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I know that the Newton-Euler equations can be proven using the center of mass as reference, but I was wondering if this is a special case, or if you can provide a counter-example. We know that when a rigid body is only translating the net torque through the center of mass is zero. Is this true when we evaluate the torque using other points too?

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  • $\begingroup$ at least I cannot understand your question, can you clarify, maybe use some more words? $\endgroup$ – trula Apr 26 '20 at 10:49
  • $\begingroup$ @trula I'm sorry I worded it poorly. I tried to clarify what I'm asking. $\endgroup$ – Giorgos Oikonomou Apr 26 '20 at 10:59
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No, it is not true; for example, consider a uniform-mass ball of radius $r$, which has a force $\boldsymbol{F}$ acting horizontally through its centre, such that it translates without rolling. The torque about its centre of mass is zero (because it is not rotating), but the torque calculated about its contact point with the ground is not zero: \begin{equation} \boldsymbol{\tau} = \boldsymbol{r}\times\boldsymbol{F} \neq 0\,. \end{equation}

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The net torque on an object can be defined as

$$\boldsymbol{\tau}=\frac{\mathrm d {\mathbf p}}{\mathrm d t}\tag{1}$$

where $\mathbf p$ is the angular momentum. The angular momentum of any body about any general point is

$$\mathbf p=I_{\text{COM}} \boldsymbol{\omega} +m\mathbf r\times \mathbf v$$

where $I_{\text{COM}}$ is the moment of inertia of the body about its center of mass, $\omega$ is its angular velocity, $\mathbf r$ is the position vector of the center of mass, $m$ is the mass of the body and $\mathbf v$ is the velocity of the center of mass of the body. Plugging this into $(1)$, we get

\begin{align} \boldsymbol{\tau}&= \frac{\mathrm d (I_{\text{COM}} \boldsymbol{\omega} +m\mathbf r\times \mathbf v)}{\mathrm dt}\\ &=\underbrace{\frac{\mathrm d (I_{\text{COM}} \boldsymbol{\omega})}{\mathrm d t}}_{\text{Term 1}} + \underbrace{\frac{\mathrm d (m\mathbf r\times \mathbf v)}{\mathrm d t}}_{\text{Term 2}} \end{align}

Since the rigid body is only translating, thus the term 1 becomes 0 (Here we have assumed that $I_{\text{COM}}$ is constant). Now applying product rule to differentiate term 2,

$$\boldsymbol{\tau}=m\left(\frac{\mathrm d \mathbf r}{\mathrm d t}\right)\times \mathbf v + m \mathbf r \times \left(\frac{\mathrm d \mathbf v}{\mathrm d t}\right )$$

Here we have assumed that $m$ is constant. Now, since our reference point is stationary, thus, $\mathrm d \mathbf r/\mathrm d t=\mathbf v$ and, also, $\mathrm d \mathbf v/\mathrm d t =\mathbf a$. Thus

$$\boldsymbol{\tau}=m\mathbf v \times \mathbf v+ m \mathbf r \times \mathbf a$$

Since the cross product of a vector with itself yields a zero vector, thus,

$$\boxed{\boldsymbol{\tau}=m\mathbf r \times \mathbf a}\quad \equiv\quad \boxed{\boldsymbol{\tau}=\mathbf r\times \mathbf F}$$

This is your final expression of torque. It is only zero when $\mathbf r$ and $\mathbf a$ are along the same line, or any one of the two of $\mathbf r$ and $\mathbf a$ is zero.

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