Pendulums — Question 6

Question 6

Buoyancy weakens a submerged pendulum without removing its inertia

A small spherical bob of density $\rho_b$ and volume $V$ hangs from a massless string of length $\ell$ measured to its center. It is fully submerged in uniform liquid of density $\rho_f<\rho_b$. Neglect drag, added liquid inertia and string buoyancy. Treat the bob as a point mass for rotation about the support.

1. Derive the small-angle period.

2. Find its ratio to the air period when $\rho_b=2500$ and $\rho_f=1000\,\mathrm{kg/m^3}$. Explain why replacing the bob mass by its apparent mass everywhere would be wrong.

Question 6 - Solution

Buoyancy weakens a submerged pendulum without removing its inertia

(a) Derivation and calculation

Ignoring fluid added inertia and drag,$$m=\rho_bV,\qquad W-B=(\rho_b-\rho_f)Vg.$$$$\rho_bV\ell^2\ddot\theta=-(\rho_b-\rho_f)Vg\ell\sin\theta.$$$$\omega^2=\frac g\ell\left(1-\frac{\rho_f}{\rho_b}\right),\quad\boxed{T=2\pi\sqrt{\frac{\ell}{g(1-\rho_f/\rho_b)}}}.$$

(b) Result and interpretation

$$\frac{T_{\mathrm{fluid}}}{T_{\mathrm{vacuum}}}=\frac1{\sqrt{1-\rho_f/\rho_b}}=\frac1{\sqrt{1-1000/2500}}=\boxed{1.29099}.$$Buoyancy reduces restoring torque while the specified bob inertia remains $\rho_bV\ell^2$. Replacing the inertial mass by a buoyancy-reduced mass would cancel a real effect.

Physical check

For a downward hanging equilibrium this model requires $\rho_b>\rho_f$; neutral buoyancy removes the gravitational restoring torque.