Prompt:
I put an ordinary rock into a glass of water. I then turn the glass of water upside down, do a little dance, and then turn the glass right side up again. Where is the rock now?

Answer:
So I have this scenario here: I put an ordinary rock into a glass of water, then I turn the glass upside down, do a little dance, and finally turn it right side up again. The question is, where is the rock now?

First, I need to visualize what's happening step by step. Let's start with the initial setup. I have a glass filled with water, and I place a rock into it. So, the rock is submerged in the water, sitting at the bottom of the glass, assuming it's denser than water and sinks.

Now, I turn the glass upside down. When I flip it, the water is going to try to spill out because of gravity, but since the glass is inverted, the water should stay inside the glass, right? But wait, if I flip it quickly, some water might spill, but mostly, the water should remain in the glass, especially if I cover the opening with my hand or something to prevent the water from pouring out.

But in this scenario, it says I turn the glass upside down and do a little dance. Hmm, doing a dance might involve moving the glass around in different directions, maybe shaking it or moving it in a circular motion. I'm not sure how that affects the water and the rock.

Let me think about the water first. When the glass is upside down, the water is held in the glass by the atmospheric pressure and the weight of the water itself. If I move the glass around while it's upside down, the water might slosh around, but it should mostly stay inside the glass, assuming I don't tilt it too much or move it too vigorously.

Now, what about the rock? The rock was at the bottom of the glass when it was right side up, which is now the top when the glass is upside down. So, when I flip the glass, the rock should now be at the top of the water column.

But, rocks are generally denser than water, so they sink in water. However, when the glass is upside down, the rock is at the top, so it's not in contact with the surface of the water anymore. Wait, no. Actually, when you flip the glass, the rock, being dense, would tend to stay at the bottom of the water column due to gravity.

But since the glass is upside down, the bottom of the water column is now at the top of the glass. So, the rock should be at the top of the water, but not necessarily outside the water because the water is contained within the glass.

Wait a minute, maybe I need to consider that when you flip the glass, the rock might fall out if it was sitting at the bottom of the glass. Let me think about that.

If the glass is filled with water and the rock is at the bottom, when you flip the glass upside down, the rock could potentially fall out of the glass if it wasn't fully submerged or if it was sitting on the edge. But if the rock was fully submerged and the glass is flipped quickly, the water should hold the rock in place.

But in reality, if you flip a glass of water with an object in it, the object might stay inside due to the water's cohesion or might fall out depending on how you flip it.

Perhaps I should consider that the rock stays inside the glass, suspended in the water, even when the glass is upside down, assuming that no water spills out.

Then, after doing a little dance, which could involve moving the glass around, maybe shaking it, the rock might move around within the water, but it should still remain inside the glass, as long as no water spills out.

Finally, when I turn the glass right side up again, the rock, being denser than water, should sink to the bottom of the glass again.

Wait, but if water spills out during any of these actions, things might be different. For instance, if some water spills out when I flip the glass or during the dance, the water level might decrease, and the rock could be partially or fully exposed.

But the scenario doesn't specify any water spilling out, so perhaps I should assume that the glass is handled in such a way that the water remains inside.

Therefore, after turning the glass right side up again, the rock should be back at the bottom of the glass, submerged in the water.

But let's consider another possibility. Maybe when I flip the glass upside down, the rock falls out onto the surface it's being flipped onto. For example, if I'm flipping the glass onto a table, the rock could drop onto the table.

Is that a possibility? Well, it depends on how I flip the glass. If I flip it quickly and the rock is stuck to the bottom of the glass due to the water's adhesion, it might stay inside. But if the rock is loose, it could fall out.

Maybe to avoid that, I should imagine that the rock is stuck to the glass by the water, so it doesn't fall out when I flip the glass.

Alternatively, perhaps the rock is small enough to be held by the water's surface tension, so it stays inside the glass even when it's upside down.

But that seems unlikely for an ordinary rock, which is probably heavier than the water's surface tension can hold.

Wait, maybe I need to think about the size of the rock. If it's a very small pebble, it might be held by the water's surface tension, but a larger rock would likely fall out.

Since the scenario refers to an "ordinary rock," I'll assume it's of a size that wouldn't be held by surface tension. Therefore, when I flip the glass upside down, the rock falls out onto the surface below.

Then, during the dance, maybe the rock moves around on that surface.

Finally, when I turn the glass right side up again, the rock is still on the surface, separate from the glass.

But this seems a bit simplistic. Maybe there's more to it.

Alternatively, perhaps the rock stays inside the glass throughout the entire process, suspended in the water, and ends up back at the bottom when the glass is right side up again.

Or maybe it sticks to the bottom of the glass when it's upside down and then falls back into the water when I turn it right side up.

I'm getting a bit confused here. Maybe I should think about real-world experiments.

I recall that if you fill a glass with water and turn it upside down, while keeping the glass inverted, the water stays in the glass due to atmospheric pressure, provided that the opening is sealed somehow.

But in this case, there's a rock in the water. If the rock is submerged and the glass is flipped quickly, the rock might stay inside the glass, suspended in the water.

If I then move the glass around while it's upside down, the rock could shift positions within the water.

Then, when I turn the glass right side up again, the rock should sink back to the bottom.

But perhaps some water spills out during these maneuvers, which could affect the position of the rock.

Alternatively, if the rock falls out when I flip the glass, it would be outside the glass, on the surface below.

Given that, I think the final position of the rock depends on whether it stays inside the glass or falls out when I flip it.

To make an educated guess, I'll assume that the rock stays inside the glass throughout the process and ends up back at the bottom when I turn the glass right side up again.

Therefore, the rock is now at the bottom of the glass, submerged in the water.

But to be thorough, I should consider other possibilities.

For example, if the rock falls out when I flip the glass, it would be on the surface below, and when I turn the glass back up, it remains there.

Alternatively, if the rock gets stuck to the bottom of the glass when it's upside down and then falls back into the glass when I turn it up, it would again end up at the bottom.

Alternatively, if water spills out during the process, the rock could be partially exposed or even fall out of the glass.

But without more specific information about how the glass is manipulated, it's hard to determine exactly what happens.

Given that, I think the most likely scenario is that the rock ends up back at the bottom of the glass, submerged in the water.

So, that's where the rock is now.

**Final Answer**

$$ \boxed{\text{The rock is at the bottom of the glass, submerged in the water.}} $$