre of mass is $\frac{a}{2}$ from the centre of the sector.

Sure?

 

[etotheipi]

Sure?

Ah. That's it. I'd assumed if you make $\delta \theta$ small enough it approaches a rectangle, but it should actually approach a triangle. I'll try that and see if it works!

Edit: It works, just need to change $\frac{a}{2}$ to $\frac{2a}{3}$. Thanks @haruspex!

 

[haruspex]

Ah. That's it. I'd assumed if you make $\delta \theta$ small enough it approaches a rectangle, but it should actually approach a triangle. I'll try that and see if it works!

Edit: It works, just need to change $\frac{a}{2}$ to $\frac{2a}{3}$. Thanks @haruspex!

Good - except that you were instructed to derive it from the arc result.
So instead, consider an arc of radius r, thickness dr.

 

[etotheipi]

Repurposing the last part for an arc of radius $r$, I obtain

$\delta S_x = r^2 \cos{\theta} \delta \theta$
$S_x = r^2 \sin{2\alpha} = 2\alpha r \bar{x} \implies d = \frac{r\sin{\alpha}}{\alpha}$

I think what you're hinting at is that we can integrate up these arcs to get a sector. I'll try that! So considering a sector with an incremental bit of arc on the end, we get

$$\int dS_x = \int_0^a \frac{r\sin{\alpha}}{\alpha} \cos{\alpha} \cdot 2\alpha r dr$$ which does indeed give

$$S_x = \frac{a^3 \sin{2\alpha}}{3} = \alpha a^2 \bar{x} = \alpha a^2 d\cos{\alpha} \implies d = \frac{2a\sin{\alpha}}{3\alpha}$$

 

[Mark44]

Thread moved from Precalc homework section.

I consider an arbitrarily small sector of width $\delta \theta$

The usual terminology used in calculus textbooks for a small change in angle is $\Delta \theta$ (\Delta, rather than lowercase \delta). In an integral, the differential $d\theta$ would be used.

 

[etotheipi]

The usual terminology used in calculus textbooks for a small change in angle is $\Delta \theta$ (\Delta, rather than lowercase \delta). In an integral, the differential $d\theta$ would be used.

That's strange! Had no idea. Is it just because the $\delta \theta$ looks a little bit funny/wonky?

 

[Mark44]

That's strange! Had no idea. Is it just because the $\delta \theta$ looks a little bit funny/wonky?

Physicists use notation like that (i.e., with lowercase delta), but I don't know why, even though I studied physics for two years. The $\Delta x$ vs. dx notation is traditional usage.

 

[etotheipi]

Physicists use notation like that (i.e., with lowercase delta), but I don't know why, even though I studied physics for two years. The $\Delta x$ vs. dx notation is traditional usage.

I remember there was a thread about this a little while ago, I can't remember exactly what the outcome was though! I like having $\Delta$ for a standard finite change in some quantity, $\delta$ for a small variation and $d$ for an infinitesimal. I suppose at the end of the day it's just notation and so long as the meaning is clear it should be okay.

 

[Mark44]

The most common use for $\delta$ in mathematics parlance is in so-called $\delta - \epsilon$ proofs that use the definition of the limit. Each of these represents a small but positive value.