Where does the '2' Factor come from in the Coriolis Force? Say we have a disk rotating at $\omega$, and an observer standing at radius $R$. The observer throws something of mass $m$ with radial velocity $v_r$, and the goal is to make this object go in a straight path along the radius. The Coriolis force says that we must exert a horizontal force of $2m\omega v_r$ - however, when I derive this myself, I get only $m\omega v_r$, as follows.
From an outsider's perspective, if we want the object to stay in a straight line, as the object proceeds towards the center, its horizontal velocity must change according to its position on the disk, such that $dv=\omega dr$. Thus the force on the object must be $m\frac{dv}{dt}=m\omega\frac{dr}{dt}=m\omega v_r$. Where did I go wrong in my derivation?
 A: The usual derivation of the Coriolis force follows from analysing rotating frames.
We define the angular velocity vector as 
$$\underline \omega = \omega \underline{\hat{z}}$$
where $\underline{\hat{z}}$ is a unit vector pointing parallel to the rotation axis.
1) Consider now a vector $\underline{r}$ that is fixed in the rotating frame (RF), meaning that, with respect to an inertial frame (IR), it is rotating at the same angular speed $\omega$. It may be easier to visualise $\underline{r}$ as being the position vector describing the position of an object rotating at $\omega$. The velocity $\underline{v_I}$ of $\underline{r}$ in the IR is given by
$$\underline{v_I}=\underline{\omega}\times\underline{r}$$
2) Now consider a vector $\underline{A}$ moving at constant (but different) velocities in both the RF and in the IF, with its motion in each frame described by the following equations:
$\underline{A}=\left\{\begin{matrix}A_i\underline{\hat{i}}+A_j\underline{\hat{j}}+A_k\underline{\hat{k}} \textrm{   in RF}
\\ 
A_x\underline{\hat{x}}+A_y\underline{\hat{y}}+A_z\underline{\hat{z}} \textrm{   in IF}
\end{matrix}\right.$
where $\underline{\hat{i}}$, $\underline{\hat{j}}$ and $\underline{\hat{k}}$ are the base vectors of the RF, and $\underline{\