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# Balancing a pencil!

Consider a pencil that stands upright on its tip and then falls over. Let’s idealize the pencil as a mass $$m$$ sitting at the end of a mass less rod of length $$l$$

Assume that the pencil makes an initial (small) angle $$\theta_{0}$$ with the vertical, and that its initial angular speed is $$\omega_{0}$$. The angle will eventually become large, but while it is small (so that $$\sin \theta \approx \theta$$), what is $$\theta$$ as a function of time?

3 years, 10 months ago

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First tell whether slipping is taking place or not, and if it is taking place , is the surface frictionless.

If slipping is not taking place , then , take moment about lowest point.

$$mglsin\theta$$ = $$ml^2\alpha$$ [Using : $$sin\theta = \theta$$]

$$\Rightarrow \frac{d^2\theta}{dt^2} = \frac{g}{l}\theta$$

Solve this 2nd order differential equation to get :

$$\Rightarrow \theta = \frac{\theta_{0} + w_{0}\sqrt{\frac{l}{g}}}{2} e^{\sqrt{\frac{g}{l}}t} + \frac{\theta_{0} - w_{0}\sqrt{\frac{l}{g}}}{2} e^{-\sqrt{\frac{g}{l}}t}$$ · 3 years, 10 months ago

How did you solve the differential equation? · 3 years, 10 months ago

Something known as wolfram :)

Actually, 2nd order diff. equations aren't taught in class 12th. · 3 years, 10 months ago

This is a second order equation with constant coefficients. So we simply need to find roots of auxiliary equation. Solution that I wrote below is equivalent to your solution after we rearrange the terms. · 3 years, 10 months ago

I did not recheck my calculation so answer that I am posting might be wrong. My answer is $$\theta(t)=\theta_{0} \cosh(\omega t)+\frac{\omega_{0}}{\omega}\sinh(\omega t)$$ for small $$\theta$$. Here $$\omega=\sqrt{\frac{g}{l}}$$ and $$\theta_{0}$$ is an initial angular displacement. · 3 years, 10 months ago