March 22, 2022

Without loss of generality consider a rigid body being rotated by some angle $d\theta$ in the $x-y$ plane. Since the rotation axis is fixed, the motion is constricted in a single plane. From geometry we can conclude that

$$ dx=-yd\theta \\ ‎\\ dy=xd\theta \tag{4.1,\quad 4.2} $$

If we calculate the work done by some force $\vec F$ to rotate this body around then

$$ dW=F_xdx+F_ydy\tag{4.3} $$

Using equations (4.1) and (4.2), we get

$$ dW=(xF_y-yF_x)d\theta\tag{4.4} $$

Here the term in brackets is defined to be $torque$ (Latin: $torquere$, to twist). Just like work is force through a displacement, it is also a torque through an angle.

Therefore in $2-D$ torque is defined as

$$ \tau=xF_y-yF_x\tag{4.5} $$

This equation can be represented using cross products as

$$ \vec\tau=\vec r \times\vec F\tag{4.6} $$

Where the right-hand rule is followed.

From equation (4.5) we can see that the torque is only caused by the component $\perp$ to radius vector. This is because $d\vec r\perp\vec r$, and because $\vec F$ must be $\parallel$ to $d\vec r$ to do work and hence to produce some torque, it follows that $\vec F \perp \vec r$. This also fits our intuition that to rotate something we should apply a force $\perp$ to radius vectors, such as when we pull on doors because a radial force would just displace an object and not rotate it.

Since the change in displacement, $d\vec r=rd\theta\hat\theta\rightarrow \vert dr\vert\propto r$, it makes sense that the greater the radius, the greater the work done, and hence the greater the torque, for the same angle $d\theta$.