A particle has velocity $\sqrt{3 r g}$ at the highest point in a vertical circle. Find the ratio of tension at the highest and lowest point.

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Solution

The correct option is B 1:4
Velocity at the topmost point is $\sqrt{3 g r}$
Conservation of energy between topmost and the bottom point gives
$0.5 (3 g r) + g (2 r) = 0.5 (v^{2})$
$\Rightarrow v^{2} = 7 g r$ , velocity at the bottom.
Ratio of the tensions
$\frac{T_{h}}{T_{l}} = \frac{m {v_{h}}^{2} / r - m g}{m {v_{l}}^{2} / r + m g}$

$\frac{T_{h}}{T_{l}} = \frac{3 g r / r - g}{7 g r / r + g}$
$\frac{T_{h}}{T_{l}} = \frac{2 g}{8 g} = \frac{1}{4}$

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A particle has velocity √3rg at the highest point in a vertical circle. Find the ratio of tension at the highest and lowest point.

The correct option is B 1:4Velocity at the topmost point is √3grConservation of energy between topmost and the bottom point gives0.5(3gr)+g(2r)=0.5(v2)⇒v2=7gr, velocity at the bottom.Ratio of the tensionsThTl=mvh2/r−mgmvl2/r+mgThTl=3gr/r−g7gr/r+gThTl=2g8g=14

A particle has velocity $\sqrt{3 r g}$ at the highest point in a vertical circle. Find the ratio of tension at the highest and lowest point.