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# Engineering

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}

 

Answers (1)
S solutionqc

 

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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