Finding centre of mass of truncated sphere using triple integrals

[amjadmuhd]

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

 

[arildno]

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.

 

[amjadmuhd]

Dear,
Please see the attached File

 

[arildno]

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

 

[amjadmuhd]

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 don't want u to wait.

 

[arildno]

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

 

[arildno]

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]

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

 

[arildno]

Okay, I'll finish this, then:

Since $z=r\cos\phi$, 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:<br /> $$r=\frac{\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}}{\cos\phi}$$<br /> <br /> Also, remember that the truncated sphere is axially symmetric, so that the x-y-coordinates must be zero.<br /> <br /> Generally, assuming unit density, the z-coordinate of the center of mass of some region R will be:<br /> $$z_{c.m}=\frac{1}{V}\int_{R}zdV$$<br /> <br /> We decompose our integral into two parts, then (with V as the ugly expression given above):<br /> $$z_{c.m}=\frac{2\pi}{V}(\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}zr^{2}\sin\phi{d\phi}dr+\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr)$$<br /> <br /> With $$Z=\sqrt{R^{2}-(\frac{\alpha}{2})^{2}}$$<br /> <br /> We now calculate the two integrals:<br /> $$\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}zr^{2}\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\int_{0}^{R}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}\frac{R^{4}}{4}\cos\phi\sin(\phi)d\phi=\frac{R^{4}}{8}\sin^{2}\phi\mid_{\sin^{-1}(\frac{\alpha}{2R})}^{\pi}=-\frac{R^{2}\alpha^{2}}{32}$$<br /> <br /> $$\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\int_{0}^{\frac{Z}{\cos\phi}}r^{3}\cos\phi\sin\phi{d\phi}dr=\int_{0}^{\sin^{-1}(\frac{\alpha}{2R})}\frac{Z^{4}}{4}\frac{\sin\phi}{\cos^{3}\phi}d\phi=\frac{Z^{4}}{8}\frac{1}{\cos^{2}\phi}\mid_{0}^{\sin^{-1}(\frac{\alpha}{2R})}=\frac{Z^{4}}{8}(\frac{4R^{2}}{4R^{2}-\alpha^{2}}-1)=\frac{Z^{4}}{8}\frac{\alpha^{2}}{4R^{2}-\alpha^{2}}$$<br /> <br /> Adding these, you should get something like:<br /> $$z_{c.m}=-\frac{\pi\alpha^{4}}{64V}$$[/itex]

 

[amjadmuhd]

dear can u tell what is alpha

 

[amjadmuhd]

Dear arildno,
Please explain about alpha ,i don't understand about it

 

[arildno]

Okay, I misread your $\gamma$ for an $\alpha$.
Sorry about that.

 

[Gregg]

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

Why isn't the answer:

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