Oscillation of a boat in still water (Metacenter and Center of Mass)

[Adesh]

Let’s say we have a boat whose longitudinal axis is the y-axis (which goes into the screen in the figure below) standing upright in a still water .


$S$ is the Center of Mass of the boat and $C$ is the Center of Mass of the displaced water.On $S$ lies the force $\mathbf W$ (the weight of the boat) and on $C$ acts the buoyant force and they exactly cancel each other (Archimedes’ Principle). Hence, the boat remains in the equilibrium.

It is a proven fact that the buoyant force acts on the Center of Mass of displaced water.

Now, imagine we rotate our boat (y-axis as the axis of rotation) towards right by some infinitesimal angle $\theta$ (with vertical), the new Center of Mass of displaced water is $C’$. Rotate the boat (from equilibrium position) towards left by the same infinitesimal angle, the new Center of Mass of displaced water is $C”$. Let $\mathscr{M}$ denote the Center of curvature of the curve $C”~ C ~ C’$, let’s call it metacenter.


(I would like to apologise for my strange looking $\mathscr{M}$ and hand made diagrams, due to lockdown I’m away from my home and hence have no access to my Mac)
Since, buoyant force $\mathbf B$ always acts on the current center of mass of displaced water, therefore, in the figure buoyant force acts on $C’$ vertically above. Now, $\mathbf W$ and $\mathbf B$ would acts as couple and hence produce a rotation. As far as I can think, the boat will rotate about $S$ and hence $\mathbf W$ will have no contribution in rotation of our boat, it’s our buoyant force $\mathbf B$ which would produce an anti-clockwise rotation (the vector $\vec{r}$ from $S$ to the line of action of $\mathbf B$, and hence by right hand rule a torque will be produced which points into the paper).

Now, my doubt is that it is written that if metacenter $\mathscr{M}$ were to lie below $S$ then the net moment produced would tend to increase the disturbance (that is it would tend to increase the angle with respect to the vertical, that is it would cause a rotation to the right) and I cannot understand how and why.

How can $\mathscr{M}$ ever affect the direction of rotation? I want to why did we designated a letter to the Center of curvature of curve $C”~C~C’$, what’s the significance of $\mathscr{M}$?

 

[willem2]

The problem with your analysis, is that you rotate the boat around the center of mass. The hull of the ship as a whole will move, and that will actually move the center of buoyancy in the opposite direction than where you think it moves.
It's much easier to rotate the ship around M. The center of buoyancy will always be directly under it, and it then only matters where S is.

 

[Adesh]

The problem with your analysis, is that you rotate the boat around the center of mass.

Are you talking about the initial disturbance that was caused to the boat? Yes I think $S$ will have a different location when the boat is rotated.

The hull of the ship as a whole will move, and that will actually move the center of buoyancy in the opposite direction than where you think it moves.

I didn’t get this part. What is center of buoyancy? Is it the same thing as Center of Mass of the displaced water? And if it is then how would it move ?

 

[Adesh]

I’m unable to understand how a rotation will be caused by the weight of the boat and the buoyant force.

 

[A.T.]

Now, my doubt is that it is written that if metacenter $\mathscr{M}$ were to lie below $S$ then the net moment produced would tend to increase the disturbance (that is it would tend to increase the angle with respect to the vertical, that is it would cause a rotation to the right) and I cannot understand how and why.

Do you understand how the metacenter is defined?
https://en.wikipedia.org/wiki/Metacentric_height#Metacentre

For the metacenter to be below the center of mass, the center of buoyancy would have to be on the other side of the center of mass (towards the elevated side of the boat). So of course it would act to increase the tilt.

 

[anorlunda]

What is center of buoyancy? Is it the same thing as Center of Mass of the displaced water?

Yes.

That single point is called the center of buoyancy. The center of buoyancy is the point where if you were to take all of the displaced fluid and hold it by that point it would remain perfectly balanced, assuming you could hold a fluid in a fixed shape. This point is also called the center of mass. The center of buoyancy for an object is the center of mass for the fluid it displaces.

It moves as the hull tips, fore/aft, side to side. That is because the shape of the hull is not vertical. Double-ended boats have less buoyancy in the stern than do boats with a squared transom in the stern. Buoyancy near the bow and stern affect the pitch stability of the hull.

If all boats were spherical, the analysis would be much easier.

 

[Adesh]

For the metacenter to be below the center of mass, the center of buoyancy would have to be on the other side of the center of mass (towards the elevated side of the boat). So of course it would act to increase the tilt.

I tried to make the diagram but didn’t get center of buoyancy towards the elevated side of the boat.


I have made $\mathscr{M}$ below $S$ (the Center of Mass of the boat) and from $\mathscr{M}$ I dropped a vertical, so by the definition of “metacenter” the new center of buoyancy will lie on the vertical line passing through the $\mathscr{M}$ . How would I get center of buoyancy on the elevated side of the boat?

 

[A.T.]

I have made $\mathscr{M}$ below $S$ (the Center of Mass of the boat) and from $\mathscr{M}$ I dropped a vertical, so by the definition of “metacenter” the new center of buoyancy will lie on the vertical line passing through the $\mathscr{M}$ .

Right, and in which direction will the moment of the buoyant force around the center of mass act then?

 

[Adesh]

Right, and in which direction will the moment of the buoyant force around the center of mass act then?

