Conservation of angular momentum on Earth If I start spinning while standing on the floor, conservation of angular momentum says something needs to start spinning in the opposite direction but what's that?
 A: Angular momentum is conserved only in closed systems. Since you're applying an external torque to start the rotation (perhaps by pushing something else to start off), it isn't a valid principle any more. When you consider a larger system, for example the combination of the planet and the person, angular momentum is conserved: if you push the ground to start the rotation, the planet's angular momentum changes (although this isn't very visible; Garf's answer provides an excellent numerical estimate), or if you push a different object next to you, it starts rotating. The same explanation applies even when you're on a different planet.
Conservation laws generally need to be applied with caution. When there's an external force, you can't always say that linear momentum is conserved, can you?
A: If I want to start spinning, I have to push on something (try imagining starting to spin in space - you won't be able to because there will be nothing to push off of).
This will most likely be the ground, or if you like the Earth itself, and so by changing my angular momentum, I also have to change the Earth's angular momentum.
However, because the Earth is so big and heavy compared to myself, me pushing on the ground has a negligible effect on the motion of the Earth.
A very simple example - say I gain angular momentum
\begin{align}
L&=mvr\\&\sim100\text{ kg }\times1\text{ m s}^{-1}\text{ }\times1\text{ m}\sim100\text{ kg m}^{2}\text{ s}^{-1}
\end{align}
then the change in angular momentum that must occur on the Earth is
\begin{align}
\Delta L&\sim100\text{ kg m}^{2}\text{ s}^{-1}\\&=M_\text{Earth}\Delta v_{\text{Earth}}R_\text{Earth}\\&\sim6\times10^{24}\text{ kg }\times\Delta v_\text{Earth}\times6\times10^{6}\text{ m}
\end{align}
and solving for $\Delta v$, we find the change in the Earth's velocity at the surface of the Earth is
$$\Delta v \sim 3\times10^{-30}\text{ m s}^{-1}$$
I think it's safe to say you can spin freely without worrying about disrupting the planet's spin.
