- #1

C. Darwin

- 18

- 0

## Homework Statement

I've got a rigid rod, length [itex]2a[/itex], held at an angle [itex]\alpha[/itex] above a smooth horizontal surface. Its end is in contact with the ground. My two generalized coordinates are [itex]x[/itex], the horizontal position of the center of the rod and [itex]\theta[/itex] the angle between the rod and the vertical.

I'm trying to find the kinetic energy of the rod. I'm not sure if I'm doing it right.

## Homework Equations

## The Attempt at a Solution

As of now, this is what I've done:

[tex]KE = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\dot{\theta}^2 = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}(\frac{1}{3}m(2a)^2)\dot{\theta}^2 = \frac{1}{2}m\dot{x}^2 + \frac{2}{3}ma^2\dot{\theta}^2[/tex]

Something doesn't seem right though, because I think that [itex]\dot{x}[/itex] should be zero. Carrying out the Lagrangian doesn't seem to show this.