Is it possible to whirl a bucket of water fast enough in a vertical circle so that the water won’t fall out? If so, what is the minimum speed? Define all quantities needed.

The normal force acting on the bucket is the difference between centripetal force and the weight of bucket. This relation helps to determine the critical speed of the bucket.

The figure below represents the free body diagram:

According to Newton’s second law of motion,

\(\begin{aligned}\sum F &= ma\\{F_N} + mg &= m\left( {\frac{{{v^2}}}{r}} \right)\\{F_N} &= m\left( {\frac{{{v^2}}}{r} - g} \right)\end{aligned}\)

The water will leave the contact with bucket, when the normal force become zero.

\(\begin{aligned}{F_N} &= 0\\m\left( {\frac{{{v^2}}}{r} - g} \right) &= 0\\{v_{critical}} &= \sqrt {rg} \end{aligned}\)

Thus, the minimum speed of whiriling is \(\sqrt {rg} \). If the bucket is rotated at this speed, then water will not fall out of the bucket.