Question

A light rod of length $1$ m is pivoted at its centre and two masses of $5$ kg and $2$ kg 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 tensions in the 2 strings will be the same.

Writing torque equation

$5g\times \frac{l}{2}-2g\times \frac{l}{2}=I\alpha$

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

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

$3\frac{gl}{2}=\frac{7l^2}{4}\alpha$

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