Average Equivalent Ring Resistance

The resistance between the two points is, for one side of the ring:

$R_1 = R \cdot \dfrac{\theta}{2\pi}$

And, for the other side is:

$R_2 = R \cdot \left(\dfrac{2\pi - \theta}{2\pi}\right)$

The equivalent resistance is:

$R_{eq} = R_1 \mathbin{\|} R_2 \ = \dfrac{R_1\cdot R_2}{R_1 + R_2} = \dfrac{R}{4\pi^2} \cdot (2\pi\omega t - \omega^2 t^2)$

The time-averaged value will be the integral of $R_{eq}$ over one period divided by one period. First evaluating the integral:

$I = \displaystyle\int_0^{2\pi/ \omega} \dfrac{R}{4\pi^2} \cdot (2\pi\omega t - \omega^2 t^2) \, dt$

$I = \dfrac{R}{4\pi^2} (\pi \omega \dfrac{4\pi^2}{\omega^2} - \dfrac{\omega^2}{3}\dfrac{8\pi^3}{\omega^3} )$

$I = \dfrac{R\pi}{3\omega}$

The time-averaged value will be:

$m = \dfrac{I}{\frac{2\pi}{\omega}}$

$m = \dfrac{R}{6}$

Then, $\color{#3D99F6} \fbox{α = 6}$

For the first cycle after time t,
$R_1 = \frac{R}{{2\pi }}\omega t$
$R_2 = R-R_1=\frac{R}{{2\pi }}(2\pi - \omega t$
$R_{eq} =\frac{R_1R_2}{R}=\frac{R}{{4{\pi ^2}}}(2\pi \omega t - {\omega ^2}{t^2})$

$R_{eq}$ vs t graph is a parabola for any one of the cycles. If A is the area of one of the parabolas as shown in the diagram, then the average resistance over n complete cycles would be
<R> = $\frac{area\,under\, R_{eq}-t \, graph}{t}$
<R> = $\frac{nA}{\frac{n2\pi}{\omega}}$
<R> = $\frac{A}{\frac{2\pi}{\omega}}$

$A = \int_0^{\frac{{2\pi }}{\omega }} {{R_{eq}}} dt$
<R> = R/6