- #1

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

haruspex said: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!

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