Lets assume that point A , when partical is situated at lowest point in vertical tube and point B , when partical is situated at highest point in vertical tube.
when partical is ,on point A
mechanical energy = kinatic energy + potational energy
= ½ m va2 + 0 …............(1) (hight is 0)
normal force = weight force
N = mg …............(2)
when partical is ,on point B
normal force = centripetal force
N = mvb2 / R put value of n in this eq.
mg = mvb2 / R
vb = (gR)1/2 …............(3)
mechanical energy = kinatic energy + potational energy
= ½ m vb2 + mg(2R)
= ½ m (gR) + 2mgR
= 5/2 (gR) ….............(4)
from law of coservation of mechanical energy
(1) & (4)
½ m va2 = 5/2 (gR)
va2 = 5gR
va = (5gR)1/2
hence a partical is complete full circle its minimum velocity is (5gR)1/2
