A railway tank wagon with its 2m diameter and 6m long horizontal cylindrical body, half full of petrol is driven around a curve of radius 100m, at a speed of 8.33 m/s. The curve runs smoothly into a straight track and the train maintains a constant speed. Find the angular amplitude and frequency of subsequent oscillation of the petrol due to this change of motion. Neglect viscosity and consider petrol as a solid semi cylinder sliding inside the tank. Given: tan^–1 (0.07) $\approx$ 4º

$\begin{array}{l}\mathrm{V}=\mathrm{\omega R}\\ \therefore \mathrm{\omega }=\frac{8.33}{100}=\frac{1}{12}\mathrm{rad}/\mathrm{s}\end{array}$
Let m = mass of petrol
Moment of inertial of petrol about an axis through O is $\mathrm{I}=\frac{1}{2}{\mathrm{mr}}^2$
where 2r = 2m 2 → r = 1m
When the wagon is driven on curved track $\mathrm{\theta }$ is such that 
$\mathrm{mgxsin}\mathrm{\theta }={\mathrm{m\omega }}^2\mathrm{xcos}\mathrm{\theta }\left[\text{ where }\mathrm{x}=\frac{4\mathrm{r}}{3\mathrm{\pi }}\right]$
$\tan \mathrm{\theta }=\frac{{\mathrm{\omega }}^2\mathrm{R}}{\mathrm{g}}={\left(\frac{1}{12}\right)}^2\times \frac{100}{9.8}=0.07$
$\therefore \mathrm{\theta }={\tan }^{-1}(0.07)=4^{\circ }$
When the wagon enters the straight segment of track, liquid oscillates about an axis through O with an amplitude of 4º.

$\mathrm{I\alpha }=-\mathrm{mgxsin}\mathrm{\theta }$
$\Rightarrow \frac{1}{2}{\mathrm{mr}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\mathrm{mgxsin}\mathrm{\theta }$
Since, $\mathrm{\theta }$ is small
$\begin{array}{l}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\frac{2\mathrm{g}}{{\mathrm{r}}^2}\left(\frac{4\mathrm{r}}{3\mathrm{\pi }}\right)\mathrm{\theta }\left[\because \mathrm{x}=\frac{4\mathrm{r}}{3\mathrm{\pi }}\right]\\ =-\frac{8\mathrm{g}}{3\mathrm{\pi r}}\mathrm{\theta }\\ \therefore \mathrm{\omega }=\sqrt{\frac{8\mathrm{g}}{3\mathrm{\pi r}}};\Rightarrow \mathrm{f}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{8\mathrm{g}}{3\mathrm{\pi r}}}=\frac{1}{2\times 3.14}\sqrt{\frac{8\times 9.8}{3\times 3.14\times 1}}\\ =\frac{2.88}{2\times 3.14}=0.46\mathrm{Hz}\end{array}$