tarponchik
Contributor
They do the same mistake/ As I mentioned, for any uniform object, the centre of buoyancy can not be located above the object's mass centre.I disavow my first link and stand by my second link.
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They do the same mistake/ As I mentioned, for any uniform object, the centre of buoyancy can not be located above the object's mass centre.I disavow my first link and stand by my second link.
OK, here is the answer. The lowest potential energy would be not when the center of gravity is as low as possible, but when the distance between the centre of mass of the object and centre of mass of the displaced water is the shortest.If the log is positively buoyant, the center of gravity in the vertical orientation would be higher than when the log is horizontal. Lowest potential energy would be when the center of gravity is as low as possible. Think of an overloaded container ship with too many containers stacked too high. It might be floating (buoyant), but it would want to heel over.
That can't be the whole story, tho'. I suppose in the vertical position you've got more of the log submerged, thus buoyant forces are pushing more of the log up. Parts of the log out of the water are only subject to the force of gravity pulling down, without any buoyant force. The most stable position would have as little of the log submerged as possible.
I don't feel like thinking about this in any great detail, but I suspect exact calculations on the vertical log would require integration of the forces working on the submerged parts of the log (buoyancy and gravity) versus the part of the vertical log sticking out of the water. Physics has never been my strong suit, but this all brings to mind long-forgotten terms such as torque, angular momentum, and formulas for all sorts of rotating objects (and lever arms).
Sorry if that's what I sound to you.I reclaim both links as I believe that the arrows represent force vectors rather than the true centers of buoyancy and mass, but am willing to listen to an alternate opinion rather than being socratically beaten into submission by a pedant.
OK, here is the answer. The lowest potential energy would be not when the center of gravity is as low as possible, but when the distance between the centre of mass of the object and centre of mass of the displaced water is the shortest.
So, going back to the log, assume the ratio of the log length (L) to diameter is 10. In the vertical position, the distance between the two centres would be 1/4 of L for a log with density of 0.5, and in the horizontal position the distance would be something like 1/60 of L. So the log floats horizontally.
Now let's make our log square in cut. We already know that it will float horizontally for the same reason as the log above, but will it turn face up or edge up? Do your math and you'll see that it will float edge up. The mass centre of the square log will be located at the water level in both cases; however, the mass centre of the displaced water will be positioned higher when the edge is up, and this makes a difference.
Finally, lets shorten our square-cut log until it becomes a cube. Following the same math as above, we conclude that the uniform cube will float vertex up.