Question

A particle of mass m tied to a string of length l and given a circular motion in the vertical plane. If it performs
the complete loop motion then prove that difference in tensions at the lowest and the highest point is 6 mg.
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Answer 1

Let

The speeds at the lowest and highest points be u and v respectively.

At the lowest point,

$Tension, \: T_{L} = mg + \frac{m {u}^{2} }{l} ...........(i)$

At the highest point

$Tension, \: T_{H} = \frac{m {v}^{2} }{l} - mg.......(ii)$

Now

By conservation of mechanical energy

$= > \frac{m {u}^{2} }{2} - \frac{m {v}^{2} }{2} = mg(2l) \\ = > \frac{m}{2}( {v}^{2} - {u}^{2}) = mg(2l) \\ = > \frac{ \cancel{m}}{2}( {v}^{2} - {u}^{2}) = \cancel{m}g(2l) \\ = > \frac{ {v}^{2} - {u}^{2} }{2} = g(2l) \\ = > 2g \times 2l = {v}^{2} - {u}^{2} \\ = > \boxed{{u}^{2} = {v}^{2} + 4gl}$

Substituting this in equation (i)

$T_{L} = mg + \frac{m( {v}^{2} + 4gl) }{l} ......(iii)$

Therefore

From equation (i), (ii) & (iii)

$T_{L} - T_{H} = 6mg$