What is minimal density?

Question

What is minimal density of soild uniform 60-degrees cone,
which can float on the surface of water with its axis vertical
and tip under the water.

60-degrees cone is made by rotating equilateral triangle
around its axis of symmetry.

1) cone pointing down, and
2) cone pointing up

are both local extrema of potiential
energy, obviously. Both can be local
minima, reagrdless of which one is
deeper.

Answer 1

A cone balancing on its point is in equilibrium. You can balance the forces and the torques. But it is in unstable equilibrium. Small perturbations induce torques away from equilibrium.

If the cone is deep enough in water, you will get stable equilibrium. Small perturbations induce torques back to equilibrium (ie, the buoyant forces induce torques that right the cone stronger than the top-heavy (and side-heavy) weight of the cone that tends to tip it over.

So you need to find the point of neutral equilibrium--the point at which tipping the cone by a small amount does not induce a net torque.

So you need to come up with an expression for the torque on your cone as a function of tip angle and how dense it is (and hence how deep it sits). At the neutral equilibrium, not only will the first derivative of that expression wrt theta be zero (equilibreum), but also the second derivative will be zero (neutral equilibrium). That will give you the density.

Have fun with that.

edit--it isn't as easy as just measuring effective center points in the vertical direction for the forces. You'll actually have to consider torques (so the horizontal components come into play) to find the equilibrium point (if there is one).

Answer 2

I'm going to assume that this is a matter of comparing the height of the center of gravity of the equilateral triangle in the water for different densities, and for the cases where it's pointed down, or pointed up. The volume of the cone is of the form k h³, where k is some irrelevant constant (yes, this problem is more general than just for the cone as described), and h is the height of the cone or pyramid. Then letting pw = density of water, and px = density variable of the object, we have for case 1) cone pointing down, and case 2) cone pointing up, assuming the height of the object being 1.

1) k h³ pw = k px, so that center of gravity is at 2/3 - (pw/px)^(1/3)
2) k (1- h³) pw = k px, so that the center of gravity is at 1/3 - (1 -pw/px)^(1/3)

Letting r = pw/px, the critical value is at

2/3 - (r)^(1/3) ≤ 1/3 - (1 -r)^(1/3)

As simple as this equation looks, it turns out to be a messy one to solve, and the numerical answer for the critical value is approximately:

r = 0.794263

At this critical density, the center of gravity of the cone would be approximately 0.258726 below the surface of the water.

I'm not checking out the case of the cone floating on its side, that's too messsy for me tonight.

Answer 3

Cat, as you noted, the Cb is variable and is the center of mass of the submerged section. Somehow you forgot the 'variable' part of your own words. The CM of the submerged section moves rapidly in the direction of tipping, so there is clearly an initial depth at which the Cb will move in a direction to cause stability. Like when the initial depth is even with the CM of the cone. It doesn't move at all when tipping occurs, but the Cb moves plenty.

Actually this is quite a problem even assuming a 2-D wedge shape. I'll have to do some pencil work to get it for a cone. See ya later.........

Answer 4

It is impossible to keep it vertical:

Because gravity acting at point Cg pulls down, and buoyancy acting at point Cb pushes up, for the cone to be in stable equilibrium when vertical, the center of gravity Cg of the cone must be lower than the center of buoyant force Cb.

Cg is fixed and is the centroid of the cone, but Cb is variable and is the centroid of the submerged part.

When the cone is vertical then Cg is the centroid of the whole cone, and Cb the centroid of the smaller submerged cone.

So Cg will always be higher than Cb when vertical and stable equilibrium cannot be attained.
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Bekki:

when the cone is vertical there is no torque because the forces are lined up with the center of mass and center of buoyancy. Do you agree?

To obtain stable equilibrium, a small deviation from vertical should cause the system to go back to vertical. This happens only if the center of mass is lower than the center of buoyancy. Then the torque will straighten the cone. Do you agree?

So please show me how to put that center of mass lower than the center of buoyancy.
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Steve,

do you mean that there is a depth where, in the event of tipping, the Cb moves higher than the Cg? Because I considered this and yes I believe it is possible. But the question requires the axis to be vertical, which I interpret as being in stable equilibrium when vertical, and that is impossible.

Of course you are welcomed to prove me wrong.