Question:

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

Updated On: Apr 24, 2026
  • $5$
  • $2$
  • $0.5$
  • $0.2$
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The Correct Option is D

Solution and Explanation

Using energy conservation: $\frac{1}{2}mv_1^2=\frac{1}{2}mv_2^2+2mgR \Rightarrow v_2^2=v_1^2-4gR$.

Thus, $\dfrac{\text{KE}_{\text{top}}}{\text{KE}_{\text{bottom}}}=\dfrac{\frac{1}{2}m v_2^2}{\frac{1}{2}m v_1^2}=\dfrac{v_1^2-4gR}{v_1^2}$.

For minimum vertical circle, $v_1^2=5gR$.

Hence, ratio $=\dfrac{5gR-4gR}{5gR}=\dfrac{1}{5}=0.2$.

Final Answer: $0.2$

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Concepts Used:

Kinetic energy

Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.