Does the Center of Mass Have Horizontal Velocity in a Frictionless System?

AI Thread Summary
In a frictionless system with two masses connected by a massless bar, the center of mass will not have horizontal velocity when the system falls due to gravity. Since gravity is the only external force acting on the system, the center of mass moves straight down, resulting in a horizontal component of velocity equal to zero. The formula for the horizontal component of velocity confirms this, as it yields a value of zero. The discussion also addresses a procedural issue regarding thread management, which was resolved by a moderator. Overall, the analysis of the center of mass's motion in this scenario is deemed correct.
cacofolius
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Hi. There are two masses connected by a massless bar, and from the unstable equilibrium position shown in the figure is slightly inclined so it falls down, being the final state of the system that both masses are in contact with the surface. There is no friction between the floor and m2. The problem is to find the horizontal component of the velocity of the center of mass.

I think that since there is no friction, and gravity being the only outside force that acts on the system, the center of mass will fall straight down, so there will be no horizontal velocity of it. So is this right?

Vx = \frac{m_{1}V_{x1}+m_{2}V_{x2}}{m_{1}+m_{2}} = 0

two  masses.jpg


Thanks in advance.
 

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I thought I was on Homework, please excuse me. I'm looking for how to move/delete the thread. Edit: I just found out I can't. All apologies.
 
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No problem, I moved it for you. Usually we require the use of a template, but you essentially provided the key elements of the template anyway. So it should be OK.

Your analysis seems correct to me.
 
Thank you very much!
 
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