Motion in Vertical Circle

Derivation:
Motion in a vertical circle:

A particle of mass m is attached to inextensible light string of length l and it is imparted a velocity u in horizontal direction lowest point. Let v be its velocity at point B of circle.

Fig (23)

Here h = R (1-Cosq) ------------ (1)

By conservation of mechanical energy

Necessary centripetal force is provided by the resultant of tension T and mgCos

Now three condition arise depending on u,

The string does not stuck at highest point if T

0 at

=

,

So substituting T = 0 and

=

for completing the circle in (B)

(At highest point)

And h = 2R

So from equation (2)

Or

the particle will complete the circle.

Substituting q = 00 and v = u =

in equation (3) we get T = 6mg. So in the critical condition tension in the string at lowest position is 6mg.

Fig (24)

Putting T = 0 before reaching highest point for

Substituting in equation (1) we get

Velocity of particle becomes zero when
0 = u2-2gh

Now the particle will leave the circle if tension in the string becomes zero but velocity is not zero or,
T = 0 but v 0 this is possible only when
h1<h2

Therefore if

the particle leaves the circle.

The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.
V = 0 but T

0 this is possible only when
h2 < h1

Further if h1 = h2,

and tension and velocity both becomes zero simultaneously.
Illustration:

A ball of mass m slides without friction down a path from height h and then moves in loop of radius