Two cylinder conected, rolling down an inclined plane

[alejandrito29]

two cylinder conected, by two roads, rolling down an inclined plane. Find equation of motion and tension of roads.

I know that the lagrangian is:

$$0.5 m \dot{x}^2+0.5 I_1 \dot{\theta}^2++0.5 I_1 \dot{\theta}^2+mgx\sin \alpha + mg(x+l)\sin \alpha$$

¿ is good my lagrangian?, ¿how is present the tension of roads at the lagrangian?

 

[BvU]

two cylinders connected, by two rods, rolling down an inclined plane. Find equation of motion and tension of rods.

I know that the lagrangian is:

$$0.5 m \dot{x}^2+0.5 I_1 \dot{\theta}^2++0.5 I_1 \dot{\theta}^2+mgx\sin \alpha + mg(x+l)\sin \alpha$$

¿ is good my lagrangian?, ¿how is present the tension of roads at the lagrangian?

I take it you missed an $0.5 m \Bigl ( {d\over dt} \left ({x+l} \right) \Bigr) ^2$ between the + + but intend it to be there ?
I take it your positive x - coordinate is to the right ?
I take it your g is a positive number ?
I take it your cylinders have the same diameters?
I take it your cylinders have the same mass?
I take it your cylinders are both massive? (So the same I as well. But then I wonder why the difference in representation in the figure!)

In that case your Lagrangian is good.

The constraint forces are not represented in the Lagrangian. That's the fun of this approach with generalized coordinates.

In fact the Lagrangian approach is only justified if there only is rolling (or if there is zero friction, so only slipping), not if there is a combination of rolling and slipping. (Constraints have to be holonomic). The "rolling only" removes one of your coordinates.

 

[alejandrito29]

I take it you missed an $0.5 m \Bigl ( {d\over dt} \left ({x+l} \right) \Bigr) ^2$ between the + + but intend it to be there ?
I take it your positive x - coordinate is to the right ?
I take it your g is a positive number ?
I take it your cylinders have the same diameters?
I take it your cylinders have the same mass?
I take it your cylinders are both massive? (So the same I as well. But then I wonder why the difference in representation in the figure!)

In that case your Lagrangian is good.

The constraint forces are not represented in the Lagrangian. That's the fun of this approach with generalized coordinates.

In fact the Lagrangian approach is only justified if there only is rolling (or if there is zero friction, so only slipping), not if there is a combination of rolling and slipping. (Constraints have to be holonomic). The "rolling only" removes one of your coordinates.

Thank, my lagrangian is then

$$0.5 m \dot{x}^2+0.5 m (\dot{x+l})^2+0.5 I_1 \dot{\theta}^2++0.5 I_1 \dot{\theta}^2+mgx\sin \alpha + mg(x+l)\sin \alpha$$ ?

and then

how I will calculate the tension?

there is a contraint (whit Lagrange multiplier) for the tension?, how i write this contraint?

 

[BvU]

$x_2 - x_1 = l$ would be a way to write the constraint. Then the Langrangian should be rewritten in terms of x₁ and x₂ as generalized coordinates. You have one Langrange equation more and a constraint equation. The corresponding Lagrange multiplier is the force of constraint (i.e. the tension in the rod).

You have not read my questions, or do you decline to answer any of them ?

What book are you using ? My Goldstein, Classical Mechanics (1980) has a nice example for a single hoop rolling down an inclined plane where the friction force of constraint is evaluated by NOT involving the $rd\theta=x$ constraint for rolling in the generalized coordinates.

 

[paranormal]

Thank you for your question. I would like to clarify a few points before providing a response. First, I would need more information about the setup of the cylinders and the roads in order to accurately determine the equation of motion and tension. For example, the mass and radius of the cylinders, the angle of the inclined plane, and the length and material of the roads would all affect the equations.

Assuming that the cylinders are connected by rigid rods and are rolling without slipping, the motion can be described using the Lagrangian formulation. The Lagrangian is a function that describes the system's dynamics and is defined as the difference between the kinetic and potential energies of the system. In this case, the kinetic energy would include the translational and rotational energies of the cylinders, while the potential energy would include the gravitational potential energy of the system.

The Lagrangian you have provided seems to be a reasonable starting point, but it is missing some key terms. Firstly, the rotational energy should include the rotational inertia of both cylinders, as well as the linear velocity of the center of mass. Additionally, the potential energy should include the distance between the cylinders (l) and the angle of the inclined plane (α). The Lagrangian should also take into account any external forces acting on the system, such as friction or air resistance.

As for the tension of the roads, it can be incorporated into the Lagrangian by considering it as an external force acting on the system. The tension would depend on the type of connection between the cylinders and the roads and would need to be taken into account in the equations of motion.

In summary, in order to accurately determine the equation of motion and tension of the roads, more information about the setup of the system is needed. Once all the relevant parameters are known, the Lagrangian formulation can be used to describe the dynamics of the system and to calculate the equations of motion and tension of the roads.