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## Homework Statement

derivation of the centre of mass of the truncated sphere

## Homework Equations

## 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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- Thread starter amjadmuhd
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derivation of the centre of mass of the truncated sphere

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.

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arildno

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i m again sorry ,i dont want u to wait.

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arildno

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

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arildno

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We have the relation:

[tex]h=R+\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]

For any particular truncation, (with [itex]\phi[/itex] 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: [tex]V_{cone}=\frac{1}{3}\pi(\frac{\alpha}{2})^{2}*\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]

wheras the volume of the "cone-less" sphere must be:

[tex]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})[/tex]

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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..

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arildno

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Okay, I'll finish this, then:

Since [itex]z=r\cos\phi[/itex], it follows that on the plane where [itex]z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex], we have that the radius follows the the curve:

[tex]r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}[/tex]

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:

[tex]z_{c.m}=\frac{1}{V}\int_{R}zdV[/tex]

We decompose our integral into two parts, then (with V as the ugly expression given above):

[tex]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)[/tex]

With [tex]Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]

We now calculate the two integrals:

[tex]\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}[/tex]

[tex]\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}}[/tex]

Adding these, you should get someting like:

[tex]z_{c.m}=-\frac{\pi\alpha^{4}}{64V}[/tex]

Since [itex]z=r\cos\phi[/itex], it follows that on the plane where [itex]z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex], we have that the radius follows the the curve:

[tex]r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}[/tex]

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:

[tex]z_{c.m}=\frac{1}{V}\int_{R}zdV[/tex]

We decompose our integral into two parts, then (with V as the ugly expression given above):

[tex]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)[/tex]

With [tex]Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}[/tex]

We now calculate the two integrals:

[tex]\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}[/tex]

[tex]\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}}[/tex]

Adding these, you should get someting like:

[tex]z_{c.m}=-\frac{\pi\alpha^{4}}{64V}[/tex]

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dear can u tell what is alpha

- #11

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Dear arildno,

Please explain about alpha ,i dont understand about it

Please explain about alpha ,i dont understand about it

- #12

arildno

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Okay, I misread your [itex]\gamma[/itex] for an [itex]\alpha[/itex].

Sorry about that.

Sorry about that.

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Why isn't the answer:

[itex]\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}[/itex]

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