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Constants of motion in Lagrangian

  1. Feb 4, 2013 #1
    There's a specific problem I'm doing, but this is more of a general question. The setup is a cylinder of mass [itex]m[/itex] and radius [itex]R[/itex] rolling without slipping down a wedge inclined at angle [itex]\alpha[/itex] of mass [itex]M[/itex], where the wedge rests on a frictionless surface. I've made the Cartesian axis centred at the initial position of the centre of mass of the cylinder. Then, [itex]x_{m}[/itex] and [itex]x_{M}[/itex] are the horizontal distances of each mass (measured from the y-axis, so both quantities are positive), [itex]y[/itex] is the vertical position of the centre of mass of the cylinder and [itex]\theta[/itex] is the angle through which the cylinder has rolled. This makes it very easy to write down the Lagrangian:
    \mathcal{L} = \mathcal{K} - \mathcal{U} = \frac{1}{2}M\dot{x}_{M}^{2} + \frac{1}{2}m(\dot{x}_{m}^{2} + \dot{y}^{2}) + \frac{1}{4}mR^{2}\dot{\theta}^2 + mgy
    As of right now, the Lagrangian "doesn't know" about the fact that the cylinder's constrained to be stuck to the wedge. Basic trigonometry gives the relation [itex]y = d \sin(\alpha)[/itex], where [itex]d[/itex] is the distance travelled by the cylinder. But then, the cylinder rolls without slipping, so [itex]d = R\theta[/itex] and therefore [itex]y = R \sin(\alpha) \theta[/itex]. But then we also have that [itex]y = (x_{m} + x_{M})\tan(\alpha)[/itex] by an identical argument.

    Here's where my question comes in. Do I have to sub in both constraints when determining the constants of motion, and does it matter which variables I eliminate? As it stands, the Lagrangian doesn't depend on [itex]\theta[/itex], [itex]x_{m}[/itex] or [itex]x_{M}[/itex], which leads one to believe that all those associated momenta are conserved, but the issue is that [itex]y[/itex] technically does depend on on those variables. It seems weird that no momenta would be conserved.
  2. jcsd
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