Slipping Tipping Rod

Relevant wiki: Problem solving 2D

Since all forces acting on the rod are vertical, the centre of mass of the rod does not move horizontally. Thus, if the rod has length $a$ , and it makes an angle of $\theta$ with the vertical, then its centre of mass must be a height $\tfrac12a\cos\theta$ above the ground. Thus the rod has kinetic energy $\tfrac12m\big(\tfrac12a\sin\theta \,\dot\theta\big)^2 + \tfrac12 \times \tfrac1{12}ma^2 \dot\theta^2 \; = \; \tfrac{1}{24}ma^2(1 +3\sin^2\theta)\dot\theta^2$ and so conservation of energy tells us that $\tfrac{1}{24}ma^2(1 + 3\sin^2\theta)\dot\theta^2 + \tfrac12mga\cos\theta \; = \; \tfrac12mga\cos\theta_0$ where $\theta_0 = 1^\circ$ is the initial inclination of the rod. Thus the time to reach the horizontal is $T \; = \; \int_{\theta_0}^{\frac12\pi} \sqrt{\frac{a(1 + 3\sin^2\theta)}{12g(\cos\theta_0 - \cos\theta)}}\,d\theta \; =\; \boxed{0.803388} \;\mathrm{s}$ with $a = 1$ m and $g = 10 \,\mbox{m s}^{-2}$ .