# Conservative forces?

1. Feb 1, 2015

### geoffrey159

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. Feb 1, 2015

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3. Feb 2, 2015

### ehild

Your result is correct, but check the formula for curl in cylindrical coordinates.

4. Feb 2, 2015

### geoffrey159

Thanks ! I've just checked on the internet, it's the same thing with 1/r factored out of the determinant