Part 1
If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position
[tex]F_g = ma[/tex]
[tex]mg = m\frac{v^2}{R}[/tex]
[tex]g = \frac{v^2}{R}[/tex]
[tex]v = \sqrt{Rg}[/tex]
given that
R = 1 m
g = 9.8 m/s^2
now from above equation we have
[tex]v = \sqrt{1(9.8)} = 3.13 m/s[/tex]
Part b)
for minimum value of angular speed we will have
[tex]\omega = \frac{v}{R}[/tex]
[tex]\omega = \frac{3.13}{1}[/tex]
[tex]\omega = 3.13 rad/s[/tex]
