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Centre of mass of truncated sphere

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    derivation of the centre of mass of the truncated sphere


    2. Relevant equations



    3. The attempt at a solution
    i tried to solve it with triple integrals but i failed to figure ut the integral limits
     
    Last edited: Apr 5, 2009
  2. jcsd
  3. Apr 6, 2009 #2

    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.
     
  4. Apr 6, 2009 #3
    Dear,
    Please see the attached File
     

    Attached Files:

  5. Apr 6, 2009 #4

    arildno

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    Please state the problem here, rather than referring to some other document I have to wait before opening.
     
  6. Apr 6, 2009 #5
    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.
     
  7. Apr 6, 2009 #6

    arildno

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    Well, I am unable to open the file prior to the mentor's approval of it.
     
  8. Apr 6, 2009 #7

    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:
    [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]
     
  9. Apr 6, 2009 #8

    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..
     
  10. Apr 9, 2009 #9

    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]
     
    Last edited: Apr 9, 2009
  11. Apr 9, 2009 #10
    dear can u tell what is alpha
     
  12. Apr 11, 2009 #11
    Dear arildno,
    Please explain about alpha ,i dont understand about it
     
  13. Apr 11, 2009 #12

    arildno

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    Okay, I misread your [itex]\gamma[/itex] for an [itex]\alpha[/itex].
    Sorry about that.
     
  14. Jun 28, 2009 #13
    [itex] \bar{z} = \frac{1}{V}\int_R z dV [/itex]

    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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