In the image the situation is illustrated, also provided is a simple proof for the circle segment area. The problem can be solved without actually knowing the lenght of the dowel.
We will only care for the circular seection of the dowel which is partially submerged. Recall that for a floating body the weight of displaced water is equal to the body's weight.
For the dowel's weight we have:
where g is the gravitational constant and m the dowel's mass.
Now for the displaced water weight:
where
is the water density, which happens to be
.
we now have the following:
but:
being V the volume of the dowel, putting the above toghether gives us:
Now
is the volume of the dowel that is submerged.The fraction of the dowel submerged will be according to the image:
So the dowel's volume that is submerged is:
Plug this into the previous expression to get:
There's only one thing missing, the
angle. Refer to the second image to get the following expression:
where r is the dowel's radius.
Plugging the above in the expression for the dowel's density we get:
![\frac{\left(\pi-Acos(\frac{1}{2r})+sin\left[ Acos\left( \frac{1}{2r}\right )\right]\right)}{\pi}.\rho_{water}\\ ](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%28%5Cpi-Acos%28%5Cfrac%7B1%7D%7B2r%7D%29%2Bsin%5Cleft%5B%20Acos%5Cleft%28%20%5Cfrac%7B1%7D%7B2r%7D%5Cright%20%29%5Cright%5D%5Cright%29%7D%7B%5Cpi%7D.%5Crho_%7Bwater%7D%5C%5C%0A)
where:
![sin\left[ Acos\left( \frac{1}{2r}\right)\right]=\sqrt{1-\left( \frac{1}{2r}\right)^2}](https://tex.z-dn.net/?f=sin%5Cleft%5B%20Acos%5Cleft%28%20%5Cfrac%7B1%7D%7B2r%7D%5Cright%29%5Cright%5D%3D%5Csqrt%7B1-%5Cleft%28%20%5Cfrac%7B1%7D%7B2r%7D%5Cright%29%5E2%7D)
We finally get:
This result is in
.
