Show that the minimum speed at dle lowest point as `sqrt(5gr)` in a vertical circle executed by the object.

Explanation

To have this minimum speed `(v_2 = sqrt(gr) )` at point 2, the body must have minimum speed also at point 1. By making use of equation we can find the minimum speed at point 2 .
` v_1^2 - v_2^2 = 4gr`
Substituting equation ` v_2 = sqrt(gr) " in " v_1^2 - v_2^2 = 4gr `
`v_1^2 - gr = 4gr`
` v_1^2 = 5gr`
` v_1 = sqrt(5gr)` ....(2)

The body must have a speed at point 1, `v_1 gt = sqrt(5gr)` to stay in the circular path. From equations `v_2 = sqrt(gr)` and `v_1 = sqrt(5gr)` , it is clear that the minimum speed at the lowest point 1 should be 5 times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle.