Line integral of a vector field (Polar coordinate)

[Lambda96]

Homework Statement

Calculate the work done, which acts on the particle

Relevant Equations

none

Hi,

I am not sure if I have solved task b correctly

According to the task, ##\textbf{F}=f \vec{e}_{\rho}## which in Cartesian coordinates is ##\textbf{F}=f \vec{e}_{\rho}= \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)## since ##f \in \mathbb{R}_{\neq 0}## is constant, ##\textbf{F}## would simply be ##f## in polar coordinates, wouldn't it?

##\dot{r}(t)## would be ##\dot{\rho}(t)## and therefore ##\dot{\rho}(t)=8 \pi \sin(4 \pi t) cos(4 \pi t)##

The line integral is:

##\int_{0}^{1} dt f \cdot 8 \pi \sin(4 \pi t) cos(4 \pi t)=0##

 

[TSny]

According to the task, ##\textbf{F}=f \vec{e}_{\rho}## which in Cartesian coordinates is ##\textbf{F}=f \vec{e}_{\rho}= \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##

Did you leave out a factor of ##f## in the expression on the far right?

For Cartesian coordinates, I would express the force in the form ##\textbf{F}=(...) \textbf e_x + (...) \textbf e_y##.
For part (b) you are staying in polar coordinates.

##\textbf{F}## would simply be ##f## in polar coordinates, wouldn't it?

##\textbf F## is a vector while ##f## is a scalar. So, they can't be equal. In polar coordinates, you would express the force in the form ##\textbf{F}=(...) \textbf e_{\rho} + (...) \textbf e_{\phi}##.

##\dot{r}(t)## would be ##\dot{\rho}(t)## and therefore ##\dot{\rho}(t)=8 \pi \sin(4 \pi t) cos(4 \pi t)##

The line integral is:

##\int_{0}^{1} dt f \cdot 8 \pi \sin(4 \pi t) cos(4 \pi t)=0##

I think your line integral expression is correct.

However, it might be good to show your instructor how the integrand ##\dot{\mathbf r} \cdot \mathbf F## reduces to your expression. Thus, how would you write the vector ##\dot{\mathbf r}## in polar coordinate form ##(...) \textbf e_{\rho} + (...) \textbf e_{\phi}##? Then you can take the dot product ##\dot{\mathbf r} \cdot \mathbf F##.

In part (a) you considered the case where ##\rho_0 = 0.1## and ##\rho_1 = 1##. But it said to let ##\rho_0## and ##\rho_1## take on general values again in parts (b) and (c).

 

[Lambda96]

Thank you TSny for your help

Did you leave out a factor of ##f## in the expression on the far right?

For Cartesian coordinates, I would express the force in the form ##\textbf{F}=(...) \textbf e_x + (...) \textbf e_y##.
For part (b) you are staying in polar coordinates.

You're right, unfortunately, I had forgotten the f on the right-hand side of the equation.

In Cartesian coordinates, ##\textbf{F}## would then be ##\textbf{F}=f \cos(\phi) \textbf{e}_x + f \sin(\phi) \textbf{e}_y ##

##\textbf{Task b}##

##\dot{\textbf{r}}## would have to be derived using the chain rule, so it would look like this

##\dot{\textbf{r} }= \dot{\rho}(t) \textbf{e}_{\rho} + \rho(t) \dot{\phi}(t) \textbf{e}_{\phi}##

I'm still a little unsure about ##\textbf{F}##, would this be as follows?

##\textbf{F}=f \textbf{e}_{\rho} + 0 \textbf{e}_{\phi}##

Then, the scalar product of ##\dot{\textbf{r}} \cdot \textbf{F}=f \dot{\rho}(t)## since ## \textbf{e}_{\rho}## and ## \textbf{e}_{\phi}## are orthogonal to each other.

Is that correct?

 

[TSny]

That all looks very good to me.

 

[Lambda96]

Thank you TSny once again for your help and explanation, which helped me a lot