Question

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

• A. 5
• B. 2
• C. 0.5
• D. 0.2

Answer 1

The correct option is D 0.2

Velocity at the lowest point of the vertical circular motion $v_h=\sqrt{5gl}$

Velocity at the highest point of the vertical circular motion $v_l=\sqrt{gl}$

where $l$ is the radius of the vertical circular motion.

Thus ratio of kinetic energy at the highest to the lowest point $\frac{K.E_h}{K.E_l}=\frac{m{v}_h^2/2}{m{v}_l^2/2}=\frac{v_h^2}{v_l^2}$

$⟹$ $\frac{K.E_h}{K.E_l}=\frac{gl}{5gl}=0.2$