Inverted physical pendulum, time taken to hit ground.

[AbigailM]

I'm currently preparing for a classical prelim and am concerned that this problem may not be correct. I'm second guessing myself due to the hint given in the problem, which I did not use. Any help is more than appreciated. A picture of the pendulum is included.

Homework Statement

A thin pencil with length 20 cm is balanced on a desktop, standing on its point so that its angle of inclination $\theta$ with respect to the vertical is nearly zero. A small perturbation is sufficient to tip it over. Suppose that initially, $\theta (0) = 1 x 10^{-17} radians$ and $\dot{\theta(0)} = 0$. Assuming that the pencil point remains fixed, in how many second will it tip over on its side through and angle of $90^{°}$?

Hint: The following integral, accurate for $\theta_{0} < 0.01$ may be useful.

$\int_{\theta_{0}}^{\pi /2} \frac{d \theta}{\sqrt{cos \theta_{0}-cos \theta}} \cong -\sqrt{2} ln \theta_{0} + 1.695$

Homework Equations

$g = 9.8m/s^{2}$
$l = 20cm = .20m$

$I = I_{cm} + mh^{2} ; h = l/2$
$= \frac{1}{12} ml^{2} + \frac{1}{4} ml^{2}$
$= \frac{4}{12} ml^{2}$

The Attempt at a Solution

$T = \frac{1}{2}I \dot{\theta^{2}} = \frac{1}{2}(\frac{4}{12}ml^{2})\dot{\theta^{2}} = \frac{1}{6}ml^{2}\dot{\theta^{2}}$
$U = mg \frac{l}{2} cos \theta$

$L = T - U = \frac{1}{6}ml^{2}\dot{\theta^{2}} - mg\frac{l}{2}cos\theta$

$\frac{\partial L}{\partial \theta} = mg\frac{l}{2}sin\theta$

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) = \frac{1}{3}ml^{2}\ddot{\theta}$

$\ddot{\theta} = \frac{3g}{2l} sin \theta = \omega^{2} sin \theta \hspace{1 in} \omega = \sqrt{\frac{3g}{2l}} = 8.58 \frac{rad}{s}$

We can use small angle approximation so $\ddot{\theta} \approx \omega^{2} \theta$

$\theta (t) = Ae^{\omega t} + Be^{-\omega t} \hspace{1 in} \dot{\theta (t)} = \omega Ae^{\omega t} - \omega Be^{-\omega t}$

$\dot{\theta} (0) = \omega A - \omega B = 0 \hspace{1 in} ; A = B$

$\theta (t) = A \left(e^{\omega t} + e^{- \omega t}\right) = 2Acosh(\omega t)$

$\theta (0) = 2A = 1 x 10^{-17} rad$

$\theta (t) = (1 x 10^{-17})cosh(\omega t)$

$\theta (t) = \frac{\pi}{2} - 1 x 10^{-17} rad \approx \frac{\pi}{2}$

$\theta (t) = \frac{\pi}{2} = (1 x 10^{-17})cosh(\omega t)$

$\frac{\pi}{2 x 10^{-17}} = cosh(\omega t)$

$t = \left(\frac{1}{\omega} \right) cosh^{-1} \left(\frac{\pi}{2 x 10^{-17}}\right) = \frac{1}{8.58} cosh^{-1}\left(\frac{\pi}{2 x 10^{-17}}\right)$

$t \approx 4.7s$

 

[gabbagabbahey]

We can use small angle approximation so $\ddot{\theta} \approx \omega^{2} \theta$

Oh? Is this true throughout the pencil's 90 degree fall?

 

[AbigailM]

Oh I see what you mean! Hrmm $\ddot{\theta} = \omega^{2} sin \theta$ can't be solved analytically can it? Maybe this is where the hint is useful. Thanks for pointing that out.

 

[gabbagabbahey]

Oh I see what you mean! Hrmm $\ddot{\theta} = \omega^{2} sin \theta$ can't be solved analytically can it?

Depends on what you consider to be a solution - $\theta(t)$ can be expressed using a special function called the Jacobi Amplitude function.

But, since what you really want to know is the time it takes for the pencil to fall, you don't actually need to calculate $\theta(t)$. Instead, you can get away with just knowing $\dot{\theta}(t)$ since

$$T=\int_0^T dt = \int_{\theta(0)}^{\theta(T)} \frac{d\theta}{\dot{\theta}}$$

To get $\dot{\theta}(t)$, just use the common trick of multiplying your DE by $\dot{\theta}$ and recognizing that

$$\dot{\theta}\ddot{\theta} = \frac{1}{2}\frac{d}{dt}\dot{\theta}^2$$

 

[AbigailM]

I noticed that your equation for T doesn't have a factor of $2 \pi$, when $\omega = \frac{2 \pi}{T}$. In the Jacobi method on Wikipedia they don't have it there either. What am I missing? Thanks again.

 

[AbigailM]

Oh hahahah nevermind. It was staring me right in the face, $\frac{dt}{d \theta}$ , move the $d \theta$ over and integrate both sides.

 

[gabbagabbahey]

Oh hahahah nevermind. It was staring me right in the face, $\frac{dt}{d \theta}$ , move the $d \theta$ over and integrate both sides.

Exactly. Of course, if treating differentials like fractions in this way bothers you, you could always just use conservation of energy to find the same result.

 

[AbigailM]

Cool, I was able to get the solution. Thanks again for all the help gabbagabbahey.