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PhysicsW.E.P.Vertical circular motionEasy2 minPYQ_2020

PhysicsEasynumerical

A particle is attached to a string and the other end of the string is attached to a fixed support. In order to just complete the vertical circular motion, what should be the ratio of the kinetic energy of the particle at its highest point to that at its lowest point?

Answer:

0.20

Solution:

K.E $= \frac{1}{2} m v^{2}$

At lowest point in vertical circular motion $V_{l} = \sqrt{5 r g}$ and at highest point $V_{h} = \sqrt{r g}$

$∴ \frac{{(K . E)}_{h}}{{(K . E)}_{l}} = \frac{1}{5} = 0.2$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion

⏱ 2mℹ️ Source: PYQ_2020

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