# Infinitely tall tree falling in a forest

• ttzhou
In summary, the tree never falls over because the angular speed (ω) diminishes as the tree gets taller and longer.
ttzhou

## Homework Statement

Suppose that there is a negligibly thin tree in the forest of infinite length that begun tipping over. Negating frictional effects from the pivoting, does the tree ever hit the ground?

## Homework Equations

My approach was to solve the problem for a tree of length $l$ and see what happens in the limit as $l \rightarrow \infty$

$\tau = I\alpha$

$\tau = Mg\cos{\theta}(l/2)\$

$I = \frac{1}{3} Ml^2\$

## The Attempt at a Solution

$I\alpha = Mg\cos{\theta} (l/2)$

$\alpha = \frac{3gl}{2} \cos{\theta}$

Now, I know how to approach and get a function for $\omega$ by using energy conservation; however, out of curiosity, would I be able to solve this DE purely mathematically?

In any case, $\omega = \sqrt{\frac{3g(1-\sin{\theta})}{l}}$

I interpret this to mean that if I take the limit as the length goes to infinity, the tree simply has no angular speed and thus cannot fall over at all. However, this seems unsatisfactory as an answer - is it possible to actually derive an explicit function for $\theta(t)$?

Thank you all for any help, and I apologize if I have made any sort of oversight or foolish mistake.

hi ttzhou!
ttzhou said:
In any case, $\omega = \sqrt{\frac{3g(1-\sin{\theta})}{l}}$

is it possible to actually derive an explicit function for $\theta(t)$?

put ψ = π/2 - θ, and then use one of the standard trigonometric identities

This is my second attempt at a reply. I decided to take a second look at my first attempt after reading tiny-tim's reply

ttzhou said:

## Homework Statement

Suppose that there is a negligibly thin tree in the forest of infinite length that begun tipping over. Negating frictional effects from the pivoting, does the tree ever hit the ground?

## Homework Equations

My approach was to solve the problem for a tree of length $l$ and see what happens in the limit as $l \rightarrow \infty$

$\tau = I\alpha$

$\tau = Mg\cos{\theta}(l/2)\$

$I = \frac{1}{3} Ml^2\$

## The Attempt at a Solution

$I\alpha = Mg\cos{\theta} (l/2)$

$\alpha = \frac{3gl}{2} \cos{\theta}$
Not that it matters much (since you approached the problem in a different way below), but I think you have your $l$ in the wrong place in the above equation.
Now, I know how to approach and get a function for $\omega$ by using energy conservation; however, out of curiosity, would I be able to solve this DE purely mathematically?

In any case, $\omega = \sqrt{\frac{3g(1-\sin{\theta})}{l}}$

I interpret this to mean that if I take the limit as the length goes to infinity, the tree simply has no angular speed and thus cannot fall over at all. However, this seems unsatisfactory as an answer

Well, for what it's worth, your approach seems pretty good to me. Your formula for ω looks correct to me, assuming a uniform gravitational acceleration, g. It's the same as what I came up with.

I'm guessing your approach is adequate for the correct answer. I mean sure, you could replace the uniform gravitational acceleration with Newton's law of universal gravitation, but that wouldn't make the problem any easier. And really, what would it mean for such a "tree" to fall down if it's already infinitely longer than the Earth's diameter? I mean, that's just getting silly. So I don't see anything wrong with your answer the way it is now.
- is it possible to actually derive an explicit function for $\theta(t)$?
I think so, but it wouldn't be a walk in the park.

Using the conservation laws that you used to find ω, you've reduced the original second order nonlinear ordinary differential equation into a first order one (but still nonlinear). So at least there's that.

Tiny-tim's advice might help quite a bit. There is a trig identity that makes it simpler -- an identity that even WolframAlpha doesn't catch on its own.

That said, "simpler" is a relative term. I'm still not making much headway myself on an a simple solution for θ as a function of t.

Last edited:
thanks to both of you, will be working on it right now; and yes, the $l$ should be in the denominator of that expression, a silly mistake!

i think you may have helped me a long time ago with an analysis problem, tiny-tim, so it's great to have your help again :)

This is what I have so far:

$\varphi = \frac{\pi}{2} - \theta$

A bit of finagling with $\sin{\theta}$ returns:

$1 - \sin{\theta} = 1 - \cos{\varphi} = 2(\sin^2{\frac{\varphi}{2}})$

Simplifying further gives me:

$\omega = \sqrt{\frac{6g}{l}} \sin{\frac{\varphi}{2}}$

$\theta' = \sqrt{\frac{6g}{l}}*\sin({\frac{\pi}{4} - \frac{\theta}{2}})$

At this point, this actually reverts back to one of my old attempts that just led to chaos. What I had originally did was let $u = \sin({\frac{\pi}{4} - \frac{\theta}{2}})$ and then force the equation into a first order DE in $u$ and $t$. I made some progress, but could not get an explicit formulation for $\theta$, because I got a pretty ugly integral at one point.

