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  • A ring of mass M and radius R is rotating about its axis with angular velocity . Two identical bodies each of mass

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 : 

Option: 1

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


Option: 2

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


Option: 3

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

 


Option: 4

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


Answers (1)

best_answer

 

 

 

 

 

By using angular momentum conservation

M R^{2} \times \omega=\left(M R^{2}+m R^{2}+m R^{2}\right) \omega^{\prime}$ \\ $ \Rightarrow \omega^{\prime}=\frac{M \omega}{M+2 m}

\text { Loss in kinetic energy } \\=\frac{1}{2} M R^{2} \times \omega^{2}-\frac{1}{2}(M+2 m) R^{2} \times \frac{M^{2} \omega^{2}}{(M+2 m)^{2}}=\frac{M m \omega^{2} R^{2}}{(M+2 m)}

Posted by

manish painkra

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