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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?

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You are right that the issue is not a small angle approximation. The mistake you made is in the constraint, which unfortunately is a bit more difficult than you were hoping. The real constraint is that:

$$ y = \frac{L\cos(\theta)}{2} $$

This forces the velocity and acceleration to be subject to the constraints:

$$ \dot{y} = -\frac{L\sin(\theta)}{2}\dot{\theta} $$

and

$$ \ddot{y} = -\frac{L}{2}\left(\sin(\theta)\ddot{\theta}+\cos(\theta)\dot{\theta}^{2}\right) $$

I am assuming $\theta$ is from the vertical based on how your question is written. There is no diagram. It should be clear that my $\ddot{y}$ is your $a_{y}$ and my $\ddot{\theta}$ is your $\alpha_{z}$.

I can try to solve the problem if you want, but this should be enough to get you back on the right track. Are you supposed to get an analytical answer? Can you do this numerically?

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  • $\begingroup$ Ah okay, I misunderstood your answer!! Yep this was the issue. Thank you. $\endgroup$ Commented Jan 24 at 6:32

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