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The center of a thin pole of length $d$ is at $(x1,y1)$. The pole is positioned so that if pushed, it would initially role in direction $\theta1$. (A vector perpendicular to the pole is pointing in direction $\theta1$.) A few moments later, we detect that the two ends of the pole have moved distances $w1$ and $w2$. Assume that in those few moments, each of the two ends of the pole moved at fixed rates. What is now the state of the pole: $(x2, y2)$ and $\theta2$?

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I'm assuming $w_1$ and $w_2$ are arc lengths rather than straight-line distances, and the pole is rolling, so at all times the velocity of the midpoint is perpendicular to the pole. I'll also assume $w_1 < w_2$ (you can do the case $w_1 = w_2$ yourself). For convenience, represent the motion in the complex plane, with $z(t)$ the midpoint and $z(t) \pm r e^{i\theta(t)}$ the endpoints. We must have $z'(t) = v(t) i e^{i\theta(t)}$ where $v(t)$ is real. The velocities at the endpoints are then $z'(t) \pm i r \theta'(t) e^{i\theta(t)} = (v(t) \pm r \theta'(t)) i e^{i \theta(t)}$, and by assumption $\dfrac{v(t) + r \theta'(t)}{v(t) - r \theta'(t)} = \dfrac{w_2}{w_1}$ is constant. Solving this, we find $v(t) = c \theta'(t)$ where $c = r (w_1 + w_2)/(w_2 - w_1)$. Then $z'(t) = i c \theta'(t) e^{i\theta(t)} = c \frac{d}{dt} e^{i\theta(t)}$. Integrate this to get $z(t) = z(0) + c (e^{i \theta(t)} - e^{i \theta(0)})$. Thus the midpoint of the pole is moving on a circle of radius $c$. The endpoints $z(t) \pm r e^{i\theta(t)} = z(0) - c e^{i\theta(0)} + (c \pm r) e^{i\theta(t)}$ are moving on circles of radius $c \pm r = \dfrac{2 w_1 r}{w_2 - w_1}$ and $\dfrac{2 w_2 r}{w_2 - w_1}$. When the first has gone through distance $w_1$, $(\theta(t) - \theta(0)) \dfrac{2 w_1 r}{w_2 -w_1} = w_1$ so $\theta(t) - \theta(0) = (w_2 - w_1)/(2r)$.

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