A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle $60^\circ$ with vertical ? [$g$ is the acceleration due to gravity]

Step 1: Define the moment of inertia.

The moment of inertia of the uniform rod of mass $M$ and length $L$ about the contact point $O$ is:

$$I_O = \frac{1}{3}ML^2$$

Step 2: Apply conservation of energy.

The decrease in gravitational potential energy as the rod rotates by an angle $\theta$ is equal to the gain in rotational kinetic energy.

$$\Delta PE = Mg\frac{L}{2}(1-\cos\theta)$$ $$KE = \frac{1}{2}I_O\omega^2$$ $$\frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 = Mg\frac{L}{2}(1-\cos\theta)$$ $$\omega^2 = \frac{3g}{L}(1-\cos\theta)$$

Step 3: Evaluate the angular speed at $\theta = 60^\circ$.

Substitute $\theta = 60^\circ$ into the angular speed equation.

$$\omega^2 = \frac{3g}{L}(1-\cos 60^\circ)$$ $$\omega^2 = \frac{3g}{2L}$$ $$\omega = \sqrt{\frac{3g}{2L}}$$

This confirms statement (1) is correct.

Step 4: Calculate the radial acceleration of the center of mass.

The radial acceleration $a_r$ of the center of mass is directed towards the pivot $O$.

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

This confirms statement (2) is correct.

Step 5: Calculate the angular acceleration.

The torque about the pivot $O$ is produced by gravity.

$$\tau_O = Mg\frac{L}{2}\sin\theta$$ $$I_O\alpha = \tau_O$$ $$\frac{1}{3}ML^2\alpha = Mg\frac{L}{2}\sin\theta$$ $$\alpha = \frac{3g}{2L}\sin\theta$$

At $\theta = 60^\circ$:

$$\alpha = \frac{3g}{2L}\sin 60^\circ$$ $$\alpha = \frac{3\sqrt{3}g}{4L}$$

This shows statement (4) is incorrect.

Step 6: Calculate the normal reaction force.

The vertical acceleration $a_y$ of the center of mass is related to the angular velocity and angular acceleration.

$$a_y = -\frac{L}{2}\left(\omega^2\cos\theta + \alpha\sin\theta\right)$$

Apply Newton's second law in the vertical direction.

$$N - Mg = Ma_y$$ $$N = Mg - \frac{ML}{2}\left(\omega^2\cos\theta + \alpha\sin\theta\right)$$

Evaluate the components at $\theta = 60^\circ$.

$$\omega^2\cos 60^\circ = \frac{3g}{4L}$$ $$\alpha\sin 60^\circ = \frac{9g}{8L}$$

Substitute these components back into the expression for $N$.

$$N = Mg - \frac{ML}{2}\left(\frac{3g}{4L} + \frac{9g}{8L}\right)$$ $$N = Mg - \frac{15Mg}{16}$$ $$N = \frac{Mg}{16}$$

This confirms statement (3) is correct.

Answer: Statements (1), (2), and (3) are correct.