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 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 :-

Rotational Motion PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q43 with Solution

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Correct answer: C

Since the rod-bead system rotates freely about the given axis and there is no external torque about that axis, angular momentum is conserved.

So, $I_i\omega_0 = I_f\omega$ where:

The rod is thin, of length $L$ and mass $M$ , rotating about an axis through its center and perpendicular to the rod.

Hence, $I_{\text{rod}} = \frac{1}{12}ML^2$

Initially, both beads are at the center, so their distance from the axis is zero. Therefore their contribution to moment of inertia is zero.

Thus, $I_i = \frac{1}{12}ML^2$

When the beads reach the opposite ends of the rod, each bead is at distance $r = \frac{L}{2}$ from the axis.

Moment of inertia of two beads: $I_{\text{beads}} = 2m\left(\frac{L}{2}\right)^2 = 2m\cdot \frac{L^2}{4} = \frac{mL^2}{2}$

So total final moment of inertia is $I_f = \frac{1}{12}ML^2 + \frac{1}{2}mL^2$

Taking $L^2$ common,

= L^2\left(\frac{M+6m}{12}\right)$$ Thus, $$I_f = \frac{(M+6m)L^2}{12}$$ --- 4. **Apply conservation of angular momentum** $$I_i\omega_0 = I_f\omega$$ $$\frac{1}{12}ML^2\,\omega_0 = \frac{(M+6m)L^2}{12}\,\omega$$ Cancel $\frac{L^2}{12}$: $$M\omega_0 = (M+6m)\omega$$ Therefore, $$\omega = \frac{M\omega_0}{M+6m}$$ --- 5. **Match with the options** This corresponds to: **Option C:** $$\boxed{\frac{M\omega_0}{M+6m}}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So they agree.