Question

A light rod of length $1m$ is pivoted at its center and two masses of $5kg$ and $2kg$ are hung from the ends as shown in figure. Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.

Answer 1

Since the rod is massless, the tension in the two string will be the same.

Writing torque equation :-

$5g\times \frac{l}{2}-2g\times \frac{l}{2}=I\alpha =[m_1\times {\left(\frac{l}{2}\right)}^2+m_2{\left(\frac{l}{2}\right)}^2]\alpha$

$\Rightarrow \frac{3}{2}gl=[\frac{{5l}^2}{4}+\frac{{2l}^2}{4}]\alpha$

$\Rightarrow \alpha =\frac{6g}{7}$