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

Show that the minimum speed at dle lowest point as sqrt(5gr)in a verti

1.	Show that the minimum speed at dle lowest point as sqrt(5gr)in a vertical circle executed by the object.
Answer» SOLUTION :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_1gt = 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.