Effective Potential with Power-Law Spiral

[DaquiriP]

Homework Statement

Given L, find the V(r) that leads to a spiral path of the form r = r₀(theta)^k.

Choose E to be zero.

The Attempt at a Solution

I know that I need to find r' as a function WITHOUT theta, but I'm not sure how to do that. I know how to find r' as a function of theta, but I don't know how to get rid of theta.

I do however know that (theta)' = L/mr², but I'm not sure if this helps me. :(

 

[mark726]

Hello,

Thank you for your question. To find the V(r) that leads to a spiral path of the form r = r0(theta)k, we can use the equation for centripetal force, F = m(r')^2/r, where F is the centripetal force, m is the mass, r' is the derivative of r with respect to time, and r is the distance from the center of the spiral path.

Since we are given that E = 0, we can set the potential energy V(r) equal to zero. This means that the only force acting on the object is the centripetal force, and we can set F = m(r')^2/r = 0.

Next, we can use the fact that r = r0(theta)k to substitute for r in the equation. This gives us m(r0(theta)k')^2/r0(theta)k = 0.

Simplifying, we get k'^2 = 0, which means that k' = 0. This means that the slope of the spiral path is constant and equal to zero.

Now, we can use the fact that (theta)' = L/mr^2 to substitute for k' in the equation. This gives us:

(theta)' = L/m(r0(theta))^2

Simplifying, we get (theta)' = L/mr0^2(theta)^2.

Since we want to find r' as a function without theta, we can rearrange this equation to get:

r' = sqrt(L/mr0^2) * 1/theta

Therefore, the potential energy V(r) that leads to a spiral path of the form r = r0(theta)k is:

V(r) = sqrt(L/mr0^2) * ln(theta) + C

I hope this helps. Let me know if you have any further questions or need clarification.