How long would it take for an upright rigid body to fall to the ground?
To complete the other answers, here's the Lagrangian of the system. Using polar coordinates centred at the base of the rod,
$$\mathcal{L}=T-V$$
$$\mathcal{L}=\frac{1}{2}I\dot{\theta^2}-\frac{1}{2}mgh\cos{\theta}$$
Using the known value for $I$, $\frac{mh^2}{3}$,
$$\mathcal{L}=\frac{1}{6}mh^2\dot{\theta^2}-\frac{1}{2}mgh\cos{\theta}$$
Using Euler-Lagrange, gives us the equation of motion:
$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{\partial\mathcal{L}}{\partial\theta}$$
$$\frac{d}{dt}(\frac{1}{3}mh^2\dot{\theta})=\frac{1}{2}mgh\sin{\theta}$$
$$\ddot{\theta}=\frac{3}{2}\frac{g}{h}\sin{\theta}$$
The answer depends on the initial conditions (initial angle, and initial angular velocity). If it starts vertical, then without an initial angular velocity it would take forever to fall.
Also note that as the rotational speed inceases, the bar might lift up from the pivot point due to the centifigual forces. Also once friction is overcome the contact is going to slip. So there are three domains for the solution.
- Contact is fixed, forces need to be calculated
- Contact is sliding, vertical force needs to be calculated
- No longer in contact, no forces applied (other than gravity).
You can try to work out the solutions from the following equations:
Equations Of Motion
$$ F=m\ddot{x} $$ $$ N=m(\ddot{y}+g) $$ $$ \frac{H}{2}\left(F\,\cos\theta+N\,\sin\theta\right)=I_{G}\ddot{\theta} $$
where: $F$ frictional (horiz.) force, $N$ normal contact force, $m$ mass of bar, ($\ddot{x}$, $\ddot{y}$) acceleration of center of gravity, $H$ the total height of the bar, $\theta$ angle of bar from vertical (+=CCW), $I_G$ mass moment of inertia of the bar at the c.g.
Velocity of contact point $$ vx_{A}=\dot{x}+\frac{H}{2}\left(\dot{\theta}\cos\theta\right) $$ $$ vy_{A}=\dot{y}+\frac{H}{2}\left(\dot{\theta}\sin\theta\right) $$
Acceleration of contact point $$ ax_{A}=\ddot{x}+\frac{H}{2}\left(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta\right) $$
$$ ay_{A}=\ddot{y}+\frac{H}{2}\left(\ddot{\theta}\sin\theta+\dot{\theta}^{2}\cos\theta\right) $$
Friction contdition $$ F\leq\mu N $$
You will find the equation of motion for case 1 being
$$\ddot{\theta}=\frac{g\sin\theta-\frac{H}{2}\dot{\theta}\cos2\theta}{\frac{I_{G}}{mH/2}+\frac{H}{2}\sin2\theta}$$
Finding the condition where you slip into case 2 and then case 3 involves monitoring the forces $F$ and $N$ and checking when $F>\mu N$ and when $N<0$.
A falling tree is basically an inverted pendulum.
The period of a pendulum of length $h$ for small oscillations is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity, about $10\ m/s$. For an inverted pendulum near the top of its arc, there is no period, but the quantity $\sqrt{h/g}$ does represent a characteristic time scale for this system. The tree will take a few of these characteristic times to fall. $h$ for a tree is an "effective height", and depends on the mass distribution of the tree. If all the mass is at the top, $h$ is the height of the tree. If the tree is uniform, $h$ is $2/3$ the true height.
For a tree with $h = 40\ m$, the characteristic time is $2\ s$. For small angles, the angle the tree makes with the vertical will be multiplied by $e$ in this time. Let's start the tree at $1^{\circ}$ so that it needs to multiply its angle by $90$ to fall. $\ln(90) = 4.5$ so the tree takes about $9\ s$ to fall.
This is mathematically an underestimate because the characteristic time increases slightly as the tree falls, but not too much. Give it a nice round $10\ s$ and you get something that matches the first YouTube video I found.