Question

A bucket of water is whirled in a circle in a vertical plane. The radius of the circle is 0.85 m. What is the minimum speed that will just prevent the water from falling out of the bucket when it is upside down at the top of the circle?

Answer 1

Answer:

v=2.88 m/s

Explanation:

Given that

R= 0.85 m

The minimum speed of the bucket to prevent the water from falling out given as

$v=\sqrt{Rg}$

Where

R=Radius of the circular arc

g=Acceleration due to gravity

We are taking g= 9.81 m/s²

Now by putting the values

$v=\sqrt{Rg}$

$v=\sqrt{0.85\times 9.81}\ m/s$

v=2.88 m/s

Therefore the speed of the bucket should be 2.88 m/s.