Masses on an inclined plane and their Lagrangian

[ShayanJ]

Consider the configuration below shown in the attached picture!
The wedge can slide on the inclined plane and the cube on the wedge.Their motion is described by $x_1$ and $x_2$ respectively. There is no friction and the inclined plane doesn't move.
Here's the Lagrangian of the system:
$\mathcal L=\frac{1}{2} M \dot{x}_1^2+\frac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2-2 \dot{x}_1\dot{x}_2 \cos\theta)+(M+m)gx_1 \sin\theta$
And the Euler-Lagrange equations are:
$M\ddot{x}_1+m(\ddot{x}_1-\ddot{x}_2\cos\theta)=(M+m)g\sin\theta\\ m(\dot{x}_2-\dot{x}_1\cos\theta)=const$
I want to know,what is the constant of motion appearing above?
One may expect it to be the component of total linear momentum parallel to the ground but it can't be that because it doesn't involve M.So what it is?
Thanks

 

[Spinnor]

Don't know the answer to your question but if there is no friction won't both masses accelerate together?

 

[ShayanJ]

Of course the motion of M may cause m to move, whether there is friction or not.

 

[dauto]

It's the horizontal component of the momentum of the mass m (The top block), obviously...

 

[ShayanJ]

It's the horizontal component of the momentum of the mass m (The top block), obviously...

No, the horizontal component of the momentum of the mass m is $\dot{x}_2 + \dot{x}_1\cos\theta$.
Anyway, still another question remains. So where is the horizontal component of the momentum of the wedge?That should remain constant too.

 

[dauto]

No, the horizontal component of the momentum of the mass m is $\dot{x}_2 + \dot{x}_1\cos\theta$.

You must've made a sign mistake somewhere than because that's the one quntity that 's conserved in this problem

Anyway, still another question remains. So where is the horizontal component of the momentum of the wedge?That should remain constant too.

No because the normal force between the wedge and the fixed incline pushes the wedge sideways.

 

[nasu]

Of course the motion of M may cause m to move, whether there is friction or not.

If there is no friction, the small block will have no horizontal motion. Unless it had some initial horizontal speed.

And are you sure you have that minus sign in the kinetic energy?

 

[dauto]

Yes, there is a sign mistake in your Lagrangian. the term with a cosine in it should be positive.

 

[ShayanJ]

You must've made a sign mistake somewhere than because that's the one quntity that 's conserved in this problem

Not, obviously both speeds have a common positive direction which means they should be added to give the speed relative to the inclined plane.

No because the normal force between the wedge and the fixed incline pushes the wedge sideways.

The normal force is equal and opposite to $(M+m)g\cos\theta$, and so is canceled out.Otherwise there could be an acceleration normal to the inclined plane which causes the wedge to fly away!

 

[dauto]

Not, obviously both speeds have a common positive direction which means they should be added to give the speed relative to the inclined plane.

The sign mistake is in the Lagrangian

The normal force is equal and opposite to $(M+m)g\cos\theta$, and so is canceled out.Otherwise there could be an acceleration normal to the inclined plane which causes the wedge to fly away!

You're getting confused. We're talking about horizontal components. $(M+m)g\cos\theta$ isn't horizontal. Gravity has no horizontal component.

 

[ShayanJ]

If there is no friction, the small block will have no horizontal motion. Unless it had some initial horizontal speed.

Why do you tend backward when the bus you're standing in accelerates forward?

And are you sure you have that minus sign in the kinetic energy?

Yes, there is a sign mistake in your Lagrangian. the term with a cosine in it should be positive.

Yeah, Looks like that should be positive.
But still the problem remains that where is the x component of the momentum of the wedge?
Or why isn't that conserved?

 

[ShayanJ]

The sign mistake is in the Lagrangian

Yeah... I got that.

You're getting confused. We're talking about horizontal components. $(M+m)g\cos\theta$ isn't horizontal. Gravity has no horizontal component.

In fact we're confusing two different horizontals!
Gravity has no component parallel to the ground. But it has a component parallel to the inclined plane. When you talk about the normal force from the inclined plane to the wedge, the weight of the wedge does make a contribution.

 

[dauto]

Why do you tend backward when the bus you're standing in accelerates forward?




Yeah, Looks like that should be positive.
But still the problem remains that where is the x component of the momentum of the wedge?
Or why isn't that conserved?

I already answered that question. You may not have believed my answer but it is in fact correct.
You need to hone your intuition a bit more.

 

[dauto]

Yeah... I got that.

In fact we're confusing two different horizontals!
Gravity has no component parallel to the ground. But it has a component parallel to the inclined plane. When you talk about the normal force from the inclined plane to the wedge, the weight of the wedge does make a contribution.

When I say horizontal I usually mean the actual horizontal.

 

[ShayanJ]

I already answered that question. You may not have believed my answer but it is in fact correct.
You need to hone your intuition a bit more.

If you mean this:

No because the normal force between the wedge and the fixed incline pushes the wedge sideways.

I already answered that too. If it wasn't canceled, it would give the wedge an acceleration normal to the inclined plane! It is canceled by the component of the weight of the wedge normal to the inclined plane.

 

[dauto]

If you mean this:

I already answered that too. If it wasn't canceled, it would give the wedge an acceleration normal to the inclined plane! It is canceled by the component of the weight of the wedge normal to the inclined plane.

That's ridiculous now. We're talking about the HORIZONTAL component of the normal which cannot be canceled by gravity. You're stuck thinking about perpendicular and parallel components to the incline. I'm talking about HORIZONTAL component.

 

[dauto]

Another way to look at it is that the normal is canceled by one component of gravity, but the other component of gravity accelerates the wedge down the ramp and that acceleration has a horizontal component. Either way there is a horizontal acceleration.

 

[ShayanJ]

That's ridiculous now. We're talking about the HORIZONTAL component of the normal which cannot be canceled by gravity. You're stuck thinking about perpendicular and parallel components to the incline. I'm talking about HORIZONTAL component.

Ohh...dammit...yeah...sorry. This was a problem with the word. It was like my mind wasn't interpreting the word horizontal correctly. That's a problem when you're not a native speaker. Sorry.You're right.

 

[nasu]

Why do you tend backward when the bus you're standing in accelerates forward?

For the horizontal direction, the small block moves (accelerates) in respect to the wedge but not in respect to the ground or the inclined plane.

 

[ShayanJ]

For the horizontal direction, the small block moves (accelerates) in respect to the wedge but not in respect to the ground or the inclined plane.

Yeah,That's the thing one of the equations is telling us. Its horizontal component of momentum is conserved. I was talking about motion w.r.t. the wedge.