Torque off center of mass rigid body

In order to obtain the equations of the motion of a rigid body, I need the applied torque about the center of mass $T_\mathrm{c}$. However, I have a torque that is applied off center of mass at the point $r_\mathrm{s}$ of the rigid body and I want to replace it with a torque about the center of mass $T_c$ and a force.

Suppose the rigid body is composed of $p$ particles, then $$T_\mathrm{s} = \sum_{i=1}^p (r_i - r_\mathrm{s}) \times F_i = \sum_{i=1}^p (r_i - r_\mathrm{c}) \times F_i + (r_\mathrm{c} - r_\mathrm{s}) \times \sum_{i=1}^p F_i$$ where $T_\mathrm{c}$ is the torque about the center of mass, $F_\mathrm{r}$ is the resultant of forces and $r_\mathrm{s} = r_\mathrm{c} + d_\mathrm{s}$. How can I determine $T_\mathrm{c}$ and $F_\mathrm{r}$? There are multiple solutions, I think, are all equivalent? For instance one can take $F_\mathrm{r} = 0$ which will make the c.o.m to stand still, while under $T_\mathrm{s}$ it is surely moving.

• Can you explain what the symbols in your question mean? $F_r$? $d_s$? – Yashas Feb 14 '17 at 10:56
• $d_s = r_s - r_c$ where $r_s$ is the coordinate in world frame of the point where the external torque is applied and $F_r = \sum_{i=1}^p F_i$. – C Marius Feb 14 '17 at 11:02
• If I assume $d_s \perp T_s$ it comes out that $d_s \perp T_c$. Search for $F_r = d_s \times T_c$. Hence $T_s = T_c - d_s \times (d_s \times T_c)$, therefore $T_s = T_c - d_s (d_s \cdot T_c) + T_c(d_s \cdot d_s)$. From here $T_c = \frac{1}{1 + \|d_s\|} T_s$, and $F_r = d_s \times T_c$ ... Is it correct ... ?? – C Marius Feb 14 '17 at 11:35
• Related question: How to choose origin in rotational problems to calculate torque – John Alexiou Feb 23 '17 at 23:32

For the equations of motion you need the net torque about the center of mass $\vec{T}_C$. If this torque is a result of an applied pure torque $\vec{\tau}$ and a force $\vec{F}$ located at a point A specified by $\vec{r}_{A/C}$ relative to the center of mass then $$\boxed{\vec{T}_C = \vec{\tau} + \vec{r}_{A/C} \times \vec{F}}$$
\begin{aligned} \sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\ \sum \vec{T}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right) \end{aligned}
In this answer $\vec{T}_A$ is the torque at a point not at the center of mass.
I will try to answer my question. Suppose $r_s -r_c = d_s \perp T_s$. We can find a system of forces $F_i$ to produce the torque $T_s$ as follows: let $r_c$ be the coordinates of the center of mass of the rigid body, $r_s$ the coordinates in world frame where the torque $T_s$ is applied, and let $r_f$ be the coordinates of a point of the body, in world frame, where the force is applied. Suppose $(r_f - r_s) \perp T_s$. We shall consider two forces: $F_1 = \alpha (r_f-r_s)\times T_s$ and $F_2 =\beta (r_f - r_s)$. Of course one can find $\alpha$ from $$T_s = (r_f - r_s) \times (F_1 + F_2) = \alpha (r_f-r_s)\times ((r_f - r_s) \times T_s) = -\alpha T_s \|r_f-r_s\|^2$$ hence $\alpha = -\frac{1}{\|r_f - r_s\|^2}$. Please note that if $\alpha$ and $\beta$ are known then the torque about the center of mass and the resultant force follow. A method for finding $\beta$ is unavailable to me at this time ... :( Of course one can take any value for $\beta$ since $F_2$ does not produce torque about $r_s$ , but I think $\beta$ should be somehow uniquely defined.