Q. In vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point is

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A

$5$

B

$2$

C

$0.5$

D

$0.2$

Solution:

Velocity at the lowest point of the vertical circular motion $v_{h} = 5 g 1$
Velocity at the highest point of the vertical circular motion $v_{1} = g l$
where 1 is the radius of the vertical circular motion.
Thus ratio of kinetic energy at the highest to the lowest point $\frac{K \cdot E _{h}}{K \cdot E _{1}} = \frac{m v _{h}^{2} /2}{m v _{1}^{2} /2} = \frac{v _{h}^{2}}{v _{1}^{2}}$
$\Rightarrow \frac{K \cdot E _{h}}{K \cdot E _{1}} = \frac{g l}{5 g l} = 0.2$