How Does Friction Affect a Rolling Cylinder Under Horizontal Force?

[KFC]

Homework Statement

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 $$\theta$$. Since the object is moving on the horizontal level, no potential. The constraint for no-slipping action is

$$f = R\theta - x =0$$

The Lagrangian is

$$L = \frac{M}{2}\dot{x}^2 + \frac{I}{2}\dot{\theta}^2$$

The Lagrangian equation is

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{x}} = \lambda \frac{\partial f}{\partial x} = -\lambda$$

and

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}} = \lambda \frac{\partial f}{\partial \theta} = R\lambda$$

where $$\lambda$$ is the Lagrange undetermined multiplier. From these equations, I get $$\lambda=F$$, 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?

 

[Jhero]

As a fellow scientist, I would like to offer some insights and suggestions on your approach to this problem. First, your use of Lagrangian mechanics is a good start, as it can provide a more elegant and comprehensive solution compared to Newton's laws. However, there are a few things that need to be addressed in your solution.

1. The constraint equation you have written, f = R\theta - x = 0, is not entirely correct. This equation implies that the contact point between the cylinder and the surface is fixed, which is not the case in this problem. The correct constraint equation should be f = R\theta - x = R\phi, where \phi is the angle between the contact point and the vertical line passing through the center of the cylinder. This will be important in your later calculations.

2. In your Lagrangian, you have only considered the kinetic energy terms, but you have not included the potential energy term. While it is true that the potential energy is zero in this problem, it is still important to include it in your Lagrangian to maintain the consistency of the equations.

3. Your Lagrangian equations are not entirely correct. The correct equations should be:

\frac{d}{dt}\frac{\partial L}{\partial\dot{x}} = \frac{\partial L}{\partial x} + \lambda \frac{\partial f}{\partial x} = -\lambda

and

\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}} = \frac{\partial L}{\partial \theta} + \lambda \frac{\partial f}{\partial \theta} = R\lambda

Note that in addition to the undetermined multiplier \lambda, there are also terms involving the partial derivatives of the Lagrangian with respect to x and \theta. These terms are necessary to account for the constraint and ensure that the resulting equations are correct.

4. Finally, to relate the undetermined multiplier \lambda to friction, you can use the fact that the friction force is equal in magnitude and opposite in direction to the component of the applied force F that is tangential to the surface. This can be written as F_{friction} = -\lambda R\phi. Therefore, solving for \lambda will give you the value of the friction force.

I hope these suggestions will help you in your solution. Keep up the good work!