Q.A bucket of water of mass $2\ \text{kg}$ is whirled in a vertical circle of radius $1.2\ \text{m}$. Find

Given: mass of water $m=2\ \text{kg}$ (does not affect either minimum speed), radius $r=1.2\ \text{m}$, $g=9.8\ \text{m/s}^2$.

(a) Minimum speed at the top. The water stays in the bucket only as long as the bucket's base can still supply some (non-negative) normal reaction $N$ on the water; the critical case is $N=0$, where gravity alone supplies the entire centripetal force: $mg = \frac{mv_{\text{top,min}}^2}{r} \quad\Longrightarrow\quad v_{\text{top,min}} = \sqrt{gr} = \sqrt{9.8\times1.2} = \sqrt{11.76} \approx 3.43\ \text{m/s}$ …

Set the normal reaction on the water to zero at the top to find $v_{\text{top,min}}=\sqrt{gr}$, then use energy conservation across the diameter $2r$ to find $v_{\text{bottom,min}}=\sqrt{5gr}$ -- identical in structure …

• Treating the water's weight as somehow different in kind from the stone-on-a-string case, rather than recognising the identical underlying physics (a normal reaction, like a tension, can only push/pull one way and can go to …