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Why do icebergs flip over? Are certain shapes of icebergs more "stable" than others, in that it's harder to flip them over? If so, why?

For example, it somehow makes intuitive sense, that a thick iceberg with a certain height (or depth, because 90% of it is below water) would be harder to flip over than a thin one with the same height/depth. But why is this the case?

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    $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Sep 4, 2021 at 13:58
  • $\begingroup$ @Community is that better? $\endgroup$ Sep 4, 2021 at 14:00
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    $\begingroup$ Community is not a real person and nobody will be notified with @community. The message was left by a reviewer in an anonymous way. This means that they won't be notified and there is no way to notify them. That is a problem with the current review system meta.stackexchange.com/a/369062 $\endgroup$
    – Prallax
    Sep 4, 2021 at 14:09
  • $\begingroup$ @Prallax ah ok, thank you. $\endgroup$ Sep 4, 2021 at 14:10
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    $\begingroup$ Draw an iceberg and see how it will float. $\endgroup$
    – rob
    Sep 7, 2021 at 14:52

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This is a nice Newtonian physics question. A very simplified analogue to this is to consider the stable positions of a 2-dimensional table shape iceberg with length $L$, width $W$ ($W < L$).

Typically the material density ratio of $\rho_{\text{ice}} / \rho_{\text{water}} = 0.9$. In a force equilibrium, this leads to the fact that 90% of the iceberg is immersed in water. As with the inverse pendulum, there are stable and unstable force equilibriums.

Algebraically, it is simpler to consider the energy equation for a table iceberg with the short side up versus the long side up. You'll find that the short side up is energetically better, and leads to a stable minimum.

Most of the energy difference is dissipated in creating water waves during the flip.

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I really don't know if this will answer your question but you have to take into account the forces acting over the iceberg. You have pressure from the sea stream, buoyancy from the Archimedes principle, pressure from wind, and weight from gravity acceleration.

Then, why would something flip? For something to flip you must have torque $$ \vec{\tau}=\vec{r}\times\vec{F} $$ When you see this equation you have to consider the object as a solid body, not a point in the center of mass. This means that if the force is applied far from the center of mass, there will be more torque. If you consider a long but thin iceberg in the vertical direction you'll see that it is less stable than in the horizontal direction. Think that horizontal forces come from wind and sea streams. The vertical forces, weight, and buoyancy only do torque when the iceberg is already angled. This effect is quite interesting because, at the moment where the iceberg tilts, the torque increases suddenly because there are more forces doing torque.

At the end of the day, the iceberg flips to reach a lower energy state (A configuration where it's more stable). Icebergs change all the time, they melt, lose pieces of ice, and therefore the stable state may change over time. That could mean flipping a couple of times.

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