Question

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 center. 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-

• A. $\frac{M{\omega }_0}{M+3m}$
• B. $\frac{M{\omega }_0}{M+2m}$
• C. $\frac{M{\omega }_0}{M+m}$
• D. $\frac{M{\omega }_0}{M+6m}$

Answer 1

The correct option is D $\frac{M{\omega }_0}{M+6m}$

Since, there is no external torque acting on the rod so we can apply conservation of angular momentum.

$I_i{\omega }_i=I_f{\omega }_f$

$\left(\frac{M{L}^2}{12}\right){\omega }_0=[\frac{M{L}^2}{12}+2m{\left(\frac{L}{2}\right)}^2]{\omega }_f$

$\therefore {\omega }_f=\left(\frac{M{\omega }_0}{M+6m}\right)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.