Normal reaction on a falling rod

Consider the instant when the rod(in blue) makes an angle $\theta$ with the vertical. Let the velocity of center of mass be $v$ , and angular acceleration of rod be $\alpha$ .Clearly, angular velocity is $\dfrac{2v}{L}$ The center of mass moves in a circle of radius $\dfrac{L}{2}$ . Thus the acceleration of center of mass is as shown(in green) in the diagram. The forces are shown in red:

Taking torque about point of contact,

$\dfrac{mL^2}{3} \alpha = mg \dfrac{L}{2} \sin \theta$ $\Rightarrow \dfrac{L}{2} \alpha = \dfrac{3g \sin \theta}{4}$

Also, using conservation of energy(taking both translational and rotational energies),

$\dfrac{1}{2} \dfrac{mL^2}{3} \dfrac{4v^2}{L^2}= mg \dfrac{L}{2} \big(1 - \cos \theta)$ $\Rightarrow \dfrac{2v^2}{L} = \dfrac{3g(1 - \cos \theta)}{2}$

Now, using Newton's second law,

$f = ma_{x} = m \bigg(\dfrac{L}{2} \alpha \cos\theta - \dfrac{2v^2}{L} \sin \theta\bigg) = \boxed{\dfrac{3mg \sin \theta}{4} \big(3 \cos \theta - 2)}$ ,

$N = mg - a_{y} = mg - m \bigg(\dfrac{L}{2} \alpha \sin\theta - \dfrac{2v^2}{L} \cos \theta\bigg) = \boxed{\dfrac{mg}{4} \big(3 \cos \theta - 1)^2}$

Clearly, friction changes its direction when $\cos \theta = \dfrac{2}{3}$ . Hence, putting this value in expression of $N$ , we get the required value as $\dfrac{mg}{4} = 2.45 \text{ N }$

$a$ is linear acceleration and $\alpha$ is angular acceleration. Gravity acts on the centre of mass of the rod, i.e., at a distance $\frac{l}{2}$ from an end.

A further question which can be added is : Physically, What is happening due to which the friction reverses its direction?

Consider the rotational acceleration about the fixed end of the rod and about the center of the rod. The respective angular accelerations of each are the same, so the ratios of the torques about each equals the ratio of the rotational inertias. The rotational inertia about the center of the rod is $(mL^{2})/12$ and the rotational inertia about the end is $(mL^{2})/3$ . The torque by gravity about the end and the torque by the normal force about the end have the same lever arm and make the same angle with the rod (at the instant friction changes direction, there is no frictional force). Thus the force of gravity on the rod must be 4 times as great as the normal force on the rod (as the rotational inertia about the end is 4 times as large). So the normal force is $mg/4 = 9.8/4 = \boxed{2.45 N}$