A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is √(2rg). What will be the speed of the body at the lowest point?

When a body moves in a vertical circle, its energy keeps changing form, but the total mechanical energy remains constant. Mechanical energy is the sum of kinetic energy (energy of motion) and potential energy (energy due to height).

At the highest point, the body is at maximum height, so it has more potential energy and less kinetic energy. At the lowest point, the height is minimum, so potential energy becomes zero and kinetic energy becomes maximum.

Given that the speed at the highest point is √(2rg), we use the law of conservation of energy:

• Total energy at top = Kinetic Energy + Potential Energy

• Total energy at bottom = Only Kinetic Energy (since height = 0)

At the top:
K.E = (1/2)mv² = (1/2)m(2rg) = mg r
P.E = mg(2r) = 2mg r

So, total energy = mg r + 2mg r = 3mg r

At the bottom:
P.E = 0
So, total energy = K.E = 3mg r

Now,
(1/2)mv² = 3mg r

Solving,
v² = 6gr
v = √(6gr)

Therefore, the speed at the lowest point is √(6gr).