Question

Given L, find the form of V (r) so that the path of a particle is given by the spiral r = Cθ^k, where C and k are constants. Hint: Obtain an expression for r ̇ that contains no θ’s

Answer 1

And what is L now?

Answer 2

lolz :D

Answer 3

Angular momentum.. Sorry bout that

Answer 4

Are you working in spherical coordinates?

Answer 5

or polar?

Answer 6

well in order for it to be in terms of r and theta it has to be polar right? Also its hard to see here, but in the hint it says rdot (ie dr/dt)

Answer 7

I can't even find the formula for angular momentum in polar coordinates online and have forgotten it, what is it?\[ L=mr(d \theta/dt)\]or something?

Answer 8

or just \[L=m{d \over dt}(r \theta)\]?

Answer 9

L = r × p. Buh the way you are supposed to work on it is to set up a lagrangian so something like Lagrangian = 1/2 m(r ̇2 + r2θ ̇2) − V (r). then youd get something like this The equations of motion obtained from varying r and θ are mr ̈ = mrθ ̇2−V′(r) and
d/dt (mr2θ ̇) = 0 so that means mr2θ ̇ is constant and then you can plug this back into the equation and stuff. I dont really get it though

Answer 10

its about "central forces" if that says anything

Answer 11

I know very little about the Lagrangian, sorry wish I could help.

Answer 12

The Lagrangian for a particle like that moving in two dimensions, I assume, would be evaluate limit algebra)
\[\mathcal{L} = \frac{1}{2}m\dot{r}^2 + \frac{1}{2}mr^2\dot{\theta}^2 - V(r)\]
So the equation of motion for theta is
\[mr^2\dot{\theta} = constant = L \]
and the equation of motion for r is
\[m\ddot{r} = mr\dot{\theta}^2 - \frac{\partial V}{\partial r} \]

Answer 13

yeah thats more or less what you do and then plug the L into the equation I think, but would you be able to show me how to get the answer?

Answer 14

for this particular question that is