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

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SUMMARY

The mass on the right side of the chain is calculated as p*(b-x)/2, where p represents the linear mass density and b is the initial length of the chain. As the chain falls a distance of x, the distribution of mass changes, with half of the chain's length contributing to each side. The confusion arises from determining when to use center of mass (CM) versus total mass in such problems. This clarification is crucial for accurately solving dynamics involving hanging chains.

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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.

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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.
 

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