Question

A pail of water is being swung in a vertical circle at the end of a 0.55-m string. How slowly can the pail go through its top positionwithout having the string go slack?

Answer 1

Given data:

* The length of the string (or the radius of the vertical circle) is,

`r=0.55\text{ m}`

Solution:

Let the tension of the string is T.

The string will go slack after zero tension.

Thus,

`T=0\text{ N}`

At the top, the vertical forces acting are,

The weight of the pail of water ,

`W=mg`

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

`F=(mv^2)/(r)`

As the weight of pail of water and tension of the string provides the centripetal force for the circular motion,

Thus,

`\begin{gathered} T+W=F \\ 0+mg=(mv^2)/(r) \\ mg=(mv^2)/(r) \\ g=(v^2)/(r) \\ v^2=gr \\ v=\sqrt[]{gr} \end{gathered}`

Substituting the known values,

`\begin{gathered} v=\sqrt[]{9.8*0.55} \\ v=2.32ms^(-1) \end{gathered}`

Thus, the velocity of the pail at the top position is 2.32 meter per second.