Discuss the motion in a vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

Hint:In a vertical circular motion, the motion is considered to be non-uniform because of gravitation force, the velocity and tension of the object will differ in magnitude at different directions.

Formulas used:
${{a}_{c}}=\dfrac{\sum{{{f}_{c}}}}{m}$
Where ${{a}_{c}}$is the circular acceleration
$\sum{{{f}_{c}}}$is the total forces applied
And m is the mass of the object

Complete step by step answer:
The motion of a body in a vertical circle when it is released from an altitude, travels vertically downwards towards the earth’s surface. This is the result of the gravitational force that is exerted by the earth. The corresponding acceleration is denoted by g and is called acceleration due to gravity. This implies that all the bodies irrespective of physical properties should take the same time to fall to the earth. In the circle, all bodies have minimum velocity at the lowest point, and the rope or string becomes slack at the topmost point of the circle.
Tension at top T=\[\dfrac{m{{V}_{T}}^{2}}{R}-mg\], where $V_T$ is the particle speed at the topmost point.
For minimum $V_T$, T=0
Hence, \[{{V}_{T}}=\sqrt[{}]{gR}\]
If ${{V}_{B}}$​ be the critical velocity of the particle at the bottom,
then from conservation of energy:
$\begin{align}
&\Rightarrow mg(2R)+\dfrac{1}{2}m{{V}_{T}}^{2}=0+\dfrac{1}{2}m{{V}_{B}}^{2} \\
& {{V}_{T}}=\sqrt{gR} \\
\end{align}$
As \[\begin{align}
& {{V}_{T}}=\sqrt{gR}>2mgR+\dfrac{1}{2}mgR \\
&\Rightarrow {{V}_{T}} =\dfrac{1}{2}m{{V}_{B}}^{2} \\
& \Rightarrow {{V}_{B}}=\sqrt{5gR} \\
\end{align}\]
Highest point H (h=2r)
\[\begin{align}
&\Rightarrow v=\sqrt{{{u}^{2}}-2gh} \\
&\Rightarrow v=\sqrt{{{(\sqrt{5gR})}^{2}}-2g(2R)} \\
&\Rightarrow v =\sqrt{5gR-4gR} \\
&\Rightarrow v=\sqrt{gR} \\
\end{align}\]
 Tension at the lowest point:
$\begin{align}
& T-mg=\dfrac{m{{v}^{2}}}{R}(\because v=\sqrt{5gR}) \\
&\therefore T=6mg \\
\end{align}$
Tension at highest point:
$
\begin{align}
& T-mg=\dfrac{m{{v}^{2}}}{R}(\because v=\sqrt{gR}) \\
&\therefore T=0 \\
\end{align}$

Note:The minimum value of the velocity when the string is horizontal i.e, h=r
We have, $u=\sqrt{5gR}$
Hence,
$\begin{align}
& v=\sqrt{{{(\sqrt{5gR})}^{2}}-2gR} \\
&\Rightarrow v =\sqrt{5gR-2gR}=\sqrt{3gR} \\
\end{align}$