Newton's second law for a body with changing mass given as $$F=ma + \frac{dm}{dt}v$$ I need the version for rotational motion. By inspection, it seems that it would be $$\tau = I\alpha + \frac{dI}{dt}\omega$$

I tried to prove this by starting from force equals the change in momentum. $$F=\frac{d}{dt}(mv)$$ $$F=\frac{d}{dt}(mr\omega)$$ $$Fr=\frac{d}{dt}(mr\omega)r$$ $$\tau=\frac{d}{dt}(mr^2\omega)$$ $$\tau = I\alpha + \frac{dI}{dt}\omega$$

Is this proof correct? I'm specifically not sure if $\frac{d}{dt}(mr\omega)r \to \frac{d}{dt}(mr^2\omega)$ is allowed.


Torque equals the rate of change of angular momentum. Differentiate the angular momentum to get the answer. And your method is not correct. Don't treat vectors like scalars. That is simply not allowed.

$$\vec{L} = I\vec{\omega} $$

Using your method, you would need to differentiate this way

$$\tau = \frac{d }{dt} (\vec{r} \times m \vec{v})$$

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    $\begingroup$ FYI - Torque equals the rate of angular momentum when summed at the center of mass, or at a fixed point. If you do that on moving joint of a robotic mechanism you will find that or some arbitrary point A $$ \overline{\tau}_A = \frac{\rm d}{{\rm d}t} \overline{L}_A + \overline{v}_A \times \overline{p} $$ where $\overline{p} = m \overline{v}_{\rm COM}$ is momentum. $\endgroup$ Sep 16 at 11:58

In general, you have fixed 3x3 mass moment of inertia tensor $\mathrm{I}_{\rm body}$ riding along the body coordinates and it needs to be rotated into the inertial directions to be used in the equations of motion. Give a 3x3 rotation matrix $R(t)$ this is done with $\mathrm{I}(t) = R\, \mathrm{I}_{\rm body} \,R^\top$

So the mass moment of inertia is changing with time due to rotation in the general rotational equation of motion summed at the center of mass and along the inertial reference directions:

$$\begin{array}{r|l} \text{momentum} & \overline{L} = \mathrm{I}\,\overline{\omega}= ( R \mathrm{I}_{\rm body} R^\top) \overline{\omega} \\ \text{torque} & \overline{\tau} = \tfrac{\rm d}{{\rm d}t} \overline{L} = \mathrm{I}\, \dot{ \overline{\omega}} + \overline{\omega} \times \overline{L} \end{array} $$

Notice that $\tfrac{\rm d}{{\rm d}t} R = \overline{\omega} \times R$

But now you want to consider a situation where the mass of the body changes, or the configuration changes, and thus $\mathrm{I}_{\rm body}(t)$, or $\dot{\rm I}_{\rm body} \neq 0$.

This changes the rate of change of momentum as follows

$$ \begin{aligned} \overline{\tau} & = \tfrac{\rm d}{{\rm d}t}\overline{L} = \mathrm{I}\, \dot{\overline{\omega}} + \left( \tfrac{\rm d}{{\rm d}t} ( R \mathrm{I}_{\rm body} R^\top ) \right)\overline{\omega} \\ & = \mathrm{I}\, \dot{ \overline{\omega}} + \overline{\omega} \times \overline{L} + \underbrace{ ( R \dot{\mathrm{I}}_{\rm body} R^\top ) \overline{\omega}}_\text{extra term} \end{aligned}$$


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