Question

A rod of negligible mass having length $l=2m$ is pivoted at its centre and two masses of $m_1=6kg$ and $m_2=3kg$ are hung from the ends as shown in figure.
(a) Find the initial acceleration of the rod if it is horizontal initially

Answer 1

REF.Image

FBD of the blocks and pivot - system :-

From equilibrium,

$T_1=m_{1}g$ & $T_2=m_{2}g\Rightarrow T_1=30N$ & $T_2=60N$

The tension provide a torque about O.

$Z_1=T_{2}r-T_{1}r=(60-30)\times 1=30N-m$

also, due to rod, an additional couple appear as,

$Z_2=2\times \frac{r}{2}\times \frac{m_{3}g}{2}=\frac{m_{3}gr}{2}=\frac{3\times 10\times 1}{2}=15N-m$

$\therefore$ Net torque, $Z_{net}=Z_1+Z_2=45N-m$

and moment of inertia about 0,

$I_0=\frac{m_3{l}^2}{12}+m_1{r}^2+m_2{r}^2=\frac{3\times 2^2}{12}+(3+6)\times 1^2=10kg{m}^2$

so angular acceleration,

We know ;-

$Z_{net}=I_0\propto$

$\propto =\frac{Z_{net}}{I_0}=\frac{45}{10}=4.5rad/s^2$