How to integral ? How can they ignore that variable?

[Outrageous]

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

 

[Mark44]

Please upload your attachment again, but oriented so people can read it.

 

[Outrageous]

This way? No sure is that correct .

 

[Mark44]

It's upside-down now. You can edit your post and try uploading the attachment again.

 

[Outrageous]

Sorry, I don't really know how to use it.

 

[LCKurtz]

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.

 

[Boorglar]

I'm not quite sure what you mean. Nobody said r was constant. But the fact that \frac{d}{dt}(r^2\theta') = 0 does imply that r^2\theta' = C is constant (why?).

 

[Outrageous]

(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.

 

[LCKurtz]

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 $\int 1\, dt$.

 

[Outrageous]

I got it. Thanks much