Understanding Spring Length in the Center-of-Mass Frame

[Saitama]

This isn't an homework question. I am trying to understand a solved example in my book. I have attached the solved example.

I am stuck at the point when it says that the instantaneous length of the spring is $r_a-r_b-l=r_a'-r_b'-l$. I can't figure out how the author got the RHS of the equation. Shouldn't it be $r_a'+r_b'-l$?

Any help is appreciated. Thanks!

 

[ap123]

How do you get
r^'_a + r^'_b - L ??

I get the same answer as the book does

 

[voko]

It follows from the definition that $r_a - r_b = r_a' - r_b'$.

However, it is not correct to call $r_a - r_b - l$ the instantaneous length of the string. The instantaneous length is $r_a - r_b$, and $r_a - r_b - l$ is the instantaneous change in the length.

 

[Saitama]

It follows from the definition that $r_a - r_b = r_a' - r_b'$.

Why did the author define $r_b'=r_b-R$? Why not $r_b'=R-r_b$? From the figure, this looks to me the correct equation.

 

[voko]

Why did the author define $r_b'=r_b-R$? Why not $r_b'=R-r_b$? From the figure, this looks to me the correct equation.

Because the author wanted to to do things in the center-of-mass frame, with all the distances measured from the C. M. This is plain if you consider all this in the vector form: $\vec{r}_b = \vec{R} + \vec{r}_b'$, the vector to point b is the sum of the vector to the C. M. and the vector displacement of point b from the C. M.

 

[Saitama]

Because the author wanted to to do things in the center-of-mass frame, with all the distances measured from the C. M. This is plain if you consider all this in the vector form: $\vec{r}_b = \vec{R} + \vec{r}_b'$, the vector to point b is the sum of the vector to the C. M. and the vector displacement of point b from the C. M.

Thanks a lot voko!