# A particle is compelled to move inside smooth vertical tube as shown in fig. The minimum speed at lowest point so that it can complete full circle is

Devang
52 Points
5 years ago
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
v= (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)
va= 5gR
va  =  (5gR)1/2
hence a partical is complete full circle its minimum velocity is  (5gR)1/2