# Centre of mass of truncated sphere

## Homework Statement

derivation of the centre of mass of the truncated sphere

## The Attempt at a Solution

i tried to solve it with triple integrals but i failed to figure ut the integral limits

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## Answers and Replies

arildno
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You need to give more information here.
HOW is the sphere trunctated?

AS for the integral limits, you should most likely shift to spherical coordinates.

Dear,
Please see the attached File

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arildno
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Please state the problem here, rather than referring to some other document I have to wait before opening.

i m sorry for that but i can not draw diagram in this page,thats why i send the attached image.
i m again sorry ,i dont want u to wait.

arildno
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Well, I am unable to open the file prior to the mentor's approval of it.

arildno
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Now, I'm not exactly sure what you mean by saying "h and R" is changing.
We have the relation:
$$h=R+\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}$$
For any particular truncation, (with $\phi$ being the angle between the upwards vertical and the position vector from the origin,) we may decompose our volume as a sphere with the appropriate cone cut out, plus that cone.

The volume of the cone is clearly: $$V_{cone}=\frac{1}{3}\pi(\frac{\alpha}{2})^{2}*\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}$$
wheras the volume of the "cone-less" sphere must be:
$$V=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{2\pi}\int_{0}^{R}r^{2}\sin\phi{dR}d\theta{d}\phi=\frac{2\pi{R}^{3}}{3}(1+\cos(\sin^{-1}(\frac{\alpha}{2R})))=\frac{2\pi{R}^{3}}{3}(1+\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{R})$$

arildno
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As for the center of mass, I forgot that that was your question. Sorry, I'll look upon it tomorrow..

arildno
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Okay, I'll finish this, then:

Since $z=r\cos\phi$, it follows that on the plane where $z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex], we have that the radius follows the the curve: $$r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}$$ Also, remember that the truncated sphere is axially symmetric, so that the x-y-coordinates must be zero. Generally, assuming unit density, the z-coordinate of the center of mass of some region R will be: $$z_{c.m}=\frac{1}{V}\int_{R}zdV$$ We decompose our integral into two parts, then (with V as the ugly expression given above): $$z_{c.m}=\frac{2\pi}{V}(\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}zr^{2}\sin\phi{d\phi}dr+\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr)$$ With $$Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}$$ We now calculate the two integrals: $$\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\frac{R^{4}}{4}\cos\phi\sin(\phi)d\phi=\frac{R^{4}}{8}\sin^{2}\phi\mid_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}=-\frac{R^{2}\alpha^{2}}{32}$$ $$\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\frac{Z^{4}}{4}\frac{\sin\phi}{\cos^{3}\phi}d\phi=\frac{Z^{4}}{8}\frac{1}{\cos^{2}\phi}\mid_{0}^{\sin^{-1}(\frac{\alpha}{2R})}=\frac{Z^{4}}{8}(\frac{4R^{2}}{4R^{2}-\alpha^{2}}-1)=\frac{Z^{4}}{8}\frac{\alpha^{2}}{4R^{2}-\alpha^{2}}$$ Adding these, you should get someting like: $$z_{c.m}=-\frac{\pi\alpha^{4}}{64V}$$ Last edited: dear can u tell what is alpha Dear arildno, Please explain about alpha ,i dont understand about it arildno Science Advisor Homework Helper Gold Member Dearly Missed Okay, I misread your [itex]\gamma$ for an $\alpha$.
Sorry about that.

$\bar{z} = \frac{1}{V}\int_R z dV$

Why isn't the answer:

$\frac{\int _0^hz r^2\text{Sin}\left[\text{ArcCos}\left[1-\frac{z}{r}\right]\right]^2dz}{\int _0^h r^2\text{Sin}\left[\text{ArcCos}\left[1-\frac{z}{r}\right]\right]^2dz}$