Balancing a pencil!

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

Assume that the pencil makes an initial (small) angle θ0\theta_{0} with the vertical, and that its initial angular speed is ω0\omega_{0}. The angle will eventually become large, but while it is small (so that sinθθ\sin \theta \approx \theta), what is θ\theta as a function of time?

#Physics #Mechanics

Note by Advitiya Brijesh
7 years, 7 months ago

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13 votes

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Comments

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.

mglsinθmglsin\theta = ml2αml^2\alpha [Using : sinθ=θsin\theta = \theta]

d2θdt2=glθ\Rightarrow \frac{d^2\theta}{dt^2} = \frac{g}{l}\theta

Solve this 2nd order differential equation to get :

θ=θ0+w0lg2eglt+θ0w0lg2eglt\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 - 7 years, 7 months ago

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How did you solve the differential equation?

Krishna Jha - 7 years, 7 months ago

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Something known as wolfram :)

Actually, 2nd order diff. equations aren't taught in class 12th.

jatin yadav - 7 years, 7 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 - 7 years, 7 months ago

I did not recheck my calculation so answer that I am posting might be wrong. My answer is θ(t)=θ0cosh(ωt)+ω0ωsinh(ωt)\theta(t)=\theta_{0} \cosh(\omega t)+\frac{\omega_{0}}{\omega}\sinh(\omega t) for small θ\theta. Here ω=gl\omega=\sqrt{\frac{g}{l}} and θ0\theta_{0} is an initial angular displacement.

Snehal Shekatkar - 7 years, 7 months ago
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