The system is a thin, uniform rod with perfect symmetry which is released from rest at some small angle $\theta>0$ from the vertical while being on a frictionless surface.

Then there are no horizontal forces acting on the pen as it falls (gravity and normal force only act vertically): \begin{equation*} \sum_\text{ext} F_x = 0 \Rightarrow a_{\text{cm,}x} = 0 \Rightarrow \boxed{x_{\text{cm},i} = x_{\text{cm},f}} \end{equation*} Choosing $+y$ pointing upwards, \begin{equation*} \sum_\text{ext} F_y = ma_y \Rightarrow \boxed{n - mg = ma_y} \end{equation*} Calling $\theta$ the angle made with the vertical and $l$ the length of the pencil, \begin{equation*} \sum_\text{ext} \tau_z = I\alpha_z \Rightarrow \boxed{\frac{1}{2}nl\sin\theta = I_\text{cm}\alpha_z} \end{equation*} Here the $z$ axis is chosen to pass through the center of mass of the pencil. Finally, there's the rigid body constraint which connects the rotational and translational motion: \begin{equation*} \boxed{a_y = +l\alpha_z/2} \end{equation*} These lead to ($\kappa = I_\text{cm}/m\left(l/2\right)^2$) \begin{equation*} \boxed{n = \frac{mg}{1-\sin\theta/\kappa}} \end{equation*} But this makes no sense to me. How can the normal force change sign and eventually diverge as $\theta$ increases from 0? Nothing physically happens at any critical angle so far as I can tell. What's going on here?

The system is a thin, uniform rod/pencil with perfect symmetry which is released from rest at some small angle $\theta>0$ from the vertical while being on a frictionless surface.

Then there are no horizontal forces acting on the pencil as it falls (gravity and normal force only act vertically): \begin{equation*} \sum_\text{ext} F_x = 0 \Rightarrow a_{\text{cm,}x} = 0 \Rightarrow \boxed{x_{\text{cm},i} = x_{\text{cm},f}} \end{equation*} Choosing $+y$ pointing upwards, \begin{equation*} \sum_\text{ext} F_y = ma_y \Rightarrow \boxed{n - mg = ma_y} \end{equation*} Calling $\theta$ the angle made with the vertical and $l$ the length of the pencil, \begin{equation*} \sum_\text{ext} \tau_z = I\alpha_z \Rightarrow \boxed{\frac{1}{2}nl\sin\theta = I_\text{cm}\alpha_z} \end{equation*} Here the $z$ axis is chosen to pass through the center of mass of the pencil. Finally, there's the rigid body constraint which connects the rotational and translational motion: \begin{equation*} \boxed{a_y = +l\alpha_z/2} \end{equation*} These lead to ($\kappa = I_\text{cm}/m\left(l/2\right)^2$) \begin{equation*} \boxed{n = \frac{mg}{1-\sin\theta/\kappa}} \end{equation*} But this makes no sense to me. How can the normal force change sign and eventually diverge as $\theta$ increases from 0? Nothing physically happens at any critical angle so far as I can tell. What's going on here?