Mass on Right: Why Not p*[(b/2)-x]?

[Shackleford]

Why is the mass on the right not p*[(b/2)-x]? I understand initially on the left and right side the distance is b/2. When it changes, it changes by x, not x/2.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20101122_203625.jpg?t=1290480153

 

[Mindscrape]

So we start off with each side hanging a distance b/2. When the chain falls a distance of x, half of it goes to the left side, and half to the right side. So at a time t, the amount of chain on the right side is

b-(b/2+x/2)=(b-x)/2

since mass is just p*(length of chain) that gives p*(b-x)/2 for the mass of the right chain. I don't know about you right now, but I remember that I used to get seriously confused about when to use CM and when I needed total mass for these types of problems.