Rotational Mechanics

Consider an infinitesimal mass located a distance $\alpha$ from the end of the rod. The rod makes an angle of $\theta$ with the ground. Write expressions for its mass, position, and velocity. The total rod mass is $m$ , and its total length is $l$ . The horizontal position of the end of the rod is $x_0$ .

$dm = \frac{m}{l} d\alpha \\ x = x_0 - \alpha \, cos\theta \\ y = \alpha \, sin\theta \\ \dot{x} = \alpha \, sin\theta \, \dot{\theta} \\ \dot{y} = \alpha \, cos\theta \, \dot{\theta} \\ v^2 = \alpha^2 \dot{\theta}^2$

Kinetic energy of infinitesimal:

$dE = \frac{1}{2} dm \, v^2 = \frac{1}{2} \frac{m}{l} d\alpha \, \alpha^2 \dot{\theta}^2 = \frac{m \dot{\theta}^2}{2l} \alpha^2 d\alpha$

Total kinetic energy:

$E = \frac{m \dot{\theta}^2}{2l} \int_0^l \alpha^2 d\alpha = \frac{m \dot{\theta}^2 l^2}{6}$

Equate the kinetic energy to the change in gravitational potential energy relative to the start:

$\frac{m \dot{\theta}^2 l^2}{6} = mg \frac{l}{2} (1 - sin\theta) \\ \dot{\theta}^2 = \frac{3g (1- sin\theta)}{l} \\ \dot{\theta} = \sqrt{\frac{3g (1- sin\theta)}{l}}$