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In the book, Feynman lectures (Part 1, Section 18; 18–2: Rotation of a rigid body), the author says: $$∆W = (xF_y− yF_x)\Delta\theta$$ where $xF_y−yF_x$ is torque. He also says:

Now let us consider a single force, and try to figure out, geometrically, what this strange thing $xF_y−yF_x$ amounts to. In Fig. 18-2, we see a force $F$ acting at a point $r$. When the object has rotated through a small angle $\Delta\theta$, the work done, of course, is the component of force in the direction of the displacement times the displacement. In other words, it is only the tangential component of the force that counts, and this must be multiplied by the distance $\Delta\theta$. Therefore we see that the torque is also equal to the tangential component of force (perpendicular to the radius) times the radius.

How can I relate (mathematically), $xF_y−yF_x$ with the tangential component of force (perpendicular to the radius) times the radius?

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Let $\vec M$ be the torque in 3D-space, $\vec r$ the position vector of some mass point $m$ and $\vec F$ the force applied to $m$, then it holds $$ \vec M = \vec r \times \vec F~. $$ Now assume the third components $r_z$ and $F_z$ are both 0, then there follows $M_x = M_y = 0$, too (just write those components of the vector product explicitly), and you are left with $$ M_z = r_x F_y - r_y F_x~, $$ so this expression really is the $z$-component of the torque.

Now to calculate the component of $\vec F$ perpendicular to $\vec r$ (which is the tangential component) in the two dimensions left (the $xy$-plane), I take a unit vector perpendicular to $\vec r$, $$ \vec e_\perp = \frac 1r \begin{pmatrix} -r_y \\ r_x\end{pmatrix}~, $$ and project $\vec F$ onto it using the scalar product, yielding $$ F_\perp = \vec F \vec e_\perp = - \frac{F_x r_y}{r} + \frac{F_y r_x}{r}~. $$ This means, I can now multiply the radius and the product is just the expression you asked about: $$ rF_\perp = r_x F_y - r_y F_x~. $$

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