Question

A uniform rod of mass $m$ and length $l$ hinged at point $\text{H}$ can rotate in a vertical plane about a smooth horizontal axis. Find the force exerted by the hinge just after the rod is released from rest, from an initial position making an angle of ${37}^{\circ }$ with horizontal.

• A. $\sqrt{1}0mg$
• B. $\frac{\sqrt{1}0}{5}mg$
• C. $\frac{3mg}{5}$
• D. $\frac{mg}{5}$

Answer 1

The correct option is B $\frac{\sqrt{1}0}{5}mg$

Just after releasing at ${37}^{\circ }$ from horizontal
$FBD$ of plank:


${\tau }_{\text{net}}=I\alpha$
$\tau$ about point $A$ ${\tau }_A=mg\cos {37}^{\circ }\frac{l}{2}=\frac{m{l}^2}{3}.\alpha$
$\Rightarrow \alpha =\frac{6g}{5l}{\text{rad/sec}}^2$
Now tangential acceleration of centre of mass
$a_t=\alpha .\frac{l}{2}=\frac{3g}{5}{\text{m/s}}^2$
Just after release, $v_{\text{cm}}=0$
$\Rightarrow$ centripetal acceleration $a_r=0$

Now resolving of $a_t$ in horizontal and vertical direction as shown in figure


From $F_{\text{net}}=ma$ in both horizontal and vertical directions
$N_2=m\left(\frac{9g}{25}\right)=\frac{9mg}{25}$
and $mg-N_1=\frac{12mg}{25}$
$\Rightarrow N_1=\frac{13mg}{25}$
Now, $R=\sqrt{N_1^2+N_2^2}$
$R=\frac{5\sqrt{1}0mg}{25}=\frac{\sqrt{1}0}{5}mg$