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Conservative forces?

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves in a horizontal plane along the parabola ##y = x^2##. At t=0, it is at the point (1,1) with speed v0. Aside from the force of constraint holding it to the path, it is acted upon by the following external forces:
    A radial force: ##\vec F_a = -A r^3\hat r##
    A force given by : ##\vec F_b = B (y^2\hat \imath - x^2 \hat \jmath)##

    where A,B are constants.
    a- Are the forces conservative?
    b- What is the speed ##v_f## of the particle when it arrives at the origin ?

    2. Relevant equations
    curl, integration on a path

    3. The attempt at a solution

    Hello, I've just started a new chapter about mathematical aspects of force and energy. It's a little hard to digest at the beginning so maybe you can check my work. Thanks !

    a - To test whether a force is conservative, I must check that ##\vec \nabla \times \vec F = \vec 0 ##, but I'm gonna need an expression of the gradient in polar coordinates for the (x,y) plane. I believe that it is ##\vec \nabla = \frac{\partial .}{\partial r} \hat r + \frac{1}{r} \frac{\partial .}{\partial \theta} \hat\theta + \frac{\partial .}{\partial z} \hat k## because if ##g = g(r,\theta,z)##, then its differential is

    ## \begin{array}{ccr}
    dg := (\nabla_{\hat r} g)\ dr + (\nabla_{\hat \theta} g)\ ds + (\nabla_{\hat k} g)\ dz
    & \text{and} &
    \begin{align}
    dg =& \frac{\partial g}{\partial r}\ dr + \frac{\partial g}{\partial \theta} \ d\theta
    + \frac{\partial g}{\partial z}\ dz \\
    =& \frac{\partial g}{\partial r}\ dr + \frac{\partial g}{\partial \theta} \ (\frac{ds}{r}) + \frac{\partial g}{\partial z}\ dz
    \end{align}
    \end{array} ##

    So if you confirm this is right,

    ##
    \begin{array}{cc}
    \vec \nabla \times \vec F_a =
    \begin{vmatrix}
    \hat r & \hat \theta & \hat k \\
    \frac{\partial .}{\partial r} & \frac{1}{r} \frac{\partial .}{\partial \theta}& \frac{\partial .}{\partial z} \\
    -A r^3 & 0 & 0
    \end{vmatrix} = \vec 0
    &
    \vec \nabla \times \vec F_b =
    \begin{vmatrix}
    \hat \imath & \hat \jmath & \hat k \\
    \frac{\partial .}{\partial x} & \frac{\partial .}{\partial y}& \frac{\partial .}{\partial z} \\
    B y^2 & -Bx^2 & 0
    \end{vmatrix} = -2B(x+y) \hat k\neq \vec 0
    \end{array}
    ##

    So that only the radial force is conservative.


    b- ##\vec F_b## does a non-conservative work on the path ##y = x^2## from x=1 to x=0, so its work is :
    ## \begin{align}
    W^{(nc)} =& B \int_{(1,1)}^{(0,0)}(y^2 dx - x^2 \ dy) \\
    =& - B\int_{0}^{1} (x^4\ dx - x^2 (2x\ dx)) \\
    =& \frac{3B}{10}
    \end{align}##

    The potential fonction of the radial force is ##U_a(\vec r) = \frac{A}{4} r^4##.

    By conservation of total energy,

    ##E_f - E_i =W^{(nc)} \Rightarrow v_f^2 = v_0^2 + \frac{A}{2m} +\frac{3B}{5m}##
     
    Last edited: Feb 1, 2015
  2. jcsd
  3. Feb 1, 2015 #2
  4. Feb 2, 2015 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your result is correct, but check the formula for curl in cylindrical coordinates.
     
  5. Feb 2, 2015 #4
    Thanks ! I've just checked on the internet, it's the same thing with 1/r factored out of the determinant
     
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