Integration in angular momentum

  • #1
Rikudo
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Homework Statement:
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Relevant Equations:
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https://www.physicsforums.com/threa...f-a-translating-and-rotating-pancake.1005990/
So,I think I posted this in the wrong place. So, I will move it to here.
Here, in post #6, it is stated that ##\int R dm = M R##. As far as I know, R change from time to time and it is not constant. Hence, isn't it incorrect to say that ##\int R dm = M R##? Or, are there any techniques that are skipped?
 

Answers and Replies

  • #2
anuttarasammyak
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Though it is not an answer to your question but a matter of taste, I am not familiar with expression dm. I prefer to say it
[tex]\mathbf{L}=\int \mathbf{r} \times \mathbf{\pi}(\mathbf{r}) d^3\mathbf{r} =\int \mathbf{r} \times \rho(\mathbf{r}) \mathbf{v}(\mathbf{r}) d^3\mathbf{r} [/tex]
where ##\pi(r)## is momentum density and ##\rho(r)## is mass density.
So now I know
[tex]dm = \rho(\mathbf{r}) d^3\mathbf{r}[/tex]
 
  • #3
haruspex
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Here, in post #6, it is stated that ##\int R dm = M R##. As far as I know, R change from time to time and it is not constant.
Right, but that integral has no moving parts. It is the integral over the mass at some instant.
 

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