- #1
Adoniram
- 94
- 6
Homework Statement
A particle of mass m moves in two dimensions under the following potential energy
function:
V(##\vec{r}##) = ½ k (x2 + 4y2)
Find the resulting motion, given the initial condition at t=0:
x = a, y = 0, x' = 0, y' = vo
Homework Equations
F = ma = -dV/dr
The Attempt at a Solution
This will obviously involve a 2nd order diff eq, and there are enough initial conditions to solve for the unknown constants. If the potential were given with the r variable instead of x and y, it would be simpler. As such, I'm not sure how to take dV/dr when V is V(x,y) not V(r)...
If I use x = r Cosθ, y = r Sinθ, I can put it as:
V(##\vec{r}##) = (3/2) k r2 Sin2θ
But now that I have θ in the formula, is it ok to take dV/dr as such and set it in F = -dV/dr?