A ring of mass M and radius R is rotating abo | vedclass

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Question

$\begin{array}{l}
Kinetic\,energ{y_{\left( {rotational} ight)}}{K_R} = \frac{1}{2}I{\omega ^2}\
kinetic\,energ{y_{(translational)}}{K_r} = \frac{

Vedclass · Accepted Answer

$\frac{{Mm}}{{\left( {M + 2m} \right)}}\,{\omega ^2}{R^2}$

Vedclass · Answer

$\begin{array}{l}
Kinetic\,energ{y_{\left( {rotational} ight)}}{K_R} = \frac{1}{2}I{\omega ^2}\
kinetic\,energ{y_{(translational)}}{K_r} = \frac{1}{2}M{v^2}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {v = R\omega } ight)\
M.{\mathop{m I}
olimits} {._{(initial)}}{I_{ring}} = M{R^2};{\omega _{initial}} = \omega \
M.I{._{\left( {new} ight)}}I{'_{\left( {system} ight)}} = M{R^2} + 2m{R^2}\
\omega {'_{\left( {system} ight)}} = \frac{{m\omega }}{{M + 2m}}
\end{array}$

Solving we get loss in $K.E.$

$ = \frac{{Mm }}{{\left( {M + 2m} ight)}}{\omega ^2}{R^2}$

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

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