Work done by gravity to fill truncated cone.

1. May 19, 2006

Dr.Brain

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. May 19, 2006

Tide

I suggest using cylindrical coordinates for your integration.

3. May 19, 2006

Dr.Brain

any reference website , for using cylindrical coordinates...its been a long time i used them//?

4. May 19, 2006

maverick280857

5. May 19, 2006

arunbg

You could also use basic trigonometric functions to make the integration simpler( use the semi vertical angle).

6. May 20, 2006

Dr.Brain

THNX MAVERICK.

7. May 21, 2006

Dr.Brain

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 ...

8. May 22, 2006

elgali

Limits!

For a given z, what are the correct limits for the integration over the radial variable?, Not r nor R.

9. May 23, 2006

Dr.Brain

I think limits for z would be 0 to H

10. May 23, 2006

Staff: Mentor

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 $\pi x^2 dy$.

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

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