Rotational Motion of Falling Uniform Rod on Rough Floor

To determine the correct statements, we analyze the rotational motion of the uniform rod about its point of contact $O$ on the rough floor.

1. Angular Speed ($\omega$)

Using the principle of conservation of mechanical energy between the vertical position ($\theta = 0^\circ$) and the position where the rod makes an angle $\theta$ with the vertical:

• Initial potential energy relative to the floor: $U_i = Mg \frac{L}{2}$
• Potential energy at angle $\theta$: $U_f = Mg \frac{L}{2} \cos\theta$
• Kinetic energy of pure rotation about the contact point $O$: $K_f = \frac{1}{2} I_O \omega^2$

The moment of inertia of the thin uniform rod about its end $O$ is $I_O = \frac{1}{3} ML^2$.

By conservation of energy: $Mg \frac{L}{2} (1 - \cos\theta) = \frac{1}{2} \left(\frac{1}{3} M L^2\right) \omega^2$ $\omega^2 = \frac{3g}{L} (1 - \cos\theta)$

At $\theta = 60^\circ$, we have $\cos 60^\circ = \frac{1}{2}$: $\omega^2 = \frac{3g}{L} \left(1 - \frac{1}{2}\right) = \frac{3g}{2L}$ $\omega = \sqrt{\frac{3g}{2L}}$

Thus, Option A is correct.

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2. Angular Acceleration ($\alpha$)

The torque about the contact point $O$ is produced solely by the gravitational force acting on the center of mass: $\tau_O = Mg \left(\frac{L}{2}\right) \sin\theta$

Using Newton's second law for rotation ($\tau_O = I_O \alpha$): $\frac{1}{3} M L^2 \alpha = Mg \frac{L}{2} \sin\theta$ $\alpha = \frac{3g}{2L} \sin\theta$

At $\theta = 60^\circ$, we have $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $\alpha = \frac{3g}{2L} \left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}g}{4L}$

Thus, Option B is incorrect.

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3. Radial Acceleration of the Center of Mass ($a_r$)

The center of mass moves in a circle of radius $r = \frac{L}{2}$ about $O$. The radial (centripetal) acceleration $a_r$ directed along the length of the rod towards $O$ is: $a_r = \omega^2 r = \omega^2 \left(\frac{L}{2}\right)$

Substituting $\omega^2 = \frac{3g}{2L}$: $a_r = \left(\frac{3g}{2L}\right) \left(\frac{L}{2}\right) = \frac{3g}{4}$

Thus, Option C is correct.

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4. Normal Reaction Force ($N$) from the Floor

The tangential acceleration $a_t$ of the center of mass perpendicular to the rod is: $a_t = \alpha r = \left(\frac{3\sqrt{3}g}{4L}\right) \left(\frac{L}{2}\right) = \frac{3\sqrt{3}g}{8}$

To find the normal reaction $N$, we resolve the radial and tangential accelerations along the vertical downward direction to find the vertical acceleration $a_y$ of the center of mass: $a_y = a_r \cos 60^\circ + a_t \sin 60^\circ$ $a_y = \left(\frac{3g}{4}\right) \left(\frac{1}{2}\right) + \left(\frac{3\sqrt{3}g}{8}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{3g}{8} + \frac{9g}{16} = \frac{15g}{16}$

Applying Newton's second law in the vertical direction: $Mg - N = M a_y$ $N = M(g - a_y) = M \left(g - \frac{15g}{16}\right) = \frac{Mg}{16}$

Thus, Option D is correct.

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Conclusion

The correct options are A, C, and D.