A uniform chain problem (Mechanics)

[assaftolko]

Homework Statement

A uniform chain with a mass of M and a length of L is put on a horizantal table in a way that half of it is hanging from the air. At the moment t=0 the chain is released from rest.

1. What is the speed of the chain as its tip will leave the table?
2. Answer question 1 again only this time the chain has a kinetic coefficients of friction q with the table.

Homework Equations

Work & energy theorem

The Attempt at a Solution

Well I tried to take the center of mass point of the half chain that's hanged in the air. If I take the height of the table to be my refference height then in t=0 you get that this cener of mass point is located at -L/4 with respect to the height of the table. When all of the chain is in the air (just as its last part leaved the table) this point is now located at -3L/4 since L/2 of chain dropped from the table during this period of time. But in the solution they say that for the final situation I had to look at the new center of mass point, which is now at the middle of the chain which means it's at -L/2 with respect to the table. I don't understand it - aren't you suppose to determine the potential energy difference between two situations through a fixed refference height and a fixed point at the body I'm looking at? How does it help me that for t=0 I looked at one point of the chain and in the final stage I looked at another point of the chain?

 

[Sourabh N]

Each point on the chain has gravitational as well as internal forces (which are rather complicated) acting. Therefore, it is better to consider the motion of the special point "center of gravity", at which there is only net force, the gravitational force.

If your book uses the term "center of mass" instead of "center of gravity", they are being sloppy. The position of "center of mass" of the uniform chain, being independent of the influence of the forces acting, is at the mid point of the chain. Whereas "center of gravity" in this question moves.

 

[assaftolko]

Each point on the chain has gravitational as well as internal forces (which are rather complicated) acting. Therefore, it is better to consider the motion of the special point "center of gravity", at which there is only net force, the gravitational force.

If your book uses the term "center of mass" instead of "center of gravity", they are being sloppy. The position of "center of mass" of the uniform chain, being independent of the influence of the forces acting, is at the mid point of the chain. Whereas "center of gravity" in this question moves.

Ok so basiclly how do I find this center of gravity point? Is it simply the "center of mass" point for the part of the chain that is in the air and is subjected to gravitational force without the ballancing normal force of the table? (and therefore in t=0 it's located at -L/4 and in the final stage it's located at -L/2 like in the solution).

Also - if you follow what you said but determine the refference height to be h=0 at the bottom of the half chain that's hanged in the air when t=0 (meaning at distance of L/2 down from the table) you get that for t=0:
E(t=0) = 0.5Mg*L/4 (with respect to the bottom point of the half chain that is hanged in the air)
E(final) = 0.5Mv^2 + Mg*0 (Because now the center of gravity point is at the middle of the chain which mean it's L/2 from the table and that's our h=0 refference height)

And as you can see we don't get the same speed as we'd get when we choose our refference height h=0 to be at the table... Why does this happen?

 

[Sourabh N]

Yes!

 

[assaftolko]

Yes!

Can you please respond the the second part of my last post?

 

[Sourabh N]

Also - if you follow what you said but determine the refference height to be h=0 at the bottom of the half chain that's hanged in the air when t=0 (meaning at distance of L/2 down from the table) you get that for t=0:
E(t=0) = 0.5Mg*L/4 (potential energy of the hanging segment) + 0.5Mg*L/2 (potential energy of the resting segment)
E(final) = 0.5Mv^2 + Mg*0 (Because now the center of gravity point is at the middle of the chain which mean it's L/2 from the table and that's our h=0 refference height)

And as you can see we don't get the same speed as we'd get when we choose our refference height h=0 to be at the table... Why does this happen?

Your E(t=0) had a term missing.
Does that answer your question?

 

[assaftolko]

Yes it does, thanks!

 

[Sourabh N]

No problem :)