Question

A uniform rod of mass $m=1.5kg$ suspended by two identical threads $l=90cm$ in length was turned through a small angle about the vertical axis passing through its middle point $C$. The threads deviated in the process through an angle $\alpha ={5.0}^o$. Then the rod was released to start performing small oscillations.
Find (a) the oscillation period; (b) the rod's oscillation energy.

Answer 1

In initial state let tension in the string be T, then

$2T=mgT=7.5N$

$\left(a\right)$ Here, $Tsin\theta$ will provide restoring torque.

So, $\tau =Tsin\theta \times \frac{L}{2}+Tsin\theta \times \frac{L}{2}\tau =\frac{7.5\times 90\times 0.08}{100}=0.54$

$\tau =I\omega \omega =\frac{0.54}{\frac{m{\tau }^2}{12}}\omega =\frac{0.54\times 12\times 100\times 100}{1.5\times 90\times 90}=5.3rad/secT=\frac{2\pi }{\omega }=1.18$

$\left(b\right)$energy of oscillation$=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{2.5\times 90\times 90}{12\times 100\times 100}\times 1.18E=0.099Joule$