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

• etotheipi
Similarly ##\Delta## may stand for small but nonzero values, e.g. for the definition of the derivative. So they are not interchangeable in mathematics, though they often used as synonyms in physics.
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!

etotheipi said:
such that the centre of mass is ##\frac{a}{2}## from the centre of the sector.
Sure?

berkeman and etotheipi
haruspex said:
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!

berkeman
etotheipi said:
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.

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

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

Last edited by a moderator:
Thread moved from Precalc homework section.
etotheipi said:
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
Mark44 said:
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?

etotheipi said:
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.

Mark44 said:
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.

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.

etotheipi

## 1. What is the centroid of an arc?

The centroid of an arc is the point at which the arc can be balanced on a pin. It is the center of mass of the arc.

## 2. How is the centroid of an arc calculated?

The centroid of an arc can be calculated by finding the midpoint of the arc's chord and drawing a line from the midpoint to the center of the circle. The point where these two lines intersect is the centroid.

## 3. What is the significance of finding the centroid of an arc?

Finding the centroid of an arc is important in engineering and physics, as it helps determine the point of balance and stability for objects with curved shapes. It is also useful in determining the center of mass for objects with irregular shapes.

## 4. How is the centroid of a sector different from the centroid of an arc?

The centroid of a sector is the point at which the sector can be balanced on a pin. It is also the center of mass of the sector. The difference is that a sector is a portion of a circle, while an arc is a portion of a curve.

## 5. Can heuristic arguments be used to find the centroid of an arc or sector?

Yes, heuristic arguments can be used to estimate the centroid of an arc or sector. These arguments rely on logical reasoning and educated guesses rather than exact calculations. They can be useful in situations where precise measurements are not available.

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