Instantaneous angular momentum of a disc 
Suppose we have a disk of radius $r$ and mass $m$ travelling at velocity $v$. I want to calculate the instantaneous angular momentum with axis through the edge of the disc (on the circumference).

Angular momentum $= I \omega$. $I = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$ by the parallel axis theorem. $\omega = \frac{v}{r}$. Therefore, angular momentum $= \frac{3mrv}{2}$.
Alternatively, angular momentum $=p\times r= m r \times v = mrv$.
Why do these two methods differ? Which, if any, are correct?
 A: From your description, I assume the disk only only translating, not rotating. Is this correct? If so, read on. If not, I'll delete.
I'm uncomfortable with the first method that uses $L=I\omega$. In this equation, it is assumed that every point on the rigid body can be characterized by the same angular velocity $\omega$. From your description of the motion of the disk, it seems like this doesn't apply here. The disk is only translating, not rotating about a point; thus, each point will have a different angular velocity. I don't have a problem with your expression for the moment of inertia $I$, but that would only be applicable if the object were rotating about a point on its edge.
I believe your second method $\vec{L}=m\vec{r}\times\vec{v}$ makes an assumption that the object is a point particle. You can see this because you treat each point in the body as being characterized by the same position vector $\vec{r}$. This may or may not lead to the correct answer. As another poster stated, the form you want to use is $L=\int \vec{r}\times d\vec{p} = \int_A \vec{r} \times \sigma \vec{v}\ dA$, where I've used a two-dimensional integral since you are treating the disk as two-dimensional. The term $\sigma$ is the two dimensional areal mass density $m/(\pi R^2)$. Let's continue with this integral to see where it leads.
$$L=\sigma\int_A \vec{r} \times \vec{v} \ dA 
=
\sigma\int_A (\vec{r}_{CM} + \vec{r}_{body}) \times \vec{v} \ dA
=
\sigma\int_A \vec{r}_{CM} \times \vec{v} \ dA
+
\sigma\int_A \vec{r}_{body} \times \vec{v} \ dA
$$
I've separated the position vector $\vec{r}$ into the sum of the vector to the center of mass and the vector from the center of mass to a general point on the body.
$$
L
=
\sigma\int_A \vec{r}_{CM} \times \vec{v} \ dA
+
\sigma \underbrace{\int_A \vec{r}_{body} \times \vec{v} \ dA}_{=0?}
\stackrel{?}{=} 
\sigma A \vec{r}_{CM}\times \vec{v}$$
I believe the integral with the underbrace is zero by symmetry. (Someone want to chime in?)
If so, this method produces your second result.
A: The first method is correct.
The second method is wrong because the equation you use only applies to point particles, not continuous masses with volume (such as a disk). You're incorrectly treating the disk as a point particle located at the disk center. If you want to use the second method, you'll need to use this equation for angular momentum of continuous masses:
$\vec{L} = \int_V dV \, \vec{r} \times \rho(\vec{r})\vec{v}$
A: Case a)
Body with uniform motion (no rotation), with C the center of the disk and A a point on the edge (for example below, at a distance $R$).
$$ \begin{aligned} 
\vec{v}_A & = (v,0,0) \\ 
\vec{\omega} & = (0,0,0) \\
\vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (v,0,0) \\
\vec{L} & = m \vec{v}_C = (m v,0,0) \\
\vec{H}_A & = I \vec{\omega} + (0,-R,0)\times \vec{L} = (0,0,R m v) \\
\end{aligned} $$
Where $\vec{L}$ is linear and $\vec{H}_A$ is angular momentum about point A.
Case b)
Body rolling with edge point A motionless, but with rotational speed $\Omega$
$$ \begin{aligned} 
\vec{v}_A & = (0,0,0) \\ 
\vec{\omega} & = (0,0,\Omega) \\
\vec{v}_C & = \vec{v}_A + (0,-R,0) \times \vec{\omega} = (\Omega R,0,0) \\
\vec{L} & = m \vec{v}_C = (m \Omega R,0,0) \\
\vec{H}_A & = I \vec{\omega} + (0,-R,0)\times \vec{L} = (0,0,\Omega (I_C+m R^2)) \\
\end{aligned} $$
The angular momentum here uses the parallel axis theorem and it expands to..
$$ v_C = \frac{\Omega}{R} \\ I_C = \frac{m}{2} R^2 $$
$$ \begin{aligned} 
L & = (m v_C,0,0) \\
H_A & = \frac{3}{2} m v_C R \\
\end{aligned} $$
So you see your two solutions correspond to two different problems.
