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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?

Note by Advitiya Brijesh
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 :

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

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@Jatin Yadav How did you solve the differential equation? Krishna Jha · 3 years, 10 months ago

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@Krishna Jha Something known as wolfram :)

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

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@Jatin Yadav 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. Snehal Shekatkar · 3 years, 10 months ago

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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. Snehal Shekatkar · 3 years, 10 months ago

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