I think the buoyant force will act on the Center of Buoyancy, since center of buoyancy lies on the vertical line (that is to the right of $C$) the moment created by the buoyant force will stabilise the disturbed position instead of increasing it.

 

[A.T.]

I think the buoyant force will act on the Center of Buoyancy, since center of buoyancy lies on the vertical line (that is to the right of $C$) the moment created by the buoyant force will stabilise the disturbed position instead of increasing it.

Draw the new center of buoyancy, the force of buoyancy and its moment around the center of mass.

 

[Adesh]

Draw the new center of buoyancy, the force of buoyancy and its moment around the center of mass.

Here is the diagram with marked “new” center of buoyancy and buoyant force $\mathbf B$.


Yes, I can see that the line of action of buoyant force $\mathbf B$ lies to the left of $S$ and hence by right hand rule a clockwise rotation will be produced, increasing the disturbance. Thank you so much sir.

But please help me once more, I’m not finding it intuitive how the force on the depressed side of the boat can ever cause a clockwise rotation. As far as I can think the buoyant force will act on the depressed side of the boat upwards and will try to make it upright. Help me is seeing how it would disturb the equilibrium even more.

 

[A.T.]

I’m not finding it intuitive how the force on the depressed side of the boat can ever cause a clockwise rotation.

To make it obvious, consider an extreme case: A boat with a huge weight on top of the mast, so that the center of mass is very high up. Will it be stable?

 

[Adesh]

To make it obvious, consider an extreme case: A boat with a huge weight on top of the mast, so that the center of mass is very high up. Will it be stable?

I’m sorry but I seriously don’t know, I haven’t experience anything like that. But I can say from my childhood experience when I use to make paper boats, if we were to stand a long stick on my paper boat it would drown.

 

[A.T.]

I’m sorry but I seriously don’t know, I haven’t experience anything like that. But I can say from my childhood experience when I use to make paper boats, if we were to stand a long stick on my paper boat it would drown.

Well, maybe you should build some boats again. It's really not that difficult to build one that flips over.

 

[Adesh]

To make it obvious, consider an extreme case: A boat with a huge weight on top of the mast, so that the center of mass is very high up. Will it be stable?

I think a little (very little) disturbance can ruin everything. It would rotate fully to the side where it gets disturbed .

 

[A.T.]

I think a little (very little) disturbance can ruin everything. It would rotate fully to the side where it gets disturbed.

Yes, and if you draw a higher center of mass in your last diagram, you can see why. What happens to the moment arm of the buoyant force around the center of mass?

 

[Adesh]

What happens to the moment arm of the buoyant force around the center of ma

The moment arm increases in length as $S$ goes higher and higher.

 

[A.T.]

The moment arm increases in length as $S$ goes higher and higher.

Yes, and so does the moment that flips the boat over.

 

[Adesh]

Yes, and so does the moment that flips the boat over.

My problem is that I’m unable to imagine how would a force on the depressed side of the boat can ever cause more depression ? In terms of torque I can see that the rotation will be clockwise, but if an upward force acts on the depressed side the depressed side must rise up and an anti clockwise rotation will be there.

 

[A.T.]

My problem is that I’m unable to imagine how would a force on the depressed side of the boat can ever cause more depression ?

It doesn't cause more depression, just rotation.

 

[Adesh]

It doesn't cause more depression, just rotation.

Yes, but it causes rotation so as to cause the depressed side to get even more submerged. But from my intuition I can think that any upward force on the depressed side would cause it to lift up.

 

[A.T.]

Yes, but it causes rotation so as to cause the depressed side to get even more submerged.

The displaced volume doesn't change, the boat just rotates within the displaced volume.

 

[Adesh]

I need to state my problem clearly (although A.T. has solved my actual problem).
Consider this image:

In the image I have divided the boat into two parts, $1$ and $2$. Ki the situation shown, the weight of the boat at $S$ is balanced by the buoyant force at $C$.

Now, let’s tilt the boat towards the side. $2$, that is side $1$ would go up and $2$ will come down.

Now, since metacenter lies below $S$ therefore buoyant force will cause a clockwise rotation and consequently side $2$ will go further down.

But from intuition, if the buoyant force were to act on the vertical line passing through $\mathscr{M}$ it would surely act on the surface of our hull on the side $2$ (I have made a big black arrow pointing upwards at bottom of the hull) and should make the side $2$ to rise up and hence an anti-clockwise rotation. Why is there a discrepancy ?

I think I have made some mistake in my diagrams. Help me out through this.

 

[A.T.]

But from intuition, if the buoyant force were to act on the vertical line passing through $\mathscr{M}$ it would surely act on the surface of our hull on the side $2$ (I have made a big black arrow pointing upwards at bottom of the hull) and should make the side $2$ to rise up and hence an anti-clockwise rotation. Why is there a discrepancy ?

You are conflating rising up and rotating:
- The rising up doesn't happen, because the weight balances the buoyant force.
- As for the rotation: If you push on something upwards, at a point left of its center of mass, then it tilts to the right. How you have labeled that part of the object is irrelevant.

 

[Adesh]

Thank you so much. You have cleared all my doubts.

As for the rotation: If you push on something upwards, at a point left of its center of mass, then it tilts to the right. How you have labeled that part of the object is irrelevant.

That part where you have said the applied force lies to the left of cente of mass killled (😁) all my doubts.