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Synopsis

In Schey’s Div, Grad, Curl, and All That, we are greeted with the following exercises (this is under the section involving surface integrals)

enter image description here

I solved problem II-6 using a surface integral and obtained the correct answer, but II-7 is where the trouble began. You see, Schey doesn't talk about moment of inertia at all in his textbook so I had to go off my AP Physics C knowledge from high school, but for some reason, I just couldn't land on a correct answer (AP Physics C was a while ago so maybe I'm rusty). I just hope to get some tips or hints on how to proceed. The answer to II-6 is $4\pi R^2 \sigma_0/3$ if that helps and the answer to II-7 is $16\pi R^4 \sigma_0/15$. Below, I give a summary of my thoughts and an attempt.

My Attempt

Noting that the distribution of mass depends only on $x^2+y^2$, which itself is dependent only on $z$ (note that $r^2=x^2+y^2=R^2-z^2$), my idea is to turn the hemisphere into a series of thin rings of radius $r$ and add up the moments of inertia $\int dI$ for each ring along the $z$ axis.

Recalling that $I = \int r^2\ dm$ (at least that's the formula I found in my old physics textbook), I explore the moment of inertia contribution from a thin ring. Since the mass distribution is constant for each ring of radius $r$ and the distance from the axis is also constant (we are considering a ring rather than a disk), we can just multiply out to find the moment of inertia. Indeed, we have $$I_{ring} = (r^2)\sigma(2\pi r) = 2\pi r^5(\sigma_0/R^2).$$ Then, recalling that the radius $r$ of each thin ring is dependent on $z$, we have the moment of inertia of the hemisphere as the integral $$\frac{2\pi\sigma_0}{R^2} \int_0^R (R^2-z^2)^{5/2}\ dz = \frac{5\pi^2 R^4}{16}$$ (courtesy of Mathematica), which is obviously wrong. I would appreciate any tips or for someone to point out what I'm getting wrong, so thank you in advance! Self studying is always a bit hard when it comes to doing the exercises.

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1 Answer 1

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I am not really familiar with how to draw here so sorry if that is what you wanted but I did solve the problem,

I think the error in your approach is how exactly you have taken the small element $dz$. If the given problem was about a solid sphere, then your method is the correct way to approach, but here as you can see in the zoomed in diagram (in my solution) , the difference between $ dz $ and $R d\theta$.

The reason for this is that when dealing with spherical shells, hollow bodies you have to integrate along their surface, but in case of solid bodies, integration should be done for body and those elements can be approximated to discs with negligible end truncations. But here, the surface has to be followed for integration, if you substitute, $z$ as $R sin(\theta) $, and change $ dz $ to $dz / (sin(\theta) $. You will land on the right answer

Here is the photo of my solution. You can see the minute difference between the above mentioned two ways in the zoomed in diagram, while you measured the verical height, the slant height needed to be measured.

enter image description here

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  • $\begingroup$ To learn how to render equations, click here. mathjax is the site standard. Images of text or equations are very strongly discouraged and can result in downvotes. $\endgroup$
    – joseph h
    Jul 19, 2021 at 7:18
  • $\begingroup$ @joseph h Thanks for the link $\endgroup$
    – Hardik
    Jul 19, 2021 at 7:35
  • $\begingroup$ you are an angel. thank you so much this question was annoying the heck out of me! $\endgroup$
    – mijucik
    Jul 19, 2021 at 7:42

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