Question

Two point mass $m$ are connected the light rod of length $l$ and it is free to rotate in vertical plane as shown. Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in vertical lane.

Answer 1

$\begin{array}{c}w=\frac{v}{2l}\\ \Rightarrow u=wL=\frac{v}{2}\\ wall=\Delta k\left[attoprestvelocity=0\right]\\ \left(-mg2l\right)+\left(-mg\cdot 4l\right)=-\left[\frac{1}{2}m{v}^2+\frac{1}{2}m{\left(\frac{v}{2}\right)}^2\right]\\ -6mgl=-\frac{m{v}^2}{2}-\frac{m{v}^2}{8}\\ \Rightarrow 6mgl=\frac{5m{v}^2}{8}\\ \Rightarrow v^2=\frac{48gl}{5}\\ \Rightarrow v=\sqrt{\frac{48gl}{5}}Answer.\end{array}$