Question

Two masses each of $m$ are attached at mid point $B$ & end point $C$ of massless rod $AC$ which is jinged at $A$. It is released from horizontal position as shown. Find the force at hinge $A$ when rod becomes vertical

Answer 1

In the given problem,

$CM=\frac{3}{4}L$ implies decrease in PE is,

$\left(2m\right)g\left(\frac{3}{4}L\right)$,

The KE gained is $\frac{1}{2}I{w}^2$

where, $I=m({\left(\frac{L}{2}\right)}^2+L^2)=\frac{5}{4}m{L}^2$

the total decrease in PE must equal the total increase in KE:

$\frac{1}{2}\left(\frac{5}{4}m{L}^2\right)w^2=\frac{3}{2}mgL$

The centripetal force required to keep the rod moving in a circle is,

$(\frac{L}{2}+L)m{w}^2=\frac{3}{2}mL\left(\frac{12g}{5L}\right)=\frac{18}{5}mg$

The force on the hinge is, sum of centripetal force and the weight of the rod

$2mg+\frac{18}{5}mg=\frac{28}{5}mg$