How to resolve impulse on a free-floating body into translation and rotation I have free-floating in 2D a long thin homogeneous rectangular body with the center of gravity at its geometric center.  It seems to me intuitively that if an impulse is applied to the middle of a long edge, parallel to the edge, there will be more translation and less rotation that if a similar impulse is applied parallel to the middle of a short edge.  Is this correct?  And more importantly, how do I quantize this?  In other words, how do I apportion translation and rotation from an impulse?
 A: The general motion of a planar rigid body is a rotation about a point. If you consider an impulse $\hat{j}$ that passes on a line a distance $a$ from the center of mass, then the center of rotation is going to be a distance $b$ on the other side of the center of mass as seen below

The relationship between the two distances is
$$ b = \frac{I}{m a}  \tag{1}$$
or if you use the radius of gyration $r$, and use $I = m r^2$ then
$$ b = \frac{r^2}{a} \tag{2}$$
For example for a rectangle of length $\ell$ and height $h$ the MMOI is $ I = \tfrac{m}{12} ( \ell^2 + h^2 )$, or $r = \tfrac{\sqrt{3}}{6}  \sqrt{ \ell^2+h^2}$
The great thing about the above relationship is that it is purely geometrical. It also has the property that the less $a$ is the larger $b$ is, and vice versa.
There are three special cases to (2)

*

*$a=0$, the impulse goes through the center of mass, and the body translates since $b=\infty$.

*$b=0$, the body is rotating about the center of mass when $\hat{j}=0$ and $a=\infty$. This situation corresponds to a pure impulsive torque on the body.

*$a=r$ the impulse is tangential to the circle of gyration, then $b=a$.

