Question

Discuss the motion of a body in a vertical circle. Find expressions for the minimum velocity at the lowest point while looping a loop and difference of tensions in the string at the lowest and highest points.

Answer 1

Motion in a vertical circle:

• When an object moves in a vertical circle, its motion is mainly governed by the conservation of mechanical energy principle.

• For the body to cross the topmost point, the minimum velocity at highest point has to be √(gR) , R is the radius of the circle.

Now applying the CONSERVATION OF MECHANICAL ENERGY between topmost point and bottom of the circle:

$\therefore \: \dfrac{1}{2} m {u}^{2} + mg(2R) = \dfrac{1}{2} m {v}^{2} + 0$

$\implies \: \dfrac{1}{2} m {( \sqrt{gR})}^{2} + mg(2R) = \dfrac{1}{2} m {v}^{2}$

$\implies \: \dfrac{1}{2} m gR + mg(2R) = \dfrac{1}{2} m {v}^{2}$

$\implies \: \dfrac{1}{2} gR + g(2R) = \dfrac{1}{2} {v}^{2}$

$\implies \: gR + 4gR = {v}^{2}$

$\implies \: {v}^{2} = 5 gR$

$\implies \: v = \sqrt{5 gR}$

So, minimum velocity at bottom has to be √(5gR).

Now , tension at bottom is :

$\therefore \: T1 - mg = \dfrac{m {v}^{2} }{R}$

$\implies \: T1 - mg = \dfrac{m {( \sqrt{5gR} )}^{2} }{R}$

$\implies \: T1 - mg = 5mg$

$\implies \: T1 = 6mg$

Tension at topmost point:

$\therefore \: T2 + mg = \dfrac{m {u}^{2} }{R}$

$\implies\: T2 + mg = \dfrac{m {( \sqrt{gR}) }^{2} }{R}$

$\implies\: T2 + mg = mg$

$\implies\: T2 = 0$

So, difference in tension = ∆T = 6mg.