JEE MAIN - Physics (2006 - No. 49)
A thin circular ring of mass $$m$$ and radius $$R$$ is rotating about its axis with a constant angular velocity $$\omega $$. Two objects each of mass $$M$$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $$\omega ' = $$
$${{\omega \left( {m + 2M} \right)} \over m}$$
$${{\omega \left( {m - 2M} \right)} \over {\left( {m + 2M} \right)}}$$
$${{\omega m} \over {\left( {m + M} \right)}}$$
$${{\omega m} \over {\left( {m + 2M} \right)}}$$
Explanation
Here angular momentum is conserved.
Applying conservation of angular momentum $$I'\omega ' = I\omega \,\,$$
$$\left( {m{R^2} + 2M{R^2}} \right)\omega \,' = m{R^2}\omega $$
$$ \Rightarrow \omega \,' = \omega \left[ {{m \over {m + 2M}}} \right]$$
Applying conservation of angular momentum $$I'\omega ' = I\omega \,\,$$
$$\left( {m{R^2} + 2M{R^2}} \right)\omega \,' = m{R^2}\omega $$
$$ \Rightarrow \omega \,' = \omega \left[ {{m \over {m + 2M}}} \right]$$
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