Question

Two identical rods each of mass m and length L,are rigidly joined and suspended in a vertical plane so as to oscillate freely about an axis normal to the plane of paper passing through 'S' (point of suspension).Find the time period of such small oscillations.

Answer 1

The movement of Inertia of the rod about the center are

$\frac{1}{3}m{L}^2$ and $\frac{1}{12}m{L}^2$ respectively.

$\Rightarrow$ using the parallel axis theorem for the lower rod,

The MI of the liquid combination is

$\begin{array}{c}I=\frac{1}{3}m{L}^2+\left(\frac{1}{12}m{L}^2+m{L}^2\right)\\ =\frac{17}{12}m{L}^2\end{array}$

The center of mass of the combination is at distance $y$ below $S$

such that

$\begin{array}{c}2my=\frac{1}{2}mL+mL=\frac{3}{2}mL\\ \Rightarrow y=\frac{3}{4}L\end{array}$

When the angular displacement from the vertical is $\theta$

measured anticlockwise then the torque is

$\tau =-2mgy\sin \theta$

The equation of motion is

$\begin{array}{c}\frac{17}{12}m{L}^2\ddot{\theta }=-2mg\left(\frac{3}{4}L\right)\sin \theta \\ \ddot{\theta }+\frac{18g}{17L}\sin \theta =0\end{array}$

For small oscillation $\left(\sin \theta \approx \theta \right)$

the angular frequency is $w\approx \sqrt{\frac{18g}{17L}}$