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Center of mass

  • Thread starter Lopina
  • Start date
  • #1
14
0

Homework Statement


We are given a bar, with length of [tex]d[/tex], and it's densitiy is given by this formula: [tex]\lambda=\lambda_{0}+2ax[/tex], where x is the distance from one side of the bar and a is a constant


Homework Equations


[tex]\lambda=\lambda_{0}+2ax[/tex]
[tex]\vec{r}_{CM}=\frac{\sum\vec{r}_{i}\Delta m_{i}}{m}[/tex]



The Attempt at a Solution


Well, I figured, if I have infinitesimal parts of the bar, I should integrate it.
So, this is what I've come up with so far:

[tex]M=\int^{d}_{0}(\lambda_{0}+2ax)dx=\lambda_{0}x+ax^{2}|^{d}_{0}=\lambda_{0}d+ad^{2}[/tex]

[tex]X_{cm}=\frac{1}{M}\int^{d}_{0}(\lambda_{0}x+2ax^{2})dx=\frac{1}{M}(\frac{\lambda_{0}x^{2}}{2}+\frac{2ax^{3}}{3})|^{d}_{0}=\frac{3\lambda_{0}d+4ad^{2}}{6(\lambda_{0}+ad)}[/tex]
 

Answers and Replies

  • #2
674
2
Where is the question? Your formulas look ok.
 
  • #3
14
0
Well, I was not sure of this because the solution looks very awful, and I had noone to verify this.
But if it seems OK to you, I guess it's correct then.
Thanks
 
  • #4
674
2
Well the math works out correctly, although you can pull a 'd' out from the numerator to slightly simplify it a bit more. Unless you misread something, then everything looks correct.
 

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