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Assume there is a rigid body in deep space with mass $m$ and moment of inertia $I$. A force that varies with time, $F(t)$, is applied to the body off-center at a distance $r$ from its center of mass. How do I calculate the instantaneous acceleration, rotational acceleration, and trajectory of this object, assuming it starts from rest?

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If the position of the center of mass is $\vec{r}_C$ and the location of the force application $\vec{r}_A$ then the Euler-Newton equations of motion for rigid body are:

$$ \vec{F} = m\,\vec{a}_C \\ (\vec{r}_A-\vec{r}_C)\times \vec{F} = I_C \vec{\alpha} + \vec{\omega}\times I_C \vec{\omega} $$

with c.g. velocity $\vec{v}_C = \dot{\vec{r}_C}$, c.m. acceleration $\vec{a}_C = \ddot{\vec{r}_C}$, $I_C$ the moment of inertia tensor about the c.m.

In 2D when $(x,y)$ is the location of the c.m. Point C this becomes

$$ \begin{vmatrix} F_x \\ F_y \\ 0 \end{vmatrix} = m \begin{vmatrix} \ddot{x} \\ \ddot{y} \\ 0 \end{vmatrix} \\ \begin{vmatrix} c_x \cos\theta \\ c_y \sin\theta \\ 0 \end{vmatrix} \times \begin{vmatrix} F_x \\ F_y \\ 0 \end{vmatrix} = \begin{vmatrix} I_x & &\\& I_y & \\ & & I_z \end{vmatrix} \begin{vmatrix} 0 \\ 0 \\ \ddot{\theta} \end{vmatrix} + \begin{vmatrix} 0 \\ 0 \\ 0 \end{vmatrix} $$

where $(c_x,c_y)$ is the position of point A from the c.m. when the body orientation is $\theta=0$ (initially).

By component then the equations are $$ \ddot{x} = F_x/m \\ \ddot{y} = F_y/m \\ \ddot{\theta} = \frac{-c_y \sin\theta F_x + c_x \cos\theta F_y}{I} $$

If the force is rotating with the body, and initially located at $(cx,0)$ pointing in the +y direction then

$$ \ddot{\theta} = \frac{c_x F_y}{I} $$

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You can always replace an off center force by the same force centered plus a torque $r\times F$.

So the trajectory and acceleration of the COM are the same you would get with the same force centered, so you can solve them independently of the rotational dynamics. The torque mentioned above is what drives the rotational dynamics of your rigid body.

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  • $\begingroup$ Please explain how a force F can be equal to said "F" plus something else, another force (torque) $\endgroup$ – Alex Doe Nov 16 '18 at 12:58
  • $\begingroup$ Torque and force are different things, e.g. the resultant force of two opposed, non-colinear forces, a "couple", is zero, but the resultant torque is not. $\endgroup$ – Jaime Nov 23 '18 at 9:19
  • $\begingroup$ Google says torque is "a twisting force that tends to cause rotation." $\endgroup$ – Alex Doe Nov 23 '18 at 17:23

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