- #1
K29
- 108
- 0
Homework Statement
I am trying to get an equation of motion for the following (seemingly simple) setup. You place on a rod on a pivot. The rod's centre of mass is precisely over the pivot. Think of balancing a ruler horizontally on your finger. Gravity, of course acts downward.
The Attempt at a Solution
The equation of motion is surely not obtained by working out the Lagrangian using the Kinetic energy as [tex]T=\frac{1}{2}I\omega^{2}[/tex]. where [tex]\omega = \dot{\alpha}[/tex] Alpha is just the angle rotated from the x-axis which I aligned along the rod when its horizontally at rest equilibrium. Clearly there is no potential energy to speak of here since the centre of mass is fixed.
After solving Lagrange's equation of motion it can quickly be seen that:
[tex]\ddot{\alpha}=0[/tex] which just can't be the case, since if you think about balancing a ruler on your finger, if you push one end down slightly it oscillates.
Thus I suppose the right form of Lagranges equation to be used is
[tex]\frac{d}{dt}\frac{\partial T}{\partial \dot{\alpha}}-\frac{\partial T}{\partial\alpha}=Q_{\alpha}[/tex]
where Q is the generalised component of force.
But I can't seem to figure out how to formulate a Force of constraint s.t. [tex]Q_{\alpha}[/tex] comes out. As you lower one side of the rod gravity acts on the centre of masses of both sides such that the net force on one side is greater than the other causing it to oscillate about the horizontal axis? Am I allowed to consider the two halves of the rod having their own centres of mass? (I have tried this and still get an answer that doesn't match reality)
This is probably not even the right way of doing things.
Last edited: