A question about linear drag force

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 1K views
shanname
Messages
8
Reaction score
0
My classical mechanics textbook says that, for a projectile, the linear drag force is given by f = -bv and the second law is written as m[itex]\ddot{r}[/itex] = mg - bv (a second order differential equation) which can be rewritten as m[itex]\dot{v}[/itex] = mg - bv (a first order differential equation) because the forces depend only on v and not on r. But I can't figure out why this is the case. Doesn't v depend on r?
 
Physics news on Phys.org
##\mathbf v ## doesn't "depend" on ##\mathbf r## in the sense that it is some (unknown) function of ##\mathbf r## and probably some other variables as well.

The point is that ##\mathbf v## is just another name for ##\mathbf{\dot r}##, (that's what "velocity" means!) and differentiating, ##\mathbf{\dot v}## is identically equal to ##\mathbf{\ddot r}##.
 
Thank you, AlephZero. I believe I understand. I mean, I know that v is just another name for [itex]\dot{r}[/itex] and the like, I just thought it could be rewritten in terms of v for that reason. No one ever explained that this is true specfically because v did not "depend" on r... is it ever the case that v does depend on r?
 
sorry, didn't realize I wasn't bolding [itex]\dot{r}[/itex].