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 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

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

Answer 1

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

As there is no external torque on system.
$\therefore$ Angular momentum of system is conserved.

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

Initially,


Finally,


So,

$\frac{M{L}^2}{12}{\omega }_0+0=(\frac{M{L}^2}{12}+2(m\left){\left(\frac{L}{2}\right)}^2\right)\omega$

So, the final angular speed of the system is

$\omega =\frac{\frac{M{L}^2}{12}{\omega }_0}{\left(\frac{M{L}^2+6m{L}^2}{12}\right)}$

$\therefore \omega =\frac{M{\omega }_0}{M+6m}$

Hence, option $\left(\text{D}\right)$ is correct.