Question

A small block is freely sliding down from top of a smooth inclined plane the block reaches bottom of inclined plane then the block describes vertical circle of radius 0.5m along smooth track the minimum vertical height of inclined plane should be
Ans:1.25m

Answer 1

radius = R = .5 m
at top, m g = m v^2/R for contact force = 0
sso
v^2/R = g
so v = sqrt (g R) at top of circle which is height 2 R
potential energy at top of plane = m g h
potential energy at 2 R = 2 m g R
loss of potential =
m g h - 2 m g R = mg (h-2R)
= Ke at top of circle = (1/2) m v^2
so
(1/2) m v^2 = m g (h-2R)
but
we know v^2 = g R
so
(1/2) m g R = m g h - 2 m g R
2.5 m g R = m g h
h = 2.5 R = 2.5 (.5) = 1.25 meters