Rod in zero gravity pushed Imagine a rod - an ideal line segment of some length $l$ - is in zero gravity and initially no force is acting on it. Then something hits it and it gets some impulse $p$ in some location along the rod at some possible angle (excluding unrealistic cases such as a perpendicular impulse that is parallel to the rod but applied not at the end).
How can we calculate the rod's movement after the impulse, meaning the speed of the centre of the rod and its rotational speed ?
 A: The equations are easy enough.  First, $momentum = impact$.
$m\ v=p$
where the velocity $v$ is in the direction of the impact. The orientation of the rod makes no difference.
Next, $angular\ momentum = angular\ impact$.
$I\ \omega=p\ d$
where $I$ is the moment of inertia about the rod center of mass about an axis perpendicular to the rod (direction of rotation).  In this case
$I=\frac{m\ l^2}{12}$
for a uniform rod. $\omega$ is the angular velocity, and $d$ is the perpendicular distance from the rod's center of mass to the impact vector.
You can eliminate $p$ if you want to get the relationship between $v$ and $\omega$.
A: 1. First is the 2D version of this problem.
1.1 Definitions
I have changed the notation to the impulse magnitude being $J$ because $p$ is usually reserved for momentum. In fact, look at the x-y components of the skew impulse as $J_x$ and $J_y$ relative to the orientation of the rod. So instead of a single impulse $J$ at some angle $\varphi$, I have resolved the impulse into x-y components.

Also important is the distance $c$ between the center of mass and the impact point. The rod is characterized by the mass $m$ and the mass moment of inertia for in-plane rotation of $I$.
In the figure above, the resulting motion is shown with the blue arrows and the impulse components in pink.
1.2 Impulse Response
Assume the body is at rest initially, so after the impact, its momentum $p$ equals the impulse. The same for angular momentum $L$.
This fact yields the resulting motion:
$$\left.\begin{aligned}p_{x}=mv_{x} & =J_{x}\\
p_{y}=mv_{y} & =J_{y}\\
L=I\omega & =c\,J_{y}
\end{aligned}
\right\} \begin{aligned}v_{x} & =\frac{J_{x}}{m}\\
v_{y} & =\frac{J_{y}}{m}\\
\omega & =\frac{c\,J_{y}}{I}
\end{aligned}$$
1.3 Effective Mass
Additionally, you can write the impulse-motion relationship in terms of the effective mass $m^\star$. Look at the motion of the impulse point A as $$\begin{aligned}
v_x^A & = v_x \\
v_y^A & = v_y + \omega c \\
\end{aligned}$$
In the horizontal direction, the effective mass equals the actual mass
$$ \begin{aligned}
 J_x & = m_x^\star v_x^A \\
   & m_x = m \\
\end{aligned} $$
But in the vertical direction, things are more interesting
$$\begin{aligned}J_{y} & =m_{y}^{\star}v_{y}^{A}\\
 & m_{y}^\star=\left(\frac{1}{m}+\frac{c^{2}}{I}\right)^{-1}
\end{aligned}$$
As you can see, the further away the impulse is (and the larger $c$ is) the less the effective mass $m_y^\star$ is, which is an effect you definitely feel for yourself if you were to punch a free-floating rod.
2. 3D Vector Equations
2.1 efinitions
Now consider the above problem, but in vector form. The impulse magnitude is $J$ but the direction is now specified by the unit vector $$\hat{n} = \pmatrix{\cos \varphi \\ \sin \varphi  \\ 0}$$
the position of the impact point A relative to the center of mass C is
$$ \vec{c} = \pmatrix{c \\ 0 \\ 0} $$
Also, note that in 3D the mass moment of inertia is a tensor
$$ {\rm I} = \pmatrix{ I_x & & \\ & I_y &\\ & & I_z } $$
where $I_x \approx 0$, and $I_y = I_z = \tfrac{m}{12} \ell^2$.
In 3D, momentum and angular momentum are defined as
$$\begin{aligned}\vec{p} & =m\,\vec{v}\\
\vec{L} & ={\rm I}\vec{\omega}
\end{aligned}$$
2.2 Impusle Response
The impulse vector $\hat{n} J$ affects the momentum vector just as above. The resulting motion is
$$ \left.\begin{aligned}\vec{p} & =\hat{n}J\\
\vec{L} & =\vec{c}\times\hat{n}J
\end{aligned}
\right\} \begin{aligned}\vec{v} & =\tfrac{1}{m}\hat{n}J\\
\vec{\omega} & ={\rm I}^{-1}\left(\vec{c}\times\hat{n}J\right)
\end{aligned} $$
2.3 Effective mass
The motion of point A is described by
$$ \vec{v}^{A}=\vec{v}+\vec{\omega}\times\vec{c} $$
Note that along the direction of the impulse the resulting motion is
$$ v_{{\rm imp}}=\hat{n}\cdot\vec{v}_{A}=\hat{n}\cdot\left(\tfrac{1}{m}\hat{n}J-\vec{c}\times{\rm I}^{-1}\left(\vec{c}\times\hat{n}J\right)\right)$$
Extracting $J$ we find the effective mass of the rod along the impulse direction
$$\begin{aligned}J & =m^{\star}v_{{\rm imp}}\\
 & m^{\star}=\left(\tfrac{1}{m}+\left(\hat{n}\times\vec{c}\right)\cdot{\rm I}^{-1}\left(\hat{n}\times\vec{c}\right)\right)^{-1}
\end{aligned}$$
When you work out the components you will find
$$m^\star = \left( \tfrac{1}{m} + \tfrac{c^2 \sin^2 \varphi}{I_z} \right)^{-1} $$
which is identical to the 2D case when $\varphi= \tfrac{\pi}{2}$.
The concept of effective mass is important because it extends the concept of reduced mass from point particles to rigid bodies. In collisions between spheres you might have encountered $m^\star = \left( \tfrac{1}{m_1} + \tfrac{1}{m_2} \right)^{-1}$ as the reduced mass. Same for two bodies interacting due to gravity.
For two bodies in contact the effective mass between them is
$$ m^\star = \left( \tfrac{1}{m_1} + \tfrac{1}{m_2} + \left(\hat{n}\times\vec{c}_1\right)\cdot{\rm I}_1^{-1}\left(\hat{n}\times\vec{c}_1\right) + \left(\hat{n}\times\vec{c}_2\right)\cdot{\rm I}_2^{-1}\left(\hat{n}\times\vec{c}_2\right) \right)^{-1}$$
References

*

*Equation above for $m^\star$ is identical to equation 8-18 in An Introduction to Physically Based Modeling:
Rigid Body Simulation II—Nonpenetration
Constraints but with only one body involved.

A: If you know the point around which the system would rotate, you can apply the Conservation of angular momentum.Now, there is only one point about which a rigid system can freely rotate, that is the Center of mass.The center of mass rotation about any other point would accelerate and give impulses both in translational and radial motion, and so this requires an external force on the system.
Firstly,
Transform into a center-of-mass frame,where the center of mass stays in the origin.
Secondly,
Determine the angular momentum L in this frame with respect to the center of mass. It will indeed be conserved even if the collision is inelastic.
Then you can easily find the angular velocity of rotation (about the origin or the COM or the center of rotation) using,
...  L=Iw
Thus, w=L/I
W=omega
I = The moment of inertia about the com or the origin