Any ideas on how to proceed? Thanks again guys!

finagle ψ = π - 2θ, and eventually you get cosecψdψ = Cdt

Interesting substitution - may I ask what the intuition was behind it?

In any case,

$\int \csc{\varphi} d\varphi = \int -2C dt$

A quick Wolfram plug, (I'm hating integrating by hand now) gives

$\ln{\tan{\frac{\varphi}{2}}} = -2Ct + K$

I got hell when I was trying to solve for K, since $\varphi = \pi$ at t = 0.

For the sake of experimentation, I continued on by letting K = 0.

A smidge of algebra later, I reached

$\theta = \frac{\pi}{2} - \arctan{(e^{-2\sqrt{\frac{3g}{l}}t})}$

Then I get gibberish when I try to interpret it, i.e: if $l \rightarrow \infty$, with t not equal to 0, then the angle becomes $\frac{\pi}{4}$... this was the point where I knew that I made a big mistake somewhere, but just couldn't find it. I think it's because I assumed the constant of integration to be zero, but oh well.

I appreciate you sticking with this for so long!

hi ttzhou!

(just got up :zzz:)
ttzhou said:
I got hell when I was trying to solve for K, since $\varphi = \pi$ at t = 0.

hmm …

might have something to do with the physical fact that if a tree starts exactly vertical, it will never move!

try again, with a realistic initial condition!

Wow... just wow at that physical interpretation error.

I wrote that post when it was 4 am and I was working on my portfolio for another course - that is the only excuse saving me from jumping off a cliff.

Ok, so for

$\ln{\tan{\frac{\varphi_0}{2}}} = K$ at t = 0.

Let's say $\theta_0 > 0$, so that $\varphi_0 < \pi, \varphi_0 = \pi - 2\theta_0$

$\theta = \frac{\pi}{2} - \arctan{(e^{-2\sqrt{\frac{3g}{l}}t + K})}$

I think I got it!

Setting $t=0$ gives the same angle as if I let $l \rightarrow \infty$ - which I interpret to mean that the tree simply does not move from its initial position.

Thus, in this case, $\theta = \theta_0$

Sigh. That was way too much work for a trivial conclusion.

Thank you so much for catching all of my stupid mistakes, this was a good review of first year.

Last edited:

## 1. What is the concept of an infinitely tall tree falling in a forest?

An infinitely tall tree falling in a forest is a thought experiment that questions the existence of sound in a hypothetical scenario where a tree falls in a forest and there is no one around to hear it. The concept raises philosophical and scientific debates about the nature of sound and its perception.

## 2. Does an infinitely tall tree make a sound when it falls in a forest?

This question is at the core of the thought experiment. The answer depends on one's definition of sound. If sound is defined as vibrations that can be perceived by the human ear, then in the absence of a listener, the tree does not make a sound. However, if sound is defined as the physical phenomenon of vibrations, then the tree does make a sound.

## 3. How does this thought experiment relate to the concept of perception?

The thought experiment raises questions about the role of perception in our understanding of the world. It challenges the idea that something must be perceived in order to exist. It also questions the reliability of our senses and how they shape our perception of reality.

## 4. What are some real-life implications of this thought experiment?

The thought experiment has implications in fields such as philosophy, physics, and psychology. It raises questions about the nature of reality, the limitations of human perception, and the role of consciousness in our understanding of the world.

## 5. Can this thought experiment be proven or disproven?

As a thought experiment, it cannot be proven or disproven in a scientific sense. It serves as a tool for critical thinking and philosophical discussions about the nature of reality and perception. However, advancements in technology and research may provide insights into the mechanisms of sound and perception, shedding light on this concept.

• Introductory Physics Homework Help
Replies
2
Views
275
• Introductory Physics Homework Help
Replies
6
Views
634
• Introductory Physics Homework Help
Replies
5
Views
865
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
20
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
558
• Introductory Physics Homework Help
Replies
2
Views
720
• Introductory Physics Homework Help
Replies
7
Views
280
• Introductory Physics Homework Help
Replies
9
Views
837
• Introductory Physics Homework Help
Replies
1
Views
275