Question

A uniform rod of mass m and length l can rotate in vertical plane about a smooth horizontal axis hinged at point H. The force exerted by the hinge just after the rod is released from rest is $\frac{mg\sqrt{x}}{5}$. Find the value of $x$.

Answer 1

Let the component of hinge force along the rod be $F_t$ and the component perpendicular to the rod be $F_n$.
Writing torque equation about the hinge :
mg($\frac{l}{2}$)cos(${37}^{\circ }$) = m$\frac{l^2}{3}\alpha$
$\alpha$ = $\frac{6g}{5l}$
Net force along the rod is zero since there is no acceleration towards the center .
$F_t$ = mg sin(${37}^{\circ }$) = mg$\frac{3}{5}$
Writing force equation perpendicular to the rod
mgcos(${37}^{\circ }$) - $F_n$ = m$\frac{l\alpha }{2}$
$F_n$ = mg(\frac{4}{5}) - m$\frac{l}{2}\frac{6g}{5l}$ = mg$\frac{1}{5}$
Net hinge force :
F = $\sqrt{F_n^2+F_t^2}$ = mg$\sqrt{{\frac{1}{5}}^2+{\frac{3}{5}}^2}$ = mg$\frac{\sqrt{10}}{5}$
$\therefore x=10$