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I have a force given by $\vec{F}=200N\cos\theta \hat{i}+200N\sin\theta \hat{j}=200N \hat{R}$ and I'm trying to figure out how to set up two line integrals to calculate work as a function of initial and final positions in terms of both$ (x,y)$ and $(R,\theta) $, and also determine whether or not this force is consevative.

For $(R,\theta) $ I have \begin{align} x&=200N\cos\theta \hat{i} & y& =200N\sin\theta \hat{j} \\ dx&=-200N\sin\theta d\theta & dy&=200N\cos\theta d\theta \end{align}

\begin{align} \int_{C}^{}\vec{F}\cdot d\vec{s} & =\int_{C}^{}\left \langle x,y \right \rangle\cdot \left \langle dx,dy \right \rangle \\ & =\int_{C}^{}200N\cos\theta (-200N\sin\theta d\theta )+(200N\sin\theta )(200N\cos\theta d\theta ) \\ & =0 \end{align}

I don't know if this is right or how I should go about writing the line integral in terms of x and y. Can someone help out?

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So a conservative force can be expressed as the gradient of a scale function: $V(r, \theta)$. Your force is:

$$ \vec F(r, \theta) = k\hat r $$

The gradient in polar coordinates is:

$$ \vec \nabla V(r, \theta) = \frac{\partial V}{\partial r}\hat r + \frac 1 r \frac{\partial V}{\partial \theta}\hat{\theta}$$

so (1) your potential can't be a function of $\theta$. (2) Let's guess a power of $r$:

$$ V(r) = ar^n $$ $$ \vec \nabla V(r, \theta) = anr^{n-1}\hat r $$

which is solved with $n=1$ and $a=k$:

$$ V(r) = kr $$

The line integral along any curve $C$ satisfies

$$ \int_C \vec F(r)\cdot d\vec s = V(r_1, \theta_1)-V(r_0, \theta_0) = k(r_1 - r_0) $$

where the $(r_i, \theta_i)$ are the endpoints of $C$.

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  • $\begingroup$ Thanks for this answer, but I'm still having a tad bit of trouble. I'm trying to find the work done by the force as a function of initial and final positions in both the $x,y$ and $(R, \theta)$ coordinate systems. $\endgroup$ – CalebWilliamsUIC Dec 2 '20 at 20:38
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To do this in polar coordinates, you'd need to express your $\hat R$ vector and $r$ as functions of some parameter. But you'd most likely end up expressing the $\hat R$ vector in terms of $x$ and $y$ positions anyway.

The force is conservative if, as you know, it can be described as the gradient of some potential. In polar coordinates, the gradient operator applied to some potential $U$ is: $$\nabla U =\hat{\pmb e}_{r}\frac{\partial U}{\partial r}+\hat{\pmb e}_{\theta}\frac{1}{r}\frac{\partial U}{\partial \theta}.$$

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At any given point you have:

$\hat{r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}$

$\hat{\theta}= -\sin{\theta} \hat{i}+ \cos{\theta}\hat{j}$

Note the dot product of either vector with itself is 1 and the product of one with the other is zero, they are orthogonal unit vectors.

The infinitesimal line element is: $d\vec{s}=dr \ \hat{r}+rd\theta \ \hat{\theta}$

From above, $\vec{F}=F_0\hat{r}$. By definition of work, Work is the line integral of the force through a distance:$W=\int {\vec{F}\cdot d\vec{s}}$.

$\vec{F}\cdot d\vec{s}=F_0\hat{r}\cdot(dr\hat{r}+rd\theta\hat{\theta})=F_0dr$. Here $d\vec{s}$ is the linear differential vector.

It follows that $W=\int F_0 dr=F_0\Delta r$. The integral only depends on the net change in $r$.

We then have two ways to prove $\vec{F}$ is conservative. You can't leave a given point, then return without there being a zero net change in r, but a zero net change in r implies no work is done along a closed path, so we have from this that $\vec{F}$ is conservative.

We also have that $\vec{F}=F_0\nabla{r}$. $\vec{F}$ is a conservative force because it can be expressed as a gradient.

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  • $\begingroup$ I don't really understand this. How can I use this to solve the problem with the line integrals? $\endgroup$ – CalebWilliamsUIC Dec 2 '20 at 21:25

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