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

Let $v_{top}$ be the speed of the particle at the highest point. By conservation of mechanical energy from the lowest point to the highest point (height $2L$), $\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}_{top}^2+mg(2L)$. Substituting $v_0=\sqrt{7gL}$, we get $\frac{1}{2}m(7gL)=\frac{1}{2}m{v}_{top}^2+2mgL$, which simplifies to $3.5gL=\frac{1}{2}{v}_{top}^2+2gL$, so $v_{top}^2=3gL$. At the highest point, the force $F_{rod}$ exerted by the rod and the gravitational force $mg$ both act downwards (towards the center). The centripetal force equation is $F_{rod}+mg=\frac{m{v}_{top}^2}{L}$. Substituting $v_{top}^2=3gL$, we get $F_{rod}+mg=\frac{m(3gL)}{L}=3mg$. Therefore, $F_{rod}=2mg$. The magnitude of the force is 2 units of $mg$.