# 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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arildno
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HOW is the sphere trunctated?

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

Dear,

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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 $\alpha$.
$\bar{z} = \frac{1}{V}\int_R z dV$
$\frac{\int _0^hz r^2\text{Sin}\left[\text{ArcCos}\left[1-\frac{z}{r}\right]\right]^2dz}{\int _0^h r^2\text{Sin}\left[\text{ArcCos}\left[1-\frac{z}{r}\right]\right]^2dz}$