How much water to fill a cylinder standing on it's end for maximum stability?

Question

I think it's half full, right?

Assume the container's walls have negligable mass.

Answer 1

I think the mass of the empty cylinder matters = M, say.
Its height H matters too.

Then mass of water, of height h, is m = kh
where k = area of base x the density of water.

Then the height of the c.g. is next to be found.
If you draw a diagram of the cylinder so that its axis is the h axis, you can now find y, the height-position of the combined c.g. of both water and cylinder, in terms of h and H.

m(y - h/2) = M (H/2 - y)

kh(y - h/2) = MH/2 - My

Differentiate y with respect to h. Find the turning point, where y is a minimum. This is the lowest the c.g. can go, and hence it is the position of maximum stability. Get the value of h at this point and this is the answer you want.

Hmm. It looks like a quadratic involving h and y.
Differentiating y with respect to h gives the expression

(M + kh) (dy/dh) = k (h - y)

The minimum height of h occurs when the derivative of y with h is zero, but this happens at h = y.

Physically, what does it mean? It means the centre of gravity goes down as water is first poured in, and finally goes down as far as the surface of the water being poured in. Here is where the stability is maximum, and the c.g. is at its lowest. Pouring more water in will make the c.g. rise again.

You can't solve this without knowing the mass of cylinder and its height and the density of water and the area of base. Substitute all this and you get a quadratic in y (or h, since h = y) and only then can you get a numerical solution.

If the container's walls have negligible mass, then there isn't this effect of lowering of the c.g. of the cylinder as water is poured in.
The c.g. of the combination is just the c.g. of the water, and that goes up and up all the way, making the cylinder more and more unstable.

Answer 2

If you are assuming the container walls have negligable mass, then putting any fluid into the cylinder would add stability. At the point where the height of the fluid excedes the diameter of the cylinder the stability of the cylinder would begin to dramatically decrease (even if it isn't half way up the cylinder yet)

Once you start counting for the mass of the walls this point will change depending upon the mass of the cylinder and the mass of the fluid.

Answer 3

A cylinder is stable if its center of gravity falls in its base when it is tilted through an angle.

The maximum angle through which a cylinder can be tilted safely before it looses its stability is given by

tan = d /h where d is the diameter of the base and h is the height of the cylinder.

But when it is having water inside, when the cylinder is tilted, the height of water surface remains the same what ever the angle through which the cylinder is tilted.

Considering the shape of water and the water surface, it can be seen easily that the center of gravity of water will fall to the base of the cylinder up to an angle of 45 degree and no more.

This means the height of water h1 is equal to 0.707 times the diameter of the cylinder.

h1 = 0.707 d.

Substituting the value of d from the above equation,

h1 = 0.707 h tanÓ¨.

If an empty cylinder is stable up to an angle Ó¨.

the same will be stable for that angle if water is filled up to 0.707 of the height of the cylinder.