A bucket containing water is revolved in vertical circle of radius r. To prevent the water from falling down, the period of revolution required is ($g$ = gravitational acceleration)

At the highest point of the vertical circle, the forces acting on the water are its weight $mg$ downwards and the normal reaction $N$ from the bucket downwards.

The necessary centripetal force is provided by these forces:

$mg + N = \dfrac{mv^2}{r}$

To prevent the water from falling down, the normal reaction must be greater than or equal to zero ($N \ge 0$). The minimum speed $v$ is obtained when $N = 0$:

$mg = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{rg}$

The angular velocity $\omega$ corresponding to this minimum speed is:

$\omega = \dfrac{v}{r} = \sqrt{\dfrac{g}{r}}$

The maximum period of revolution $T$ required to prevent the water from falling is:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{r}{g}}$

Answer: $2\pi\sqrt{\dfrac{r}{g}}$