# Velocity of Center of Mass

1. Oct 22, 2014

### cacofolius

• Moved from general physics forum
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?

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

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2. Oct 22, 2014

### cacofolius

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.

Last edited: Oct 22, 2014
3. Oct 22, 2014

### Staff: Mentor

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.

4. Oct 24, 2014

### cacofolius

Thank you very much!