Question

A bucket containing water is whirled in a vertical circle at arms length. Find the minimum speed at top to ensure that no water spills out. Also find corresponding angular speed. (Assume r = 0.75 m)(Ans : 2.711 m/s. 3.615 rad/s)

Answer 1

A bucket containing water is whirled in a vertical circle at arms length.
then, radius of vertical circle = arms length
it is given that radius of vertical circle , r = 0.75

so, minimum speed v at the top to ensure that no water spills out is given by $v=\sqrt{gr}$
where g is acceleration due to gravity.

so, v = √(9.8 × 0.75) = 2.711 m/s

we know, $\omega=\frac{v}{r}$

where $\omega$ is angular speed

so, $\omega$ = 2.711/(0.75) =3.615 rad/s