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Is it possible to whirl a bucket of water fast enough in a vertical circle so the water won't fall out? If so, what is the minimum speed?

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Answered 2 years ago
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In order to whirl a bucket of water so that water doesn't fall out the centripetal force on it at the highest point of its trajectory should be equal to its weight. Assuming a water of mass mm we then have

mv2r=mg    v=rg\begin{aligned} m \frac{v^2}{r} = mg\implies \boxed{v = \sqrt{rg}} \end{aligned}

where rr is the radius of the circle.

Note that the above speed is the minimum speed required.

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