Part A)
If water will not spill from the bucket then at minimum speed condition the value of normal force on water will be zero at the top position
so now we will have
[tex]F_g = \frac{mv^2}{R}[/tex]
[tex]mg = \frac{mv^2}{R}[/tex]
[tex]gR = v^2[/tex]
[tex]v = \sqrt{Rg}[/tex]
now we know that
R = 1 m
g = 9.8 m/s^2
[tex]v = \sqrt{1(9.8)}[/tex]
[tex]v = 3.13 m/s[/tex]
PART B)
for minimum angular speed we know the relation between angular speed and linear speed as
[tex]v = r\omega[/tex]
now we know that
r = 1 m
[tex]3.13 = 1(\omega)[/tex]
so angular speed will be
[tex]\omega = 3.13 rad/s[/tex]
