Question

Two blocks, each of mass m, are attached to the ends of amassless rod which pivots as shown in the figure (Figure 1).Initially the rod is held in the horizontal position and thenreleased.(A) Calculate the magnitude of the net torque on this system when it is first released. Use the fact that L2>L1

Answer 1

The force acting on each block is gravitational force given as,

`F=mg`

The torque acting due to first block is,

`\tau_1=\hat{l_1i}* mg(-\hat{j})`

This can be solved as,

`\tau_1=l_1mg(-\hat{k})`

The torque acting due to second block is,

`\tau_2=\hat{l_2i}* mg(-\hat{j})`

This can also be solved as,

`\tau_2=l_2mg(-\hat{k})`

Therefore, the net torque acting on system is,

`_{}\tau=\tau_1-\tau_2`

Substituting values,

`\begin{gathered} \tau=l_1mg(-\hat{k})-l_1mg(-\hat{k}) \\ =mg(l_2-l_1)\hat{k} \end{gathered}`

Thus, the net torque acting on system is

`\tau=mg(l_2-l_1)\hat{k}_{}_{}`