Help understanding an example

  • Thread starter Saitama
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  • #1
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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!
 

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Answers and Replies

  • #2
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How do you get
r'a + r'b - L ??

I get the same answer as the book does
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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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.
 
  • #6
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92
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! :smile:
 

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