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 center. Two beads of mass m and negligible size are at the center of the rod initially.The beads of mass $$m$$ and negligible size are at the center 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:

We have a thin, smooth rod of length $$L$$ and mass $$M$$ which is rotating about an axis passing through its centre and perpendicular to the length of the rod. Two identical beads, each of mass $$m$$, initially sit at the centre of the rod. Because the surface is smooth, no external torque acts on the system while the beads slide, so the total angular momentum of the “rod + beads” system must remain constant.

The principle we shall use is the conservation of angular momentum, stated as

$$\text{Initial angular momentum} = \text{Final angular momentum}\,,$$

or algebraically

$$I_i\,\omega_0 = I_f\,\omega_f,$$

where $$I_i,\,I_f$$ are the moments of inertia of the whole system before and after the beads move, and $$\omega_0,\,\omega_f$$ are the corresponding angular speeds.

First we compute the initial moment of inertia. For a thin rod of length $$L$$ about an axis through its centre and perpendicular to its length, the standard formula is

$$I_{\text{rod}} = \frac{1}{12} M L^2.$$

Both beads are exactly at the centre initially, so each has distance $$r=0$$ from the axis, giving for a point mass $$I = m r^2 = 0$$. Hence the total initial moment of inertia is simply

$$I_i = \frac{1}{12} M L^2.$$

Now let the beads slide outward to the opposite ends of the rod. Each end is at a distance $$\dfrac{L}{2}$$ from the centre. The moment of inertia of a single bead at that position is

$$I_{\text{one bead}} = m\left(\frac{L}{2}\right)^2 = m\,\frac{L^2}{4}.$$

Because there are two identical beads, their combined contribution is

$$I_{\text{both beads}} = 2 \times m\,\frac{L^2}{4} = m\,\frac{L^2}{2}.$$

The rod itself has not changed, so its moment of inertia is still $$\dfrac{1}{12} M L^2.$$ Thus the final moment of inertia of the whole system is

$$I_f = \frac{1}{12} M L^2 + \frac{1}{2} m L^2.$$

Applying conservation of angular momentum, we write

$$\Bigl(\frac{1}{12} M L^2\Bigr)\,\omega_0 \;=\;\Bigl(\frac{1}{12} M L^2 + \frac{1}{2} m L^2\Bigr)\,\omega_f.$$

We can cancel the common factor $$L^2$$ from every term, obtaining

$$\Bigl(\frac{1}{12} M\Bigr)\,\omega_0 = \Bigl(\frac{1}{12} M + \frac{1}{2} m\Bigr)\,\omega_f.$$

Solving for the unknown final angular speed $$\omega_f$$, we get

$$\omega_f = \frac{\frac{1}{12} M}{\frac{1}{12} M + \frac{1}{2} m}\;\omega_0.$$

To clear the fractions, we multiply numerator and denominator by 12:

$$\omega_f = \frac{M}{M + 6m}\;\omega_0.$$

That is,

$$\boxed{\;\omega_f = \dfrac{M\omega_0}{M + 6m}\;}.$$

Comparing with the given options, we see this expression matches Option C.

Hence, the correct answer is Option C.