# 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.)