Infinitely tall tree falling in a forest

[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

<br /> \tau = I\alpha<br />

<br /> \tau = Mg\cos{\theta}(l/2)\<br />

<br /> I = \frac{1}{3} Ml^2\<br />

The Attempt at a Solution

<br /> I\alpha = Mg\cos{\theta} (l/2)<br />

<br /> \alpha = \frac{3gl}{2} \cos{\theta}<br />


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.

 

[tiny-tim]

hi ttzhou!

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

 

[collinsmark]

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

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

<br /> \tau = I\alpha<br />

<br /> \tau = Mg\cos{\theta}(l/2)\<br />

<br /> I = \frac{1}{3} Ml^2\<br />

The Attempt at a Solution

<br /> I\alpha = Mg\cos{\theta} (l/2)<br />

<br /> \alpha = \frac{3gl}{2} \cos{\theta}<br />

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.

 

[ttzhou]

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 :)

 

[ttzhou]

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&#039; = \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!

 

[tiny-tim]

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

 

[ttzhou]

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!

 

[tiny-tim]

hi ttzhou!

(just got up :zzz:)

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!

 

[ttzhou]

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 &gt; 0, so that \varphi_0 &lt; \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.