First, consider that Earth's spin is at an angle to the orbital axis.
Here $$\begin{array}{r|c|c|c}\\ \text{Quantity} & \text{Symbol} & \text{Value} & \text{Units} \\ \hline
\text{orbital distance} & R & 1 & \text{AU} \\
& & 1.496\cdot 10^{11} & \text{m} \\
\text{orbital speed} & \Omega & 1 & \text{rev/year} \\
& & 1.991\cdot 10^{-7} & \text{rad/s} \\
\text{spin} & \omega & 1 & \text{rev/day} \\
& & 7.2921\cdot 10^{-5} & \text{rad/s} \\
\text{axial tilt} & \theta & 23.4 & \deg \\
& & 0.4084 & \text{rad}
\end{array}$$
The combined rotation (given the title about the negative x-axis from above) is
$$ \vec{w} = \pmatrix{0 \\ 0 \\ 1.991 \cdot 10^{-7}} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{bmatrix} \pmatrix{0\\0\\7.2921 \cdot 10^{-5}
} = \pmatrix{0 \\ 2.8961\cdot 10^{-5} \\ 6.7123\cdot 10^{-5} }\; \text{[rad/s]} $$
which can be translated into
$$ \vec{w} = \pmatrix{0 \\ 5.9735 \\ 13.845 } \; \text{[deg/hr]}$$
What is interesting is that you can calculate the instant center of rotation of the earth relative to the earth $(c_y,c_z)$ ($c_z$ shown negative below). This is the point which the earth is actually spinning about.
To find the point calculate the orbital speed (positive x-axis is out the page)
$$ \vec{v} = \vec{\Omega} \times \pmatrix{0\\-R\\0} = \pmatrix{ 2.9785\cdot 10^{4} \\ 0 \\0} \;\text{[m/s]}$$
and then the center of rotation
$$ \pmatrix{ 0 \\ c_y \\ -c_z} = \frac{ \vec{w} \times \vec{v}}{ \| \vec{w} \|^2} = \pmatrix{0 \\ 3.7410\cdot 10^{8} \\ -1.6141\cdot 10^{8} }\;\text{[m]} $$
which is interesting considering in lunar distance units (1 LD = 384402000
m)
$$ \pmatrix{ 0 \\ c_y \\ -c_z} = \pmatrix{ 0 \\ 0.9732 \\ -0.4199 }\;\text{[LD]} $$
which is almost one LD towards the sun always, and one-half LD under the earth in the summer solstice, and one-half LD over the earth in the winter solstice.
Now that the kinematics of the earth are established, we can talk about dynamics.
The earth is rotating with $\vec{w}$ and so its angular momentum at the center of the earth is $$\vec{L}_E = \mathrm{I}_E\, \vec{w}$$ where ${\rm I}_E$ is the mass moment of inertia of the earth.
But since the earth is translating also, it has linear momentum $$ \vec{p} = m_E \vec{v}$$.
To calculate the angular momentum of the earth about the sun, then we combine both quantities with the following rule
$$ \vec{L}_S = \vec{L}_E + \pmatrix{0\\-R\\0} \times \vec{p} $$
If you do the calculation you will find the majority of angular momentum along the z-axis, with a small component along the y-axis.
What is interesting is that you can find the location in space where the axis of percussion of the earth passes through. In a similar fashion to above, this point is
$$ \pmatrix{0\\h_y\\h_z} = \frac{ \vec{p} \times \vec{L}_E}{ \| \vec{p} \|^2} $$
The significance of this point in space is that if you were to apply an equal and opposite momentum $\vec{p}$ to the earth through the center of percussion, the earth would not only stop orbiting but also stop spinning. You can remove all the kinetic energy of the earth with one impulse through this point. It would stop the earth on its tracks.
