# How long it will take for a upright rigid body to fall on a ground

Let's suppose there is a straight rigid bar with height $h$ and center of mass at the middle of height $h/2$. Now if the bar is vertically upright from ground, how long will it take to fall on the ground and what is the equation of motion of the center of mass (Lagrangian)?

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This is basic mechanics. Far too low level for this site. –  user346 Dec 30 '10 at 7:39
Why will it fall if the bar is vertically upright? The system is in (unstable) equilibrium. –  kennytm Dec 30 '10 at 9:09
@KennyTM: just assume infinitesimal horizontal force. –  Marek Dec 30 '10 at 11:53
@space_cadet: I strongly disagree. At the least, it's a difficult problem in first year undergraduate mechanics. The Lagrangian bit brings it to last year undergraduate (potentially). –  Noldorin Dec 30 '10 at 13:00
–  kennytm Dec 30 '10 at 15:13

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.

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i am thinking that angular acceleration is harmonic in nature. if 'θ' be the instantaneous angle made my tree with the vertical upright, then the θ¨ = g*sinθ. I think if i solve this equation from 0 to π/2 i think i will get the answer, but i, however, am not sure. –  Santosh Linkha Dec 31 '10 at 7:19
@explorex You should have $\ddot{\theta} = \frac{g}{l}\sin\theta$. If you start with $\theta = 0$, the solution is simply $\theta(t) = 0$, meaning the tree doesn't go anywhere. If you start from $\theta = \epsilon$, you will not get an exact expression. My answer is an approximate solution to this last equation. –  Mark Eichenlaub Dec 31 '10 at 7:45
thanks mark for correction, but could you give expression for time interms of θ –  Santosh Linkha Dec 31 '10 at 7:58
@explorex Roughly, $t = \sqrt{h/g}\ln\left(\frac{\theta}{\theta_0}\right)$ with $\theta_0$ the initial angle of the tree from vertical. –  Mark Eichenlaub Dec 31 '10 at 8:06
-1, small angle aproximations do NOT apply here. Also a prediod is only defined when $\ddot{\theta}\propto-\theta$ (notice the minus sign). When positive you have an imaginary natural frequency. –  ja72 Jan 1 '11 at 21:31

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:

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=\frac{\partial\mathcal{L}}{\partial\theta}$$

$$\frac{d}{dt}(\frac{1}{3}mh^2\dot{\theta})=\frac{1}{2}mgh\sin{\theta}$$

$$\ddot{\theta}=\frac{3}{2}\frac{g}{h}\sin{\theta}$$

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Your differential equation is not dimensionally correct. I think you have the moment of inertia wrong, and the potential energy is proportional to $h$, not to $h^2$. –  Mark Eichenlaub Jan 1 '11 at 19:08
@Mark Thank you, corrected. –  Sklivvz Jan 1 '11 at 19:29

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.

1. Contact is fixed, forces need to be calculated
2. Contact is sliding, vertical force needs to be calculated
3. 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 $N<0$.

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can you do this without initial friction and also the time please. –  Santosh Linkha Dec 31 '10 at 4:10
i am thinking that angular acceleration is harmonic in nature. if 'θ' be the instantaneous angle made my tree with the vertical upright, then the θ¨ = g*sinθ. I think if i solve this equation from 0 to π/2 i think i will get the answer, but i, however, am not sure. –  Santosh Linkha Dec 31 '10 at 7:20