# To prove mathematically that a tip standing vertically will fall exponentially as soon as it deviates from its original position. (Homework Problem) [closed]

This is one of my questions from Homework for Quantum Mechanics. I know the answer quantitatively but cannot find any mathematical explanation.

Question: Consider a thin needle of total mass m and length L. If it has an “atomically thin” endpoint but is otherwise macroscopic in size, you would not expect to be able to stand the needle exactly on its endpoint. Describe the classical motion of the needle when |θ| << 1 (here you should define θ = 0 to be when the needle is perfectly upright). Show that the needle will begin to tip over exponentially quickly if it deviates even slightly from θ = 0. In your derivation, you can assume there’s friction where the needle connects to the ground, such that the needle rotates rigidly about its endpoint.

My Approach: I know that needle will fall as soon as it deviates slightly from θ = 0 because its center of mass will deviate and gravity will pull it down. How do I show this mathematically?

If you follow my comment's advice, you obtain an equation of motion of the form $$\ddot{\theta}=\omega^2\sin\theta$$, where $$\omega>0$$ is a multiple of $$\sqrt{g/L}$$ you can compute as an exercise. Its value is conceptually unimportant, though you might be expected to determine it due to its appearance in the exponential tipping discovered below.
Although $$\theta(t)\equiv0$$ is a solution, it's unstable because of the way a small deviation evolves. While $$\theta$$ is small, $$\ddot{\theta}\approx\omega^2\theta$$. This approximation implies $$\theta=Ae^{\omega t}+Be^{-\omega t}$$ for constants $$A,\,B$$. Taking $$\theta=0$$ at $$t=0$$, $$A=-B\ne0$$, so eventually $$\theta\approx Ae^{\omega t}$$. This exponential growth is what you were asked to prove.