0
$\begingroup$

The tank has the shape of a horizontal cylinder with radius $r$ and length $l$. The water exists through a small opening on the top right side of the cylinder. It seems that there are two ways to solve this problem:

  • integrate horizontally
  • integrate vertically

My first attempt was done by integrating vertically. Slicing the cylinder horizontally produces a rectangle of length $l$ and width $2x$. Using Pythagorean theorem we obtain $x=\sqrt{r^2-y^2}$.

The volume can be found by adding up small rectangular prims, i.e., $V = 2xl\Delta y = \int_{-r}^{r}2l(\sqrt{r^2-y^2})dy$.

To make sure this is correct let $l=5$ and $r=2$. The volume of this cylinder is $V=\pi r^2h = 62.83$. The volume obtained using the integral is also $62.83$.

I encounter problems when I try to find the work required to pump the water.

The mass of the water is the volume times the density $$m = 2000l\sqrt{r^2-y^2}\Delta y$$

The force due to gravity is $$F = (9.8)\left(2000l\sqrt{r^2-y^2}\Delta y\right)=19600l\sqrt{r^2-y^2} \Delta y$$

The water must travel a distance of $-y+r$ to exit through the small hole, thus the work required to pump the water is $$W = 19600l\sqrt{r^2-y^2}(r-y) \Delta y$$

Which is $$19600l\int_{-r}^{r}\sqrt{r^2-y^2}(r-y)dy$$

After taking this (nasty) integral we get $$W = 9800\pi lr^3$$.

For some reason I seem unable to understand the horizontal integration.

I know the volume can be expressed as $V = \pi r^2 \Delta l$. Thus, the mass is $1000\pi r^2 \Delta l$, force is $9800 \pi r^2 \Delta l$ and the work is $$9800 \pi r^2 \int_0^l dl = 9800\pi r^2 l$$

Where does the other $r$ come from?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

In short, all you've computed in the last integral is the weight of the water, not the work applied to pump it out of the tank. The missing factor is the vertical displacement to the opening.

Honestly, I'm not sure if there's a general way to integrate horizontally and include the vertical displacement without adding an additional step where we integrate each "slice" vertically.

That said, in this case, it's simple enough to see that the center of gravity of the water in a filled tank lies $r$ below the level of the opening. This vertical displacement from the center of gravity to the opening is where you find the missing $r$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .