A bucket full of water is tied with a rope $1.6 m$ long and revolved in a vertical circle. What would be the minimum speed of the bucket at the highest point so that the water may not fall? ( $g = 10 m / s^{2}$ )

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Solution

For water not to spill out of bucket
$\begin{matrix} \frac{m v^{2}}{r} = m g v = \sqrt{r g} = \sqrt{1.6} \times \sqrt{10} v = \sqrt{16} = 4 m s^{- 1} \end{matrix}$

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A bucket full of water is tied with a rope 1.6m long and revolved in a vertical circle. What would be the minimum speed of the bucket at the highest point so that the water may not fall? (g=10m/s2)

For water not to spill out of bucket mv2r=mgv=√rg=√1.6×√10v=√16=4ms−1

A bucket full of water is tied with a rope $1.6 m$ long and revolved in a vertical circle. What would be the minimum speed of the bucket at the highest point so that the water may not fall? ( $g = 10 m / s^{2}$ )

For water not to spill out of bucket
$\begin{matrix} \frac{m v^{2}}{r} = m g v = \sqrt{r g} = \sqrt{1.6} \times \sqrt{10} v = \sqrt{16} = 4 m s^{- 1} \end{matrix}$