Stability of a floating cylinder

[etotheipi]

Homework Statement

Discussion of stability of a floating cylinder, spun off from previous homework thread (https://www.physicsforums.com/threads/depth-to-which-the-sphere-is-submerged.992438/post-6380911)

Relevant Equations

N/A

A tougher set of problems concerns stable 'postures'. For a uniform solid cylinder radius r, length h, in which orientations can it float?

Now that is an evil question! The more interesting case looks like when the cylinder is upright, so that is what I focused on. I defined two coordinates; $\theta$, the angular displacement of the cylinder from the vertical and $d$, the distance between the point on the cylindrical axis which is currently at the water level and the point on the cylindrical axis which is on the lower face. Let the radius and height of the cylinder be $r$ and $h$ respectively.

The volume of displaced water is enclosed by a cylindrical segment, for which I found some formulae here pertaining to the centroid and volume. In terms of a body fixed coordinate system with origin at the centre of the lower face of the cylinder with the $\hat{z}'$ axis pointing along the cylindrical axis and the $\hat{x}'$ axis parallel to the lower face, the centre of buoyancy is at$$x'_{b} = - \frac{r[(d + r\tan{\theta}) - (d - r\tan{\theta})]}{4[(d+r\tan{\theta}) + (d-r\tan{\theta})]} = -\frac{r^2 \tan{\theta}}{4d}$$ $$z'_b = \frac{5(d-r\tan{\theta})^2 + 6(d+r\tan{\theta})(d-r\tan{\theta}) + 5(d+r\tan{\theta})^2}{32d} = \frac{4d^2 + r^2 \tan^2{\theta}}{8d}$$Also, the volume submerged is$$V = \frac{1}{2}\pi r^2 [(d + r\tan{\theta}) + (d-r\tan{\theta})] = \pi r^2 d$$which implies that the buoyant force is$$B = \pi r^2 d \rho_w g$$Now we want to find the torque of the buoyant force about the centre of mass, and to do this we need to find the horizontal distance between the line of action of the buoyant force and the centre of mass. I converted the rotated, body fixed coordinates of the centre of buoyancy to coordinates w.r.t. a basis aligned with the water and the initial vertical, with origin at the centre of the lower face. The horizontal distance is then$$\begin{align*} \Delta x = x_b -x_{cm} &= - \left[ \left(\frac{r^2 \tan{\theta}}{4d}\right)\cos{\theta} + \left(\frac{4d^2 + r^2 \tan^2{\theta}}{8d}\right) \sin{\theta} \right] - \left (-d\sin{\theta} \right) \\
&= \frac{1}{2}d\sin{\theta} \left ( 1 - \frac{r^2(2+ \tan^2{\theta})}{4d^2} \right)
\end{align*}$$Then the torque of the buoyant force about the centre of mass is$$\tau = B\Delta x = \frac{\pi r^2 d^2 \rho_w g \sin{\theta}}{2}\left ( 1 - \frac{r^2(2+ \tan^2{\theta})}{4d^2} \right)$$That's I think sufficient to deduce what we need, but for clarity it might help to explicitly write out how it relates to $\ddot{\theta}$:$$2d^2 \rho_w g \sin{\theta} \left ( 1 - \frac{r^2(2 + \tan^2{\theta})}{4d^2} \right) = h \rho (r^2 + \frac{1}{3}h^2)\ddot{\theta}$$Take the small angle approximation, and disregard the $\tan^2{\theta}$ term,$$\frac{2d^2 \rho_w g (1- \frac{r^2}{2d^2})}{h\rho (r^2 + \frac{1}{3}h^2)} \theta = \ddot{\theta}$$From this we cannot deduce SHM, because $d = d(\theta)$ is still some as of yet undetermined function of $\theta$, however it seems reasonable to assert that in order for the cylinder to be stable to small disturbances the LHS must be negative; if we assume that $d'(\theta)|_{\theta = 0}$ is also fairly small, then $d \approx d_0$ for small disturbances and $$1 - \frac{2r^2}{4d_0^2} < 0$$ $$\sqrt{2} d_0 < r$$At first glance, that seems along the right lines considering that we'd expect short fat cylinders to be more stable than tall thin cylinders. I hope I haven't made any major errors!

 

[haruspex]

Homework Statement:: Discussion of stability of a floating cylinder, spun off from previous homework thread.
Relevant Equations:: N/A

Now that is an evil question! The more interesting case looks like when the cylinder is upright, so that is what I focused on. I defined two coordinates; $\theta$, the angular displacement of the cylinder from the vertical and $d$, the distance between the point on the cylindrical axis which is currently at the water level and the point on the cylindrical axis which is on the lower face. Let the radius and height of the cylinder be $r$ and $h$ respectively.

