Question

Review: A long, cylindrical rod of radius r is weighted on one end so that it floats upright in a fluid having a density $\mathit{\rho }$. It is pushed down a distance $\mathit{x}$ from its equilibrium position and released. Show that the rod will execute simple harmonic motion if the resistive effects of the fluid are negligible, and determine the period of the oscillations.

Short answer

The period of the oscillations is$T=\frac{2}{r}\sqrt{\frac{\pi M}{\rho g}}$

Step-by-step solution

A concept:

According to Newton’s second law is given by:

$\underset{}{\overset{}{\sum F_y}}=0$

Let  L represent the length below water at equilibrium and M the tube’s mass:

According to Newton’s second law is given by:

$\underset{}{\overset{}{\sum F_y}}=0$

Gives,

$-Mg+\rho \pi r^{2}Lg=0$

Now with any excursion $x$ from equilibrium:

$-Mg+\rho \pi r^2\left(L-x\right)g=Ma$

Subtracting the equilibrium equation gives:

$$\begin{gathered} -\rho \pi r^{2}gx=Ma \\ a=-\left(\frac{\rho \pi r^{2}g}{M}\right)x \\ a=-{\omega }^{2}x \end{gathered}$$

The opposite direction and direct proportionality of $a$to $x$imply SHM with angular frequency

$$\begin{gathered} \omega =\sqrt{\frac{\rho \pi r^{2}g}{M}} \\ T=\frac{2\pi }{\omega } \\ T=\frac{2}{r}\sqrt{\frac{\pi M}{\rho g}} \end{gathered}$$