JEE MainPhysicsModerateWork Power & Energy

A frictionless track ABCDE ends in a circular loop of radius R. A
body slides down the track from point A which is at height $h = 82cm$ .
The maximum value of R for a body to complete the loop successfully is

Show answer and solution

Using energy conservation:

\frac{1}{2}mv_{B}^{2} = mgh

V_{B} = \sqrt{\frac{2mgh}{m}}

v_{B} = \sqrt{2hg}\ \ldots(1)

Also to complete vertical circle

v_{B} = \sqrt{5gR}

...(2)

\therefore R = \frac{2}{5}h = 32.8cm

Source: Grass-G / P-06 Physics paper 14 October.docx

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