1. Model the motion

Since the floor has large friction, the lower end of the rod does not slip. So the rod rotates about the contact point with the floor as a fixed pivot.

For a uniform rod of length $L$ pivoted at one end:

• Moment of inertia about the pivot: $I_O = \frac{1}{3}ML^2$
• Distance of center of mass from pivot: $r_{CM} = \frac{L}{2}$

Let $\theta$ be the angle the rod makes with the vertical. We are asked for $\theta = 60^\circ$.

2. Angular speed using energy conservation

Initially, the rod is vertical and at rest.

When the rod makes angle $\theta$ with the vertical, the center of mass has fallen by $\Delta h = \frac{L}{2} - \frac{L}{2}\cos\theta = \frac{L}{2}(1-\cos\theta)$

Loss in gravitational potential energy = gain in rotational kinetic energy: $Mg\cdot \frac{L}{2}(1-\cos\theta) = \frac{1}{2}I_O\omega^2$

Substitute $I_O = \frac{1}{3}ML^2$: $Mg\cdot \frac{L}{2}(1-\cos\theta) = \frac{1}{2}\cdot \frac{1}{3}ML^2\omega^2$

$\frac{MgL}{2}(1-\cos\theta) = \frac{1}{6}ML^2\omega^2$

$3g(1-\cos\theta)=L\omega^2$

At $\theta=60^\circ$, $\cos60^\circ=\frac12$: $L\omega^2 = 3g\left(1-\frac12\right)=\frac{3g}{2}$

So, $\omega^2=\frac{3g}{2L}$ $\omega=\sqrt{\frac{3g}{2L}}$

Therefore, Option D is correct.

3. Angular acceleration

Torque of gravity about the pivot is $\tau = Mg\cdot \frac{L}{2}\sin\theta$

Using $\tau = I_O\alpha$: $Mg\cdot \frac{L}{2}\sin\theta = \frac{1}{3}ML^2\alpha$

$\alpha = \frac{3g}{2L}\sin\theta$

At $\theta=60^\circ$: $\alpha = \frac{3g}{2L}\cdot \frac{\sqrt3}{2} = \frac{3\sqrt3}{4}\frac{g}{L}$

This is not equal to $\frac{2g}{L}$.

Therefore, Option A is incorrect.

4. Radial acceleration of the center of mass

For rotation about a fixed point, radial (centripetal) acceleration of the center of mass is $a_r = \omega^2\left(\frac{L}{2}\right)$

Using $\omega^2 = \frac{3g}{2L}$: $a_r = \frac{3g}{2L}\cdot \frac{L}{2} = \frac{3g}{4}$

Therefore, Option C is correct.

5. Normal reaction from the floor

Let us find acceleration of the center of mass in vertical direction.

For a rod making angle $\theta$ with vertical, taking pivot as origin, $x = \frac{L}{2}\sin\theta, \qquad y = \frac{L}{2}\cos\theta$

Differentiate twice: $\ddot y = -\frac{L}{2}(\alpha\sin\theta + \omega^2\cos\theta)$

At $\theta=60^\circ$:

• $\sin60^\circ = \frac{\sqrt3}{2}$
• $\cos60^\circ = \frac12$
• $\alpha = \frac{3\sqrt3}{4}\frac{g}{L}$
• $\omega^2 = \frac{3g}{2L}$

So, $a_y = \ddot y = -\frac{L}{2}\left(\frac{3\sqrt3}{4}\frac{g}{L}\cdot \frac{\sqrt3}{2} + \frac{3g}{2L}\cdot \frac12\right)$

$a_y = -\frac{L}{2}\left(\frac{9g}{8L} + \frac{3g}{4L}\right)$

$a_y = -\frac{L}{2}\left(\frac{15g}{8L}\right) = -\frac{15g}{16}$

Now apply Newton's second law in vertical direction: $N - Mg = M a_y$

$N - Mg = -\frac{15Mg}{16}$

$N = Mg - \frac{15Mg}{16} = \frac{Mg}{16}$

Therefore, Option B is correct.

6. Final evaluation of options

• A: Incorrect
• B: Correct
• C: Correct
• D: Correct

So the correct choices are: $\boxed{B,\ C,\ D}$

7. Comparison with stored correct answer

Stored correct answer: B, C, D

My derived answer: B, C, D

They match exactly.