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constaint to a rolling object |
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| Jan4-09, 05:03 PM | #1 |
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constaint to a rolling object
1. The problem statement, all variables and given/known data
A cylinder, radius R, mass M and moment of inertia I, is rolling on a horizontal surface without slipping and a constant force F is exerted on the center of the cylinder horizontally to the right. Try to find the friction by solving the Lagrangian equation. 2. The attempt at a solution Since the object is rolling without slip, there must be a friction force exerted at the contact point to the left. When we go to the Lagrangian mechanics, let's call the center of the cylinder (x, y) and the angle made by a specific point on the edge of the cylinder to the vertical line passing through center [tex]\theta[/tex]. Since the object is moving on the horizontal level, no potential. The constraint for no-slipping action is [tex]f = R\theta - x =0[/tex] The Lagrangian is [tex] L = \frac{M}{2}\dot{x}^2 + \frac{I}{2}\dot{\theta}^2[/tex] The Lagrangian equation is [tex] \frac{d}{dt}\frac{\partial L}{\partial\dot{x}} = \lambda \frac{\partial f}{\partial x} = -\lambda [/tex] and [tex] \frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}} = \lambda \frac{\partial f}{\partial \theta} = R\lambda [/tex] where [tex]\lambda[/tex] is the Lagrange undetermined multiplier. From these equations, I get [tex]\lambda=F[/tex], but how do I relate undetermined multiplier to friction? If we start from newton's law, it is very easy to solve for the friction, which is obvious written in terms of F. But from my solution above, seems cannot get the same form. What's going on with my solution? |
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