Question

A student swings a 5.0 Kg pail of water in a vertical circle of radius 1.3m. What is the minimum speed at the top of the circle if the water is not to spill from pail.

Answer 1

Answer:

v = 3.57 m / s

Explanation:

Let's use Newton's second law in the upper part of the circle, with the minimum speed so that the water does not rest on the circle, that is, the normal is zero

          W = m a

   

acceleration is centripetal

          a = v² / r

let's substitute

         mg = m v² / r

         v = $\sqrt{r g }$

let's calculate

        v = $\sqrt{1.3 \ 9.8}$ra 1.3 9.8

        v = 3.57 m / s