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(Figure). The rod is free to rotate in a vertical plane about a horizontal axis passing through $P$ . The density $d_{1}$ of the material of the rod is smaller than the entity $d_{2}$ of the liquid. the rod is displaced by a small angle $θ$ from its equilibrium position and then released. If the angular frequency in terms of the given parameters is $w = \sqrt{\frac{x g}{2 L} (\frac{d_{2} - d_{1}}{d_{1}})}$ . Find $x$

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Solution

Let $S$ be the cross sectional area of the rod.
In the displaced position as shown in the figure, weight W and upthrust $F_{B}$ both pass through the centre of gravity.
$W = v_{r} d_{r} g = S L d_{1} g$
$F_{B} = v_{l} d_{l} g = S L d_{2} g$
Given $d_{1} < d_{2}$
$∴ W < F_{B}$
Therefore, net force acting at G will be
$F = F_{B} - W = (S L g) (d_{2} - d_{1})$ upwards. Restoring torque of this force about point P is
$τ = F$ x $r_{⊥} = (S L g) (d_{2} - d_{1}) (Q G)$
$∴ τ = - (S L g) (d_{2} - d_{1}) (\frac{L}{2} sin θ)$
For small values of $θ$ , $sin θ = θ$
$∴ τ = - (S L g) (d_{2} - d_{1}) (\frac{L}{2} θ)$
Hence, motion of the rod will be simple harmonic.
But $τ = I \frac{d^{2} θ}{d t^{2}} = - (S L g) (d_{2} - d_{1}) (\frac{L}{2} θ)$ ....(1) where I is the moment of inertia of the rod about point P.
$I = \frac{M L^{2}}{3} = \frac{(S L d_{1}) L^{2}}{3}$
Substituting I in eqn(1) we get
$\frac{d^{2} θ}{d t^{2}} = - (\frac{3}{2} g \frac{(d_{2} - d_{1})}{d_{1} L}) θ$
Comparing this equation with standard differential equation of SHM, $\frac{d^{2} θ}{d t^{2}} = - ω^{2} θ$
we get,
$ω = \sqrt{\frac{3 g}{2 L} (\frac{d_{2} - d_{1}}{d_{1}})}$

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