To calculate the centre of gravity of a spherical cap

AI Thread Summary
The discussion revolves around calculating the center of gravity of a spherical cap, where the original book answer is questioned. The user proposes an alternative method involving a different integral limit and an area formula for the cap. Initial calculations yield a different result than the book, prompting a review of the area formula used. After identifying a misinterpretation of the area formula, the user corrects it and confirms that the new approach leads to the correct expression. The conversation concludes with the user expressing gratitude for the clarification.
gnits
Messages
137
Reaction score
46
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:

IMG_20210330_164738_415.jpg


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:

IMG_20210330_164704_521.jpg


And here is the answer:

IMG_20210330_164731_518.jpg


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:

I1.JPG


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.
 
Physics news on Phys.org
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##.
 
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.
 
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.
 
So with that, are you now able to find the correct expression?
 
Yes, I am. Thanks very much.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top