Conservative Forces - Maths of Force & Energy

[geoffrey159]

Homework Statement

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 ?

Homework Equations

curl, integration on a path

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 going to 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}##

 

[geoffrey159]

up

 

[ehild]

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

 

[geoffrey159]

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