- #1

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!

##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!