The volume of displaced water is enclosed by a cylindrical segment, for which I found some formulae here pertaining to the centroid and volume. In terms of a body fixed coordinate system with origin at the centre of the lower face of the cylinder with the $\hat{z}'$ axis pointing along the cylindrical axis and the $\hat{x}'$ axis parallel to the lower face, the centre of buoyancy is at$$x'_{b} = - \frac{r[(d + r\tan{\theta}) - (d - r\tan{\theta})]}{4[(d+r\tan{\theta}) + (d-r\tan{\theta})]} = -\frac{r^2 \tan{\theta}}{4d}$$ $$z'_b = \frac{5(d-r\tan{\theta})^2 + 6(d+r\tan{\theta})(d-r\tan{\theta}) + 5(d+r\tan{\theta})^2}{32d} = \frac{4d^2 + r^2 \tan^2{\theta}}{8d}$$Also, the volume submerged is$$V = \frac{1}{2}\pi r^2 [(d + r\tan{\theta}) + (d-r\tan{\theta})] = \pi r^2 d$$which implies that the buoyant force is$$B = \pi r^2 d \rho_w g$$Now we want to find the torque of the buoyant force about the centre of mass, and to do this we need to find the horizontal distance between the line of action of the buoyant force and the centre of mass. I converted the rotated, body fixed coordinates of the centre of buoyancy to coordinates w.r.t. a basis aligned with the water and the initial vertical, with origin at the centre of the lower face. The horizontal distance is then$$\begin{align*} \Delta x = x_b -x_{cm} &= - \left[ \left(\frac{r^2 \tan{\theta}}{4d}\right)\cos{\theta} + \left(\frac{4d^2 + r^2 \tan^2{\theta}}{8d}\right) \sin{\theta} \right] - \left (-d\sin{\theta} \right) \\
&= \frac{1}{2}d\sin{\theta} \left ( 1 - \frac{r^2(2+ \tan^2{\theta})}{4d^2} \right)
\end{align*}$$Then the torque of the buoyant force about the centre of mass is$$\tau = B\Delta x = \frac{\pi r^2 d^2 \rho_w g \sin{\theta}}{2}\left ( 1 - \frac{r^2(2+ \tan^2{\theta})}{4d^2} \right)$$That's I think sufficient to deduce what we need, but for clarity it might help to explicitly write out how it relates to $\ddot{\theta}$:$$2d^2 \rho_w g \sin{\theta} \left ( 1 - \frac{r^2(2 + \tan^2{\theta})}{4d^2} \right) = h \rho (r^2 + \frac{1}{3}h^2)\ddot{\theta}$$Take the small angle approximation, and disregard the $\tan^2{\theta}$ term,$$\frac{2d^2 \rho_w g (1- \frac{r^2}{2d^2})}{h\rho (r^2 + \frac{1}{3}h^2)} \theta = \ddot{\theta}$$From this we cannot deduce SHM, because $d = d(\theta)$ is still some as of yet undetermined function of $\theta$, however it seems reasonable to assert that in order for the cylinder to be stable to small disturbances the LHS must be negative; if we assume that $d'(\theta)|_{\theta = 0}$ is also fairly small, then $d \approx d_0$ for small disturbances and $$1 - \frac{2r^2}{4d_0^2} < 0$$ $$\sqrt{2} d_0 < r$$At first glance, that seems along the right lines considering that we'd expect short fat cylinders to be more stable than tall thin cylinders. I hope I haven't made any major errors!

Yes, that looks good for the case where one end is fully submerged, the other end is clear of the water, and we want the cylinder on a vertical axis... though it would be more complete expressed using the density ratio and the overall cylinder length instead of d.
But what about stable on a tilted axis? Or one or both ends partly submerged?!
This problem gives and gives!

 

[etotheipi]

But what about stable on a tilted axis? Or one or both ends partly submerged?!
This problem gives and gives!

My instinct tells me that the only stable positions would be upright with one end fully submerged and horizontal; the analysis for the horizontal case would probably be quite similar, except we'd now expect a relation something like $h > \alpha r$ for some factor $\alpha$.

For it to be stable on a tilted axis, would we need to account for a non-uniform density distribution?

 

[Lnewqban]

Is the cylinder solid or, being a vessel, is half full with liquid?
Does that make a difference about the position the cylinder adopts while floating?

 

[etotheipi]

Is the cylinder solid or, being a vessel, is half full with liquid?
Does that make a difference about the position the cylinder adopts while floating?

I took the cylinder to be of homogenous density $\rho$. It would also be interesting to look at a hollow container partially filled with liquid; I don't think the equilibrium depths would be any different for a fixed mass of (container + water) vs a homogenous solid, but the dynamics and the stability could be very different!

 

[haruspex]

My instinct tells me that the only stable positions would be upright with one end fully submerged and horizontal

Then your childhood was deficient in time spent at the shore, observing floating logs and blocks of wood.
As the cylinder length increases, the vertical posture becomes unstable and it will float at a jaunty angle, then later shift to a more horizontal axis.
I'm not sure whether any parameter combinations (density ratio, length to diameter ratio) support more than one stable posture, or whether, once flipped onto its side, perfectly horizontal is the only option.

 

[Frodo]

A disk and a stick are both cylinders.

A uniform disk floats with its axis vertical.
A uniform stick floats with its axis horizontal.

So, as a cylinder's ratio of length to diameter changes, it will transition from one mode to the other.

Googling metacentre / metacentric height will probably help.

 

[haruspex]

A uniform disk floats with its axis vertical.
A uniform stick floats with its axis horizontal.

So, as a cylinder's ratio of length to diameter changes, it will transition from one mode to the other.

Is that based on analysis or intuition?

To see what happens for rectangular blocks, take a look at http://datagenetics.com/blog/june22016/index.html