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

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.

• You could make your question a bit clearer by emphasizing the difference between the accelerations $\vec{a}_i$ and the angular accelerations $\vec{\alpha}_i$, which the casual reader might not spot (the symbols look quite similar). Are we to assume that there are some internal forces of constraint which act in the tangential direction? So it's like a set of mass points sliding along the spokes of a (massless) bicycle wheel? – user197851 Dec 4 '18 at 12:27
• Assume a pump at the center or whatever. I think the bottom line starts after "such that". – Sameer Baheti Dec 4 '18 at 12:52
• Look into defining the total angular momentum first, and then differentiating to get torque. Trying to work with $\vec{F} = m \vec{a}$ is far more complex than working with $\vec{p}= m \vec{v}$. – John Alexiou Dec 6 '18 at 13:28

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.
• No, that's not what I am asking. $\vec \alpha$ is constant. Read the last line or paragraph again. – Sameer Baheti Dec 6 '18 at 9:13
• @SameerBaheti - maybe you need to give a more concrete example of the velocity vector field $\vec{v}( \vec{r} )$ if it is not equal to $\vec{\omega} \times \vec{r}$ with constant rotational velocity. – John Alexiou Dec 6 '18 at 13:26