. A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position (Take g = 10 m/sec²)

To find the minimum speed at which the bucket must be whirled in a vertical circle so that water does not spill at the highest point, we need to analyze the forces acting on the water when the bucket is at the very top of its path.

At the highest position, two main forces act on the water inside the bucket:
1. The force of gravity, acting vertically downward:
$F{g}_{}=mg$
2. The normal reaction force exerted by the bottom of the bucket on the water, also acting downward:
$N$

Together, these downward forces provide the necessary centripetal force required to keep the water moving in a circular path of radius $r$ with velocity $v$:
$N+mg=\frac{m{v}^2}{r}$

To prevent the water from spilling, the bucket must maintain a minimum speed such that the water remains in contact with the bottom of the bucket. This means the normal reaction force $N$ must be greater than or equal to zero ($N��0$).

For the minimum speed ($v_{\text{min}}$), the normal reaction force becomes exactly zero ($N=0$) at the highest point:
$mg=\frac{m{v}_{\text{min}}^2}{r}$

We can cancel the mass $m$ from both sides of the equation:
$g=\frac{v_{\text{min}}^2}{r}$
Solving for $v_{\text{min}}$:
$v_{\text{min}}^2=rg$
$v_{\text{min}}=\sqrt{rg}$

Given:
Radius of the path (length of the string), $r=1.6\text{m}$
Acceleration due to gravity, $g=10{\text{m/s}}^2$

Substitute these values into the formula:
$v_{\text{min}}=\sqrt{1.6\times 10}$
$v_{\text{min}}=\sqrt{16}$
$v_{\text{min}}=4\text{m/sec}$

Thus, the minimum speed required to prevent the water from spilling is 4 m/sec.