I read online that rotational work is defined as $$W =\int \tau \mathrm{d} \vartheta$$ which is not exactly the same as $$W = \int \vec{F} \cdot \mathrm{d}\vec{r}$$ So i was wondering if it possible to give a little more general formula for rotational work that resembles the one with the dot product. If I consider a force not perpendicular to the axis of rotation, can I use the infinitesimal displacement vector $\mathrm{d} \vec{\vartheta}$ and write this? $$\mathrm{d}\vec{r} = \mathrm{d}\vec{\vartheta} \times \vec{r}$$ Doing so would allow me to write $$W = \int \vec{F} \cdot \mathrm{d}\vec{r} = \int \vec{F} \cdot (\mathrm{d}\vec{\vartheta} \times \vec{r}) = \int (\vec{r} \times \vec{F}) \cdot \mathrm{d}\vec{\vartheta} = \int \vec{\tau}\cdot \mathrm{d}\vec{\vartheta}$$ and keep a perfect symmetry with the other definition of work. Where is my mistake?

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    $\begingroup$ What makes you think you made a mistake? Especially if you realize that ${\rm d}\vec{\vartheta} = \vec{\omega} {\rm d}t$ just as ${\rm d}\vec{r} = \vec{v} {\rm d}t$. In the end you have $W = \int P {\rm d}t$ where $P=\vec{F}\cdot\vec{v} = \vec{\tau} \cdot \vec{\omega}$ is power. $\endgroup$ Commented Apr 1, 2018 at 18:05
  • $\begingroup$ @ja72 Well, online I didn't found this formula, only the one without the dot product and I found it very strange because of the symmetry with the non rotational case, so I thought that there might be some mistake in my reasoning. $\endgroup$
    – Julian
    Commented Apr 1, 2018 at 18:28
  • $\begingroup$ The key world here is symmetry. Actually it is called duality usually in rigid body mechanics. There is something intriguing in this symmetry between linear and rotational quantities. Some quantities in pairs represent lines in space (line of action, line of rotation, percussion axis) and the rules that govern them are identical. (See included link) $\endgroup$ Commented Apr 1, 2018 at 18:33
  • $\begingroup$ Also read this answer on angular momentum, to see that geometry (and symmetry) in mechanics. $\endgroup$ Commented Apr 1, 2018 at 18:40
  • $\begingroup$ @ja72 Thank you for the answer and the links. $\endgroup$
    – Julian
    Commented Apr 1, 2018 at 18:43

1 Answer 1


It is easier to derive 3D power, and then assert that work is the time integral of power.

Consider a moving rigid body, with a force $\boldsymbol{F}$ applied and a torque $\boldsymbol{\tau}_C$ applied at the center of mass. The body is instantaneously rotating about some point such that the velocity of the center of mass is $\boldsymbol{v}_C = \boldsymbol{\omega} \times \boldsymbol{r}_C$. Additionally the equipollent torque at the rotation center is $\boldsymbol{\tau} = \boldsymbol{\tau}_C + \boldsymbol{r}_C \times \boldsymbol{F}$.

Let us look at power at the center of mass and at the center of rotation. The two should be identical because power should be coordinate invariant.

$$\require{cancel} \begin{aligned} P & = \boldsymbol{F} \cdot \boldsymbol{v}_C + \boldsymbol{\tau}_C \cdot \boldsymbol{\omega} & &\mbox{at center of mass} \\ \hline P & =\boldsymbol{F} \cdot \cancelto{0}{\boldsymbol{v}} + \boldsymbol{\tau} \cdot \boldsymbol{\omega}=(\boldsymbol{\tau}_C + \boldsymbol{r}_C \times \boldsymbol{F}) \cdot \boldsymbol{\omega} & &\mbox{at rotation center} \\ & =(\boldsymbol{r}_C \times \boldsymbol{\omega}) \cdot \boldsymbol{F} + \boldsymbol{\tau}_C \cdot \boldsymbol{\omega} = \boldsymbol{F} \cdot \boldsymbol{v}_C + \boldsymbol{\tau}_C \cdot \boldsymbol{\omega} & &\checkmark \end{aligned} $$

So the rule for power (and hence work) is:

At any arbitrary location A, dot product the force with the linear velocity and the torque with rotational velocity to calculate power. Work is the time integral of power.

$$ P = \boldsymbol{F} \cdot \boldsymbol{v}_A + \boldsymbol{\tau}_A \cdot \boldsymbol{\omega} $$ $$ W = \int P {\rm d}t $$

Pick the center of rotation and $\boldsymbol{v}_A \rightarrow 0 $, or the line of action of the force and $\boldsymbol{\tau}_A \rightarrow 0$ and the above simplifies to just one inner product.


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