1
$\begingroup$

I wanted to know if it is somehow possible to express the magnitude of the torque vector about some axis using some dot product of other vectors in the system (the position of the object, the axis of rotation, the force applied, the angular momentum etc.).

Naturally, I know the vector itself can be written as a cross product. But if I happen to only care about the magnitude of the torque, is there some way I can express it only using dot product? (Of course, I could just find the magnitude of the torque vector after taking the cross product, but I'm looking for interesting new ways to express it that might be useful while problem solving).

Thanks a lot!

$\endgroup$

1 Answer 1

1
$\begingroup$

The important quantity is the moment arm $d$ of a force vector $\boldsymbol{F}$. That is the minimum distance between the force line of action and the point of measuring torque. This will give you the magnitude of the torque vector

$$ \tau = \| \boldsymbol{\tau} \| = \| \boldsymbol{r} \times \boldsymbol{F} || = \sin\theta\, \| \boldsymbol{r} \| \| \boldsymbol{F} \| = d \, \| \boldsymbol{F} \| $$

Where $\boldsymbol{r}$ is the location of the force, $\boldsymbol{F}$ is the force vector and $\theta$ is the angle between these two vectors.

In order to get $d$, you can apply trigonometry (like above), or if you know the direction $\boldsymbol{e}$ and a point $\boldsymbol{r}$ on the force line of action then any point on the line of action is described by

$$\boldsymbol{p} = \boldsymbol{r}+t\ \,\boldsymbol{e}$$ and the closest point to the origin has $t =-\frac{\boldsymbol{e}\cdot \boldsymbol{r}}{\| \boldsymbol{e} \|^2}$ . Note that the moment arm is $$d = \| \boldsymbol{p} \| = \sqrt{ \boldsymbol{p} \cdot \boldsymbol{p}} = \sqrt{ (\boldsymbol{r}\cdot \boldsymbol{r})+2 (\boldsymbol{r}\cdot \boldsymbol{e}) t + (\boldsymbol{e}\cdot \boldsymbol{e}) t^2}$$

Using the $t$ value from above then

$$ d = \sqrt{ (\boldsymbol{r}\cdot \boldsymbol{r}) - \left( \frac{ (\boldsymbol{e}\cdot \boldsymbol{r})}{ ( \boldsymbol{e}\cdot \boldsymbol{e})} \right)^2 } $$

and

$$ \tau = d \, F$$

Note: $F=\| \boldsymbol{F} \|$ is the magnitude of the force.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.