To calculate the centre of gravity of a spherical cap

[gnits]

Homework Statement

To calculate the centre of gravity of a spherical cap

Relevant Equations

comparison of moments

Could I please ask for help as to why I disagree with a book answer on the following question:

Answer given is book is $$\frac{1}{2}(a+b)$$

Here's my proposed method:

Prior to this question there is an example of a similar question:

And here is the answer:

So, to solve my question I propose to solve the same integral but instead of the limits being $$0\,\,to\,\,\frac{\pi}{2}$$ I will use $$arcsin(b/a)\,\,to\,\,\frac{\pi}{2}$$

And for the area of the whole cap I will use the formula $$Area_{cap} = \pi(h^2+a^2)$$ where h is the height of the cap, so in my case h = a - b and so I have $$A_{cap} =\pi((a-b)^2+a^2)$$

Using Wolfram Alpha to solve the integral (for now, to see if I agree with book answer, will derive by hand if it works) have:

So equating moments, this would lead me to:

$$\pi\,a\,w\,(a^2-b^2)\,=\,\pi((a-b)^2+a^2)\,w\,\bar{x}$$

which gives:

$$\bar{x} = \frac{a(a^2+b^2)}{a^2+(a-b)^2}$$

Which is not the book answer.

Thanks for any help,
Mitch.

 

[Orodruin]

You can easily see that your answer must be wrong as the limit when $b \to a$ should be $a$.

Where did you get the expression for the cap area? It does not seem correct as it is non-zero when $h \to 0$.

 

[gnits]

Thanks very much. Yep, the formula for the surface area is wrong. I got it from https://mathworld.wolfram.com/SphericalCap.html. I may have misinterpreted it. I should have noticed that it didn't tend to zero as h tended to zero. Thanks for seeing that. I used instead a different version of the formula from the same page and it checks out in Wolfram Alpha, will derive by hand now.

$$Area_{cap} = 2\,\pi\,r\,h$$

Thanks,
Mitch.

 

[gnits]

I see that I did misinterpret the first version of the formula. The a in that version is the radius of the base of the cap , not the radius of the whole sphere.

 

[Orodruin]

So with that, are you now able to find the correct expression?

 

[gnits]

Yes, I am. Thanks very much.