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Classical Mechanics: Central Force Problem

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m moves under an attractive central force of [itex]Kr^4[/itex] with an angular momentum [itex]L[/itex]. For what energy will the motion be circular? Find the frequency of the radial oscillations if the particle is given a small radial impulse.

    2. Relevant equations

    [itex]E=\frac{1}{2}μ\dot{θ}^2+\frac{L^2}{μr^2}+U(r)~and~\vec{F(r)}=-\vec{\nabla} U(r)=-Kr^4\hat{e_r}[/itex]

    3. The attempt at a solution
    Part a)

    First I need to find the potential from the central force, also because its a central force (acting only antiparallel to the radial vector) [itex]\frac{\partial L}{\partial t} = 0~\Rightarrow~L=C [/itex]

    When I integral the force I get that [itex]U(r)=\frac{Kr^5}{5}+C[/itex] I set C=0 then [itex] U(r) = \frac{Kr^5}{5}[/itex]

    Now, [itex] U_s (r)= \frac{L^2}{μr^2}+\frac{Kr^5}{5}[/itex]

    Next I seek the zero(s) of Us(r):

    [itex]0=\frac{L^2}{μr^2}+\frac{Kr^5}{5}~→~r=-\sqrt[7]{\frac{5L^2}{μK}}[/itex]

    because, a radial distance can't be negative this implies for C=0 that Us>0

    Secondly, I want to know the extrema of Us:

    [itex] \frac{\partial U_s}{\partial r} = \frac{-2L^2}{μr^3}+Kr^4 = 0~→~r=\sqrt[7]{\frac{2L^2}{μK}}[/itex]

    Finally, I need to verify whether this is a minimum or maximum:

    [itex] \frac{\partial}{\partial r} (\frac{\partial U_s}{\partial r}) = \frac{6L^2}{μr^4}+4Kr^3~→~∂^2_rU_s(r= \sqrt[7]{\frac{2L^2}{μK}})>0~\Rightarrow~r=\sqrt[7]{\frac{2L^2}{μK}}[/itex] is a min of Us

    So since the minimum of energy occurs at [itex]r= \sqrt[7]{\frac{2L^2}{μK}}[/itex] the energy of the particle in a circular orbit will be (for [itex]\dot{r}=0[/itex]) [itex] E = \frac{L^2}{μ[\sqrt[7]{\frac{2L^2}{μK}}]}+\frac{K[\sqrt[7]{\frac{2L^2}{μK}}]}{5}[/itex]

    That was part A) I'm not quite sure if I'm justified in setting C=0 in this case; however I am dealing with a central force that increases with r4 so I don't immediately see anything wrong with my answer. Now for part B) I'm not quite sure on how to do this one, but this is what I have so far:

    A radial impulse implies that [itex]\dot{r}≠0[/itex] which means that its radial position will change with time so its orbit will no longer be just circular. Find the frequency of the radial oscillations When I read this part of the question it had me thinking that particle will oscillate along the radial direction like a mass-sping system so that the motion of this particle is like a sinusoidal wave wrapped around a circle (think of sin(x) except that the x-axis is bent into a circle and the oscillations occur perpendicular to the circumference of the circle in the plane of the circle. Am I correct in my description and way of thinking?

    As for calculating the radial frequency I'm not sure how to go about this either, but here is what I do know:

    [itex]\vec{L}=\vec{r}X\vec{p}~→~\frac{\partial \vec{L}}{\partial t} = \dot{\vec{r}}X\vec{p}+\vec{r}X\dot{\vec{p}}[/itex] I'm also going to take a slight leap of faith here and say that [itex] \dot {\vec{L}} = 0 [/itex] so that [itex]\dot{\vec{r}}X\vec{p} = -\vec{r}X\dot{\vec{p}}[/itex] beyond this however, I am at a loss. Thank you for any guidance you can provide in advance!
     
  2. jcsd
  3. Feb 23, 2014 #2
    The second term in your original energy expression is missing a factor of 2, since it is coming from [itex]\mu r^2\dot{\phi}^2/2=L^2/(2\mu r^2)[/itex], where [itex]L=\mu r^2 \dot{\phi}[/itex]. I think the radius can also be calculated easier by just taking [itex]\dot{r}=0[/itex] in the radial equation of motion:
    [tex]
    0=\frac{d}{dt}\left(\mu \dot{r}\right) = \mu r \dot{\phi}^2 - \frac{\partial U}{\partial r} =\mu r \left(\frac{L}{\mu r^2}\right)^2 - Kr^4
    [/tex]
    which gives you [itex]r=\left[L^2/(\mu K)\right]^{1/7}[/itex].

    The potential energy is defined up to an additive constant, so the choice of [itex]C[/itex] is not an issue.

    That is a good way of thinking of what is happening.

    Think about what you said regarding thinking of the radial oscillations like a mass-spring system. If you take the effective potential and expand it in a power series about its minimum [itex]r_{\text{min}}[/itex], the second order term looks just like the potential of a spring-mass system, with spring constant [itex]\left.\partial^2 U_s/\partial r^2\right|_{r=r_{\text{min}}}[/itex]. How is this "effective spring constant" related to the oscillation frequency it would have?
     
  4. Feb 23, 2014 #3
    Ah yes, I see I now: so [itex](ω_0)^2 = \frac{k}{m}[/itex] with k being the second derivative evaluated at r_(min). Thank you jk86!
     
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