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A ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega $. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be
$\frac{{m\left( {M + 2m} \right)}}{M}\,{\omega ^2}{R^2}$
$\frac{{Mm}}{{\left( {M + m} \right)}}\,{\omega ^2}{R^2}$
$\frac{{Mm}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}$
$\frac{{\left( {M + m} \right)M}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}$
Solution
$\begin{array}{l}
Kinetic\,energ{y_{\left( {rotational} \right)}}{K_R} = \frac{1}{2}I{\omega ^2}\\
kinetic\,energ{y_{(translational)}}{K_r} = \frac{1}{2}M{v^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {v = R\omega } \right)\\
M.{\mathop{\rm I}\nolimits} {._{(initial)}}{I_{ring}} = M{R^2};{\omega _{initial}} = \omega \\
M.I{._{\left( {new} \right)}}I{'_{\left( {system} \right)}} = M{R^2} + 2m{R^2}\\
\omega {'_{\left( {system} \right)}} = \frac{{m\omega }}{{M + 2m}}
\end{array}$
Solving we get loss in $K.E.$
$ = \frac{{Mm }}{{\left( {M + 2m} \right)}}{\omega ^2}{R^2}$
