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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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Check out Center of Percussion. –  Mike Dunlavey Nov 2 '12 at 0:00
Does the force rotate with body, or is always in the same orientation? Also initially what is the angle between the force axis and the orientation of the body? –  ja72 Nov 2 '12 at 17:02

2 Answers 2

up vote 3 down vote accepted

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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Do you intend to say center of mass when you say c.g.? Because center of gravity according to the gravitational field changing inside the body. (eg. Large body nearby will cause the c.g. to shift to a different place than if that body was not there.) –  Pradyun Jun 11 at 11:54
Yes, I edited the question. –  ja72 Jun 11 at 12:51

You can always repalce 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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