Vertical Circular Motion

At the highest point of a vertical circle, both gravity and any normal or tension force point downward toward the centre, and together they supply the centripetal force $\frac{m{v}^2}{r}$. The critical minimum speed occurs when the contact force just vanishes, so gravity alone provides the centripetal requirement: $mg=\frac{m{v}_{top}^2}{r}$. Cancelling mass and solving, $v_{top}=\sqrt{gr}=\sqrt{10\times 1.6}=\sqrt{16}=4\text{ m/s}$. The 2 m/s value would leave gravity stronger than the centripetal demand, so the water would fall out. The 8 m/s value uses $\sqrt{gr}$ but with the radius mistakenly doubled or the result squared incorrectly. The 16 m/s value confuses $gr$ with the speed itself, skipping the square root. The mass cancelling from the critical condition reveals a deep point: the threshold speed depends only on the radius and gravitational field, not on how heavy the bucket and water are, so the same minimum speed protects a small cup or a large pail of equal radius. One can also connect this to energy by using the higher speed needed at the bottom, found from energy conservation across the height $2r$, but the top point alone fixes the limiting condition. This is the classic NCERT vertical-circle threshold condition. As a plausibility check, below 4 m/s the required inward force is less than gravity, the string slackens, and the water spills, confirming 4 m/s as the genuine minimum.