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

Let the mass of the sphere be $m$ and the lowest point be the reference level for potential energy ($U=0$).

When the sphere reaches the height of suspension, the thread becomes horizontal, so the height gained by the sphere is $l$.

For the minimum velocity $v$ at the lowest point, the velocity of the sphere at the height of suspension will just become zero.

Applying conservation of mechanical energy between the lowest point and the height of suspension:

$\frac{1}{2}m{v}^2+0=0+mgl$

$v^2=2gl$

$v=\sqrt{2gl}$

To ensure the string does not slack before reaching the horizontal position, we check the tension $T$ at an angle $\theta$ with the vertical:

$T=\frac{m{v}^{\prime 2}}{l}+mg\cos \theta$

Using energy conservation, $v^{\prime 2}=v^2-2gl(1-\cos \theta )=2gl\cos \theta$.

Substituting $v^{\prime 2}$, we get $T=3mg\cos \theta$. Since $\cos \theta >0$ for $\theta <90^{\circ }$, the tension remains positive and the string does not slack.

Thus, the minimum velocity required is $\sqrt{2gl}$.

Answer: $\sqrt{2gl}$