Ice melting in water in a gravity free fall

[dharavsolanki]

Here is the original question: -

consider a gravity free hall in which a tray of mass M , carrying a cubical block of ice of mass m is at rest in middle . if the ice melts ,by what distance does the center of mass of "the tray plus ice" system descends?

Well, the answer that I came up with is, that, since there is no eternal force on the system, the centre of mass cannot be displaced, no matter what.

Now, my doubts are regarding this itself. When the ice melts, it turns into water of lesser volume. In the absence of gravity, the cube of ice turns into a cube of water, but the water does not spread out evenly on the tray.

In this scenario, how is it possible that the water, whose volume is bound to be lesser than that of the ice, still maintains the same position of the centre of mass? If the volume is different, the geometry also has to be different to maintain the same position of the ice. How does it all happen?

 

[Stonebridge]

Here is the original question: -



Well, the answer that I came up with is, that, since there is no eternal force on the system, the centre of mass cannot be displaced, no matter what.

Now, my doubts are regarding this itself. When the ice melts, it turns into water of lesser volume. In the absence of gravity, the cube of ice turns into a cube of water, but the water does not spread out evenly on the tray.

In this scenario, how is it possible that the water, whose volume is bound to be lesser than that of the ice, still maintains the same position of the centre of mass? If the volume is different, the geometry also has to be different to maintain the same position of the ice. How does it all happen?

The water isn't going to spread out on the tray if there is no gravity.*
The ice will melt and, because of the surface tension the water has, will form a perfect sphere. The centre of mass of that sphere will be in the same place as the c of m of the cube was.
(*I am ignoring the minute gravitational attraction between the tray and the water and the possible adhesion of water to the tray. If this is meant to be included, the question get a whole lot trickier!)
But as a general principle, you are correct when you say that in the absence of any external force, the c of m of the tray-water system will not move. This would guide any answer involving the more complex scenario I mentioned.

 

[hamster143]

The centre of mass of that sphere will be in the same place as the c of m of the cube was.

What you're forgetting is that the diameter of the sphere will be larger than the size of the original ice block.

 

[kcdodd]

Conservation of momentum. The melting ice cannot push itself in any direction to displace the center of mass even if it rearranges itself. Put yourself in free fall with the ice-tray system and it should be clear.

 

[Stonebridge]

What you're forgetting is that the diameter of the sphere will be larger than the size of the original ice block.

It's irrelevant as regards where the centre of mass is.
Can you explain why it should matter.

 

[Borek]

*I am ignoring the minute gravitational attraction between the tray and the water and the possible adhesion of water to the tray. If this is meant to be included, the question get a whole lot trickier!

Absolutely nothing tricky here. Question is about the position of the center of the mass of whole system (that is, ice plus tray). No matter what happens to water (what shape it takes), answer is exactly the same.

 

[Stonebridge]

Question is about the position of the center of the mass of whole system (that is, ice plus tray). No matter what happens to water (what shape it takes), answer is exactly the same.

Which is what I said.

But as a general principle, you are correct when you say that in the absence of any external force, the c of m of the tray-water system will not move.

What would be tricky would be the final shape of the water. I am aware that the question was not asking about this, but the OP was concerned that this might have an effect.

 

[hamster143]

It's irrelevant as regards where the centre of mass is.
Can you explain why it should matter.

Centre of mass of the sphere, or centre of mass of the (sphere+tray) system?

Assuming that you mean centre of mass of sphere+tray:

If the whole system is suspended in the center of the room, I might be convinced that it is irrelevant.

If the tray is standing on the table which is standing on the floor of the room, or the tray is standing directly on the floor, it's not.

 

[dharavsolanki]

Hi Guys. Thanks for the energetic discussion. NOW I get the point of considering the whole system - to avoid more complex situations.

However, the point here is - "How is the centre of mass maintained at the same position if the water changes its shape? Also, what shape does water take?"

Plus, what if the system is kept over an altar connected rigidly to the gravity free hall? Will there be any normal force from the altar?

 

[Stonebridge]

Centre of mass of the sphere, or centre of mass of the (sphere+tray) system?

Assuming that you mean centre of mass of sphere+tray:

If the whole system is suspended in the center of the room, I might be convinced that it is irrelevant.

If the tray is standing on the table which is standing on the floor of the room, or the tray is standing directly on the floor, it's not.

I think we might be in danger of talking at cross purposes here.
In the case of A cube of ice turning into a sphere of water - if it doesn't get drawn to the tray by adhesion - the volume of the sphere, or the change in volume between ice and water, is of no relevance to the centre of mass question. The centre of mass of the sphere will be at the same place as the cube's was in the absence of any external force. The centre of mass of the tray-water system will also not change.
The centre of mass of the tray-cube system would be just above the surface of the tray, just inside the cube if the cube is in the centre of the tray and the tray is uniform. (It depends on the relative mass of the two.)
If the melting water adheres to the tray, there is still no change in the centre of mass of the tray-water system. However, as the water will, presumably, form a thin layer, and if the layer is so thin that it lies below the position where the c of m was originally, the tray would move slightly towards the water and the water towards the tray. The result being that the new position of the tray, with thin layer, would include the c of m within it.
The force that would effect this movement of the tray would be the force of adhesion between the tray and water.
In a room with no gravity, by the way, how do you define "standing on the floor" or "standing on a table". These are terms that rely on a gravitation force to give them meaning.
Anyway, the conclusion is, whatever the fine details, the c of m of the tray-water/ice system does not change its position in the absence of any external forces.

