Archimedes' Principle — Question 7

Question 7

Neutral buoyancy alone does not guarantee rotational stability

A rigid body is wholly submerged in a uniform liquid and has exactly neutral buoyancy, $B=W$. In its initial vertical orientation, its center of buoyancy C lies a distance $d$ above its center of mass G. Displace it through a small angle $\theta$ from vertical about G; the surrounding fluid remains hydrostatic. Ignore added dynamic forces for the static torque question.

1. Find the buoyancy torque about G and determine whether the vertical orientation is stable.

2. Compare the cases C below G and C coincident with G. Why is equality of buoyancy and weight insufficient to decide orientation?

Question 7 - Solution

Neutral buoyancy alone does not guarantee rotational stability

(a) Derivation and calculation

Take clockwise tilt $\theta$ as positive. For C above G, the buoyancy lever arm is $d\sin\theta$:$$\boxed{\tau_G=-Bd\sin\theta\simeq-Bd\theta}.$$$$\left.\frac{d\tau_G}{d\theta}\right|_0=-Bd<0.$$The torque opposes a small tilt, so the vertical orientation is stable.

(b) Result and interpretation

$$\text{C below G:}\quad\tau_G\simeq+Bd\theta\quad\text{(unstable)};$$$$\text{C=G:}\quad\tau_G=0\quad\text{(neutral)}.$$$$B-W=0\ \not\Rightarrow\ \tau_G=0.$$Equal forces can still form a couple when their lines of action are separated.

Physical check

This is a static torque conclusion; no angular frequency is inferred without rotational inertia and fluid dynamic effects.