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Work done by gravity to fill truncated cone.

  1. May 19, 2006 #1
    Ok i need to calculate the work done by gravity , while filling a truncated cone of bottom radius R and upper radius r (R>r) and height H , with sand of density 'd' , if we start filling the cone from bottom..

    What i did was , I considered a disc of radius 'x' as a part of the cone and with thickness 'dy' , this dy is not vertical but slanting as per the contour of the cone , so the force on this disc would be 'd(pie)(x^2)(dy)g' and the total height to be raised against gravity is H //

    So i guess work done should be : (d)(pie)(x^2)(dy)g .H , now because dy is slanting and not vertical ,should I integrate this expression from 0 to the 'total slanting height of the cone' ??? ..hwo shud i integrate ..??
  2. jcsd
  3. May 19, 2006 #2


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    I suggest using cylindrical coordinates for your integration.
  4. May 19, 2006 #3
    any reference website , for using cylindrical coordinates...its been a long time i used them//?
  5. May 19, 2006 #4
  6. May 19, 2006 #5
    You could also use basic trigonometric functions to make the integration simpler( use the semi vertical angle).
  7. May 20, 2006 #6
  8. May 21, 2006 #7
    Ok i used the cylindrical coordinates ... The work done would be

    /dW = / g.dm.z (because z is the distance to be transversed in vertical direction while filling the sand)

    then dm = d.dV where d is the density of the sand

    in cylindrical coordinates:

    /dV = / r dr dO dz

    i integrated dr from R to r
    and dz from 0 to H ...

    I got the wrong answer..
  9. May 22, 2006 #8

    For a given z, what are the correct limits for the integration over the radial variable?, Not r nor R.
  10. May 23, 2006 #9
    I think limits for z would be 0 to H
  11. May 23, 2006 #10

    Doc Al

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    While the edges of your discs are slanted, that does not mean "dy" is slanted. Since the discs are infinitesimally thin, the slant of the edges is irrelevant: The volume of each disc is still [itex]\pi x^2 dy[/itex].

    Now consider that each disc is raised by a height of "y", not H, so your integral becomes:
    [tex]\int_0^H \rho \pi x^2\,gy \,dy[/tex]

    Evaluating this is easy. First find an equation expressing x as a function of y. (No need for cylindrical coordinates.)
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