# You are rotating a bucket of water in a vertical circle. Assuming that the radius of the rotation of the water is 0.95 m, what is the minimum velocity of the bucket at the top of its swing if the water is not to spill?

Jul 24, 2018

Minimum velocity $v = 3.05 m {s}^{-} 1$

#### Explanation:

The acceleration required to keep the water following the circumference of the circle ${a}_{c}$ (called centripetal acceleration) must be $\ge g$ (acceleration due to gravity) to prevent the water from spilling.

The minimum centripetal acceleration must therefore $= g$

Formula for centripetal acceleration:

${a}_{c} = {v}^{2} / r$ where $v =$ velocity ($m {s}^{-} 1$) and $r$ = radius ($m$)

At minimum velocity required, ${a}_{c} = g$, so
${v}^{2} = g r$

${v}^{2} = 9.8 \cdot 0.95 {m}^{2} {s}^{-} 2$

$\text{ } = 9.31 {m}^{2} {s}^{-} 2$

$v = 3.05 m {s}^{-} 1$