- #1
kevinet
- 1
- 0
So ya, the question that I'm trying to solve is to derive a=v^2/r using calculus.
I am very intrigued when in class my teacher derive angular velocity [tex]\omega[/tex]
he did it very elegantly by starting off with the equation
s=r[tex]\theta[/tex] (s is the distance traveled)
then
ds/dt = d (r[tex]\theta[/tex])/dt (differentiating both sides by d/dt)
hence
v/r = d[tex]\theta[/tex]/dt (which is how we derive angular velocity :D)
So I am intrigued to how they come up with a=v^2/r and try randomly any calculus methods that I have at my disposal (high school calculus)
Because I don't know what equation they started with, I decided to backtrack from the already established equation a= v^2/ r
if a=v^2/r
dv/dt = (v) (v/r) substituting v/r into d[tex]\theta[/tex]/dt
dv/dt = (v) (d[tex]\theta[/tex]/dt) cancelling out dt from both equation and rearranging
(1/v) dv = d[tex]\theta[/tex] integrating both sides (separable differential equation)
ln v = [tex]\theta[/tex] rearranging
v = e^[tex]\theta[/tex] Which is a relationship that doesn't make any sense?
Because I'm doubtful of that equation I decided to differentiate back with respect to dt
checking answer:
v= e^[tex]\theta[/tex]
dv/dt = d (e^[tex]\theta[/tex])/dt
a= (e^[tex]\theta[/tex]) (d[tex]\theta[/tex]/dt) chain rule
a= (v) (v/r) substituting e^[tex]\theta[/tex]=v and d[tex]\theta[/tex]/dt = v/r
hence we came up with the original equation a=v^2/r which mathematically works?
I am sure that I have made a mistake somewhere along the way, but I checked and rechecked, and I can't figure out where, the math works, so it should be okay?
Lots of thanks for those who could point out where it went wrong, even more thanks if you could show me how a=v^2/r is derived using calculus...
Cheers ^o^
Kevin
I am very intrigued when in class my teacher derive angular velocity [tex]\omega[/tex]
he did it very elegantly by starting off with the equation
s=r[tex]\theta[/tex] (s is the distance traveled)
then
ds/dt = d (r[tex]\theta[/tex])/dt (differentiating both sides by d/dt)
hence
v/r = d[tex]\theta[/tex]/dt (which is how we derive angular velocity :D)
So I am intrigued to how they come up with a=v^2/r and try randomly any calculus methods that I have at my disposal (high school calculus)
Because I don't know what equation they started with, I decided to backtrack from the already established equation a= v^2/ r
if a=v^2/r
dv/dt = (v) (v/r) substituting v/r into d[tex]\theta[/tex]/dt
dv/dt = (v) (d[tex]\theta[/tex]/dt) cancelling out dt from both equation and rearranging
(1/v) dv = d[tex]\theta[/tex] integrating both sides (separable differential equation)
ln v = [tex]\theta[/tex] rearranging
v = e^[tex]\theta[/tex] Which is a relationship that doesn't make any sense?
Because I'm doubtful of that equation I decided to differentiate back with respect to dt
checking answer:
v= e^[tex]\theta[/tex]
dv/dt = d (e^[tex]\theta[/tex])/dt
a= (e^[tex]\theta[/tex]) (d[tex]\theta[/tex]/dt) chain rule
a= (v) (v/r) substituting e^[tex]\theta[/tex]=v and d[tex]\theta[/tex]/dt = v/r
hence we came up with the original equation a=v^2/r which mathematically works?
I am sure that I have made a mistake somewhere along the way, but I checked and rechecked, and I can't figure out where, the math works, so it should be okay?
Lots of thanks for those who could point out where it went wrong, even more thanks if you could show me how a=v^2/r is derived using calculus...
Cheers ^o^
Kevin