Question

what is the ratio of the kinetic energy of a particle at the bottom to the kinetic energy at the top when it just loops a vertical loop of radius of r.

Answer 1

As velocity at the top. vT = √gR ,

velocity at the bottom vB = √5gR

And kinetic energy is proportional to the square of the velocity ,

The required ratio is 5.

hope it helps u....

Answer 2

The ratio of the kinetic energy of a particle at the bottom to the kinetic energy at the top when it just loops a vertical loop of the radius of r is 5

Given:

The radius of the vertical loop is r

Find:

The ratio of the kinetic energy of a particle at the bottom to the kinetic energy at the top

Solution:

Velocity at the bottom = $\sqrt{5gl}$

Velocity at the highest point =$\sqrt{gl}$

The ratio of kinetic energy at the highest to bottom point =$\frac{KEh}{Keb}$= $\frac{mvh^{2}/2 }{mvb^{2}/2 }$= $\sqrt{5gl} ^{2}/\sqrt{gl^{2} }$

= 5

So the ratio of kinetic energy at the bottom point to the highest point is 5

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