Question

A thin uniform rod of mass $M$ and length $L$ is hinged at its upper end, and released from rest in a horizontal position. The tension at a point located at a distance $L/3$ from the hinge point, when the rod becomes vertical is $xmg$. Find $x$.

Answer 1

Taking, energy equation . we get,

$\frac{1}{2}I{\omega }^2=Mg\frac{L}{2}$

$\omega \frac{1}{2}\frac{M{L}^2}{3}{\omega }^2=Mg\frac{L}{2}$

$\Rightarrow \omega =\sqrt{3g}L$

let ,T be tension at hinge point and $T^{{}^{\prime }}$ be tension at $\frac{L}{3}$ position.

From fbd,

$T=T^{{}^{\prime }}+\frac{M}{3}g+M{\frac{L}{6}}^2\omega +\frac{Mg}{3}$

$T=Mg+M\frac{L}{2}\frac{3g}{L}=\frac{5Mg}{2}$

by substituting,

$T^{{}^{\prime }}=2Mg$

$\Rightarrow x=2$