Can torque about an axis be defined for non-rigid bodies?

Question

I've assumed a body in which points are radially moving outward(say).$$\vec F_i=m_i\vec a_i$$ where $\vec a_i = \vec \alpha_i \times \vec r_i$.

Considering a point $O$ about which torque will be calculated. Having defined torque as $\vec \tau_i = \vec r_i\times \vec F_i$, I want to calculate it for the whole body as follows: $$\sum (\vec r_i\times \vec F_i)=\sum\big(m_i\vec r_i\times(\vec \alpha_i\times\vec r_i)\big)$$ Cancelling the internal torques in pair and writing $$\vec \tau=\sum(\vec r_i\times\vec F_{i\ applied})=\sum(m_iR_i^2\vec \alpha_i)$$ where $R_i$ is the radial distance from the axis.

Now, my question: Can I take the $\vec \alpha$ out of the sum assuming a body that may not be rigid in the sense that it is like a rotating fluid flowing radially outwards at the same time such that $\vec \omega$ is the same along all radial directions and also on each radii.

Answer 1

I think what you are asking is if you have rotational acceleration as a function of radial distance only $\vec{\alpha}(r)$, if you have collect the terms outside the sum. The answer is no. Only constant values can be collected outside the sum.

$$ \sum_i m_i \vec{\alpha} \times \vec{r}_i \neq \vec{\alpha} \times \sum_i m_i \vec{r}_i $$ unless $\vec{\alpha}$ is a constant value.

Answer 2

In such a case energy, momentum, angular momentum must be accounted for by distributions or densities. The time derivative of the angular momentum density would be a valid candidate for the torque density.