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.