Why Doesn't Constant Center of Mass Velocity Reduce Degrees of Freedom?

[deuteron]

Homework Statement

What are the constraints of the system?

Relevant Equations

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Consider the above system, where both the wedge and the mass can move without friction. We want to get the equations of motion for the both of them using Lagrangian formalism, where the constraints in the solution sheet are given as:

$$y_2=0$$
$$\tan \alpha=\frac {y_1}{x_1-x_2}$$

However, since there is no net force on the system in the $x-$direction, can't we also say that the center of mass has constant velocity, where we can choose an inertial frame in which $\dot x_{cm}=0$, and thus

$$\dot x_{cm} =\frac {m_1\dot x_1 +m_2\dot x_2}{m_1+m_2}=0\ \Rightarrow\ m_1\dot x_2 = -m_2\dot x_2$$

which would reduce the number of degrees of freedom to $1$?

 

[PeroK]

Okay, but you were asked for the equations of motion.

 

[PeroK]

Homework Statement: Find the equations of motion
Relevant Equations: .

View attachment 332662

Consider the above system, where both the wedge and the mass can move without friction. We want to get the equations of motion for the both of them using Lagrangian formalism, where the constraints in the solution sheet are given as:

$$y_2=0$$
$$\tan \alpha=\frac {y_1}{x_1-x_2}$$

However, since there is no net force on the system in the $x-$direction, can't we also say that the center of mass has constant velocity, where we can choose an inertial frame in which $\dot x_{cm}=0$, and thus

$$\dot x_{cm} =\frac {m_1\dot x_1 +m_2\dot x_2}{m_1+m_2}=0\ \Rightarrow\ m_1\dot x_2 = -m_2\dot x_2$$

which would reduce the number of degrees of freedom to $1$?

Not really. The system has two degrees of freedom, but under gravity alone, the motion of the small block down the wedge is related to the motion of the wedge.

 

[deuteron]

Okay, but you were asked for the equations of motion.

The constraints were a sub-question, I changed it now, sorry :')

 

[deuteron]

Not really. The system has two degrees of freedom, but under gravity alone, the motion of the small block down the wedge is related to the motion of the wedge.

Why is the relationship not a constraint that reduces the number of degrees of freedom? Wouldn't we know the position of the wedge if we knew the position of the mass, using $\dot x_{cm}=0$?

 

[PeroK]

Why is the relationship not a constraint that reduces the number of degrees of freedom? Wouldn't we know the position of the wedge if we knew the position of the mass, using $\dot x_{cm}=0$?

Suppose you put a small block on a larger rectangular block on a smooth surface. The system, under gravity alone, does not move. That doesn't mean it has zero degrees of freedom. The blocks are still free to move independently.