A bucket of water is tied to one end of a rope $8 \mathrm{~m}$ long and rotated about the other end in a vertical circle. Find the number of revolutions per minute such that water does not spill.
Q 60.16
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[Important note: The circular motion of the bucket in a vertical plane under gravity is not a uniform circular motion. Assuming the critical case of the motion such that the bucket has the minimum speed at the highest point required for the water to stay put in the bucket, we can find the minimum frequency of revolution. ]
Data $: r=8 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \pi=3.142$
Assuming the bucket has a minimum speed $v=\sqrt{r g}$ at the highest point, the corresponding angular speed is
$
\omega=2 \pi f=\frac{v}{r}=\frac{\sqrt{r g}}{r}=\sqrt{\frac{g}{r}}
$
The minimum frequency of revolution,
$
\begin{aligned}
f & =\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \\
& =\frac{1}{2 \times 3.142} \sqrt{\frac{9.8}{8}} \\
& =\frac{1}{6.284} \sqrt{1.225}=0.1761 \mathrm{rps} \\
& =0.1761 \times 60 \mathrm{rpm}=10.566 \mathrm{rpm}
\end{aligned}
$
Data $: r=8 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \pi=3.142$
Assuming the bucket has a minimum speed $v=\sqrt{r g}$ at the highest point, the corresponding angular speed is
$
\omega=2 \pi f=\frac{v}{r}=\frac{\sqrt{r g}}{r}=\sqrt{\frac{g}{r}}
$
The minimum frequency of revolution,
$
\begin{aligned}
f & =\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \\
& =\frac{1}{2 \times 3.142} \sqrt{\frac{9.8}{8}} \\
& =\frac{1}{6.284} \sqrt{1.225}=0.1761 \mathrm{rps} \\
& =0.1761 \times 60 \mathrm{rpm}=10.566 \mathrm{rpm}
\end{aligned}
$
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