Question

A thin rod of length L and area of cross section S is pivoted at its lowest point P inside a stationary, homogeneous and non viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density $d_1$ of the rod is smaller than the density $d_2$ of the liquid. The rod is displaced by a small angle $\theta$ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.

Answer 1

If the rod is displaced through an angle a from its equilibrium position i.e., vertical position then it will be under the thrust U and its own weight mg. There will be a net torque, which will try to restore the rod to its equilibrium position.
Weight of the rod acting downward mg = SL$d_1$g
Buoyant force acting upwards U = SL$d_2$g
Net force acting on the rod in the upward direction.
= SL($d_2$ - $d_1$)g .
Where the area of cross-section of the rod is S
Net.restoring torque $\tau$ = SL ($d_2$ - $d_1$)g x L/2 $\alpha$ ($sin\alpha =\alpha$)
If a is in clockwise direction, then t will be in anticlockwise direction. Thus
$\tau$ = -1/2 $S{L}^2$/($d_2$ - $d_1$)g$\alpha$
$\tau$ 1 $\alpha$ = ($M{L}^2$/3) $d_2$$\alpha$/$d{t}^2$ = ($SL{d}_1\times L^2$/3) $d_2$$\alpha$/$d{t}^2$
$d_2$$\alpha$/$d{t}^2$ = -3g/2L ($d_2$ - $d_1$/$d_1$) $\alpha$
This is the equation of angular S.H.M.
For which the angular frequency co is given by
${\omega }^2$ = $\frac{3g}{2L}$ ($d_2$ - $d_1$/$d_1$)

The angular frequency $\omega$ =$\sqrt{\frac{3g}{2L}(d_2-d_1)/d_1}$

Therefore time period T = $2\pi$/$\omega$ = 2$\pi$ $\sqrt{2L/3g(d^1/d_2-d_1)}$