Circular motion of a body in vertical plane: A body of mass m is moving in a circular path of radius r in a vertical plane using an inextensible massless string. The velocity of the body at highest point (point P) is v_{1} and velocity at lowest point (point Q) of its motion is v_{2}. The tension of the string at point P and Q are T_{1} and T_{2} respectively.

The force acting on the body at point P towards the centre of the circle is

mg + T_{1} = (centripetal force). But at point P, T_{1} = 0. So, v_{1} = ——(1)

At point Q the force towards the centre of the circle is T_{2} – mg = (centripetal force) ——(2)

The energy at point P is E_{P} = mg2r + and the energy at point Q is E_{Q} = . As the energy is conserved, then E_{P} = E_{Q}.

So, mg2r + =

Or, 4gr + v_{1}^{2} = v_{2}^{2}

Or, 4gr + gr = v_{2}^{2} [using equation (1)]

v_{2} = ——(3)

From equation (2) and (3), we get, T_{2} – mg = 5mg

T_{2} = 6mg.

So, the tension on the string at the lowest point is 6 time the weight of the body.

When the body is at point R, the speed of the body is v and tension of string is T, then = T – mgcos.

So, T = + mgcos.

The energy at point R is E_{R} = .

Using conservation of energy at point P and R we get, = mg(r + rcos) +

Or, = 2gr(1 + cos) + rg = rg(3 + 2cos)

v = .

Concept on vertical circular motion: A ball is released from the top of an inclined plane. The lower part of the inclined plane is connected with a vertical circular path of radius r. Find the minimum height from which the ball is released so that it can move through the circular path. [Consider all the surfaces are frictionless]

To move through the circular path, the required velocity at point B is v = .

Using conservational of mechanical energy at point A and B we get, potential energy at point A = kinetic energy at point B

Or, mgh =

h = = 2.5r.