A uniform thin ring ABCD rolls without slipping on a rough horizontal surface. The line BD is horizontal and line AC is vertical. The kinetic energies of section AB, BC, CD and DA are respectively $K_{1}, K_{2}, K_{3}$ and $K_{4} .$ Relation among $KE' s \frac{K_{1}}{K_{4}} = a, \frac{K_{2}}{K_{3}} = b$ then ab is

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Detailed Solution

Consider the following ring :

By symmetry, $θ_{1} = θ_{4}; θ_{2} = θ_{3}$

$v_{1} = \sqrt{v^{2} + {(ω r)}^{2} + 2 (v) (ω r) (\frac{1}{\sqrt{2}})}$

$v_{2} = \sqrt{v^{2} + {(ω r)}^{2} - 2 (v) (ω r) (\frac{1}{\sqrt{2}})}$

$v_{3} = \sqrt{v^{2} + {(ω r)}^{2} - 2 (v) (ω r) (\frac{1}{\sqrt{2}})}$

$v_{4} = \sqrt{v^{2} + {(ω r)}^{2} + 2 (v) (ω r) (\frac{1}{\sqrt{2}})}$

Thus, $K_{1} = K_{4} and K_{2} = K_{3}$

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