Given data:
* The length of the string (or the radius of the vertical circle) is,
[tex]r=0.55\text{ m}[/tex]Solution:
Let the tension of the string is T.
The string will go slack after zero tension.
Thus,
[tex]T=0\text{ N}[/tex]At the top, the vertical forces acting are,
The weight of the pail of water ,
[tex]W=mg[/tex]where m is the mass, g is the acceleration due to gravity,
The centripetal force acting on the apil of water moving in vertical circular motion is
[tex]F=\frac{mv^2}{r}[/tex]As the weight of pail of water and tension of the string provides the centripetal force for the circular motion,
Thus,
[tex]\begin{gathered} T+W=F \\ 0+mg=\frac{mv^2}{r} \\ mg=\frac{mv^2}{r} \\ g=\frac{v^2}{r} \\ v^2=gr \\ v=\sqrt[]{gr} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{9.8\times0.55} \\ v=2.32ms^{-1} \end{gathered}[/tex]Thus, the velocity of the pail at the top position is 2.32 meter per second.
