A cylindrical bucket filled with water is whirled around in a vertical circle of radius r. What can be the minimum speed at the top of the path if water does not all out from the bucket ? If it continues with this speed, what normal contact force the bucket exerts on water at the lowest point of the path?

Explanation

consider water as the system. At the top of the cfircle its acceleration towards the centre is vertically downward with magnitude `v^2/r`. The forces on water asre ure.
a. weight Mg downward and
b. normal force by the bucket, aslso downward

So, from Newton's second law
`Mg=N=Mv^2/r`
For water not to fall out from the bucket `Nge0`
Hence, `Mv^2/rgeMg or, v^2rarr=rg`
the minimum speed at the top must be `sqrt(rg)`
If the bucket continues on the circle with this minimum speed `sqrt(rg)` the forces at the bottom of theh path are
a. weight Mg downward and
b. normal contact force N by the bucket upward.
The acceleration is twoards the centre which is vertically upward, so
`N'-Mg=Mv^2/g`
`or, N'=M(g+v^2/r)=2Mg`