Physics - W.E.P. Question with Solution | TestHub

For the particle to complete the vertical circle, the tension in the string must be non-negative at the highest point. At the highest point, the minimum speed $v_H$ required is such that the centripetal force is provided by gravity, i.e., $\frac{m{v}_H^2}{L}=mg$, which gives $v_H=\sqrt{gL}$. Using conservation of mechanical energy between the lowest point (speed $v_L$) and the highest point (speed $v_H$), we have $\frac{1}{2}m{v}_L^2+0=\frac{1}{2}m{v}_H^2+mg(2L)$. Substituting $v_H^2=gL$, we get $\frac{1}{2}m{v}_L^2=\frac{1}{2}m(gL)+2mgL=\frac{5}{2}mgL$. Thus, $v_L=\sqrt{5gL}$.