A particle of mass m starts from rest and slides on a polished track around a vertical circular loop of radius a as shown. Show that the minimum starting height h above the bottom of the loop in order that the particle will leave the track at any point is h=5a/2.

Answer 1

at top of loop centripetal acceleration = g or it leaves the track

g = v^2/a

height at top of loop = 2 a
potential energy at top of loop = m g h = 2 m g a
total energy at top of loop = .5 m v^2 + 2 m g a
= .5 m g a + 2 m g a
= 2.5 m g a
so
h = 2.5 a