Today, a friend and I solved a classical mechanics problem using group theory. The problem was the following:
Around a circumference, there are $N$ children evenly spaced. In the center, there is a tire. Each children pulls the tire by a rope with equal forces. Is the resultant force always zero?
My friend associated each force vector with a complex root of unity, and using the fact that the group of the $N$-roots of unity is cyclic, showed the identity: $$1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1} = 0$$ which equals saying that the resultant force is zero. I considered the set $\Omega$ of all permutations of the children, with a group action $\pi\colon \mathbb{Z}/N\mathbb{Z}\times \Omega \to \Omega$ by cyclic permutations. I considered the "force" function $f\colon \Omega \to \mathbb{R}^2$ which associated each system with it's resultant force vector. I also considered the group action $\varphi\colon \mathbb{Z}/N\mathbb{Z} \times \mathbb{R}^2 \to \mathbb{R}^2$ by rotations of the plane. Then, I argumented the following identities for all $S \in \Omega, x \in \mathbb{Z}/N\mathbb{Z}$: $$f(\pi(x, S)) = \varphi(x, f(S))$$ $$f(\pi(x, S)) = f(S)$$ which implied $f(S)$ is a fixed point of $\varphi$, and thus is always the zero vector.
I never expected a application of group theory to classical mechanics (though I know about it's uses in crystallography). Are there other well know examples of this?