 

[Stonebridge]

Hi Guys. Thanks for the energetic discussion. NOW I get the point of considering the whole system - to avoid more complex situations.

However, the point here is - "How is the centre of mass maintained at the same position if the water changes its shape? Also, what shape does water take?"

Plus, what if the system is kept over an altar connected rigidly to the gravity free hall? Will there be any normal force from the altar?

How? To move the centre of mass you need an external resultant force. No force? No change. Newton's Laws.

Water drops in a gravity free environment form into spheres. This is because of the surface tension/cohesion forces within the water.
If the water in contact with the tray is subject to adhesive forces (to the tray) that are stronger than the cohesive forces between water and water, then the water will stick to the tray and form a thin layer on it. However, I guess that the cohesive forces in the water will tend to produce a circular pool. It depends how big the tray is and how much water there was!
There would be no "normal" force in a gravity free environment, if by normal reaction you mean the reaction force on a surface caused the weight of an object pushing down.
There is no "down"!

 

[dharavsolanki]

Awesome replies. Too good a discussion. However noob like this might sound, but you guys rock.

:)

 

[YellowTaxi]

Which is what I said.

What would be tricky would be the final shape of the water. I am aware that the question was not asking about this, but the OP was concerned that this might have an effect.

shape can become the shape of bridgett bardot or the eifel tower and the centre of mass will always be in the same place as where it started.

 

[dharavsolanki]

shape can become the shape of bridgett bardot or the eifel tower and the centre of mass will always be in the same place as where it started.

Let me rephrase the question. The centre of mass will always be in the same place as where it started, but what will be the shape of the water?

 

[Stonebridge]

Let me rephrase the question. The centre of mass will always be in the same place as where it started, but what will be the shape of the water?

If it doesn't stick to the tray - a perfect sphere floating very near the surface of the tray. If it does stick, a pool of water on the surface the tray. The appearance of the pool will depend on
- the shape of the tray
- the depth of the tray
- the size of the tray
- the amount of water
My guess is that with a tray that is large enough and deep enough, the pool will be circular.

 

[Borek]

The appearance of the pool will depend on
- the shape of the tray
- the depth of the tray
- the size of the tray
- the amount of water

- wettability of the tray material

 

[hamster143]

Assuming no "wettability"/adhesion of the tray material (let's say it's oiled or something), the ice will melt and surface tension will try to make it into a sphere. Since the radius of that sphere would be greater than 1/2 the size of the cube, that will mean that the sphere will push against the tray, and (3rd Newton's law) the tray will push against the sphere.

This is where it becomes important whether the tray is attached to anything. If the tray is standing on the table, forces between the tray and the sphere would cause the c of m of the sphere to shift, and the tray would stay put, and thus the c of m of the combined system would shift as well.

If the tray is suspended in the air, c of m's of water and of tray would move in opposite directions till water is allowed to assume spherical shape.

But that's not all! During this melting process, forces between two constituents will impart some momentum to both of them. Since that momentum does not go anywhere, the sphere and the tray would continue moving away from each other, till they hit the ceiling and the floor of the room, respectively.

 

[Stonebridge]

I agree with that, Hamster.
I may have misunderstood your earlier post and overlooked the fact that the water sphere formed by the melted ice block - assuming no sticking to the tray - would not be able to form "perfectly" because its radius would be greater than half the side of the cube. Yes, the surface tension forces would attempt to create the sphere and push on the tray everso slightly. The result would be as you say. The OP says that it is floating freely so the two would drift apart: the c of m of the water-tray system staying at the same place.
It would be interesting to predict what happens subsequently to the sphere as it floats across the room. It would no doubt oscillate in some way as it distorts and stretches. I'm not even going to think about that problem.
And then what happens to the tray? Depends on whether the force from the water acts through its c of m or is slightly off centre? Hmm.
Interesting question and a lot of physics in there if you look closely at it. (Reminds himself to look more closely in future!)

 

[dE_logics]

IMO, the geometry of the structure will change...that means the center of gravity too will.

The fluid will contract (or will it?...sorta forgot), thus more mass will be concentrated towards the lower part of the tray lowering the center of gravity.

 

[dharavsolanki]

Excellent discussion!

Thank God I asked this question!

Great reading!

@Stonebridge
Can you explain why you think the sphere should oscillate?

 

[Stonebridge]

Excellent discussion!

Thank God I asked this question!

Great reading!

@Stonebridge
Can you explain why you think the sphere should oscillate?

Maybe something like this