Question

A smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size, each of mass $m$, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity ${\omega }_0$ about an axis perpendicular to the rod and passing through the mid-point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is

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

Answer 1

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

Consider the rod and two beads together as a system, there are no external torques on the system about the fixed fixed point $O$ (because there are no external forces ). Hence, angular momentum of the system about the fixed point $O$ is conserved. Initially, two beads of mass $m$ each are at the centre of the rod of mass $M$ and length $L$, and angular velocity ${\omega }_0$.


Initial angular momentum of the system about axis of rotation is,
$L_0=I_0{\omega }_0$
where $I_0=\frac{M{L}^2}{12}+m(0{)}^2+m(0{)}^2$
i.e $L_0=\frac{M{L}^2}{12}{\omega }_0$

Finally, the two beads are at the extreme ends of the rod rotating with an angular velocity $\omega$. Thus, the final angular momentum of the system about the axis of rotation is
$L=I\omega$
where $I=[\frac{M{L}^2}{12}+m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2]$
i.e $L=[\frac{M}{12}+\frac{m}{2}]L^2\omega$
From conservation of angular momentum, about the fixed point $O$
$L_0=L$
$\Rightarrow \frac{M{L}^2}{12}\times {\omega }_0=[\frac{M}{12}+\frac{m}{2}]L^2\omega$
$\Rightarrow \omega =\frac{M{\omega }_0}{M+6m}$