# A thin smooth rod of length L and mass M is rotating freely with angular speed  $\omega _{0}$ about an axis perpendicular to the rod and passing through its centre . Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod . The angular speed of the system , when the beads reach the opposite ends of the rod will be : Option 1) $\frac{M\:\omega _{0}}{M+m}$ Option 2) $\frac{M\:\omega _{0}}{M+3m}$ Option 3) $\frac{M\:\omega _{0}}{M+6m}$ Option 4) $\frac{M\:\omega _{0}}{M+2m}$

Applying conservation of angular momentum

$\left ( \frac{ML^{2}}{12} \right )\omega_{0}=\left [ \frac{ML^{2}}{12}+2m\left ( \frac{L}{2} \right )^{2} \right ]\omega^{1}$

$\Rightarrow ML^{2}\omega _{0}=\left ( ML^{2}+6mL^{2} \right )\omega ^{1}$

$\Rightarrow \omega^{1}=\frac{\omega_{0}M}{M+6m}$

Option 1)

$\frac{M\:\omega _{0}}{M+m}$

Option 2)

$\frac{M\:\omega _{0}}{M+3m}$

Option 3)

$\frac{M\:\omega _{0}}{M+6m}$

Option 4)

$\frac{M\:\omega _{0}}{M+2m}$

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