How to integral ? How can they ignore that variable?

  • Thread starter Outrageous
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Homework Statement


I don't understand case 1. In case 2 that is normal integration , isn't?
Then why case 1 can simply ignore the r ? I don't think r is constant.


Homework Equations





The Attempt at a Solution


Please guide. Thanks
 
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  • #2
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Please upload your attachment again, but oriented so people can read it.
 
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  • #3
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This way? No sure is that correct .
 

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  • #4
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It's upside-down now. You can edit your post and try uploading the attachment again.
 
  • #5
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Sorry, I don't really know how to use it.
 

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  • #6
LCKurtz
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Homework Statement


I don't understand case 1. In case 2 that is normal integration , isn't?
Then why case 1 can simply ignore the r ? I don't think r is constant.

Are you talking about the equation$$
\frac m r \frac{d(r^2\dot \theta)}{dt}=0$$giving ##r^2\dot \theta=l##? If so, they didn't ignore the ##\frac m r##. They multiplied both sides by its reciprocal before integrating.
 
  • #7
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I'm not quite sure what you mean. Nobody said r was constant. But the fact that [itex]\frac{d}{dt}(r^2\theta') = 0 [/itex] does imply that [itex]r^2\theta' = C [/itex] is constant (why?).
 
  • #8
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(1/r)(dr/dt)= 0 , (dr/dt)= 0
∫ dr = ∫ 0 dt
The answer should be r = constant.
Isn't ?
Can the answer be ∫(1/r)dr= ∫dt .
Then ln(r/ro)= t. Integral from ro to r , and integral from 0 to t. ro is original value r.
 
  • #9
LCKurtz
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I thought you said you were puzzled about 1, not 2.


(1/r)(dr/dt)= 0 , (dr/dt)= 0
∫ dr = ∫ 0 dt
The answer should be r = constant.
Isn't ?

Yes, that's right.

Can the answer be ∫(1/r)dr= ∫dt .
Then ln(r/ro)= t. Integral from ro to r , and integral from 0 to t. ro is original value r.

##\int \frac 1 r\, dr = \int 0\, dt##, not ##\int 1\, dt##.
 
  • #10
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I got it. Thanks much
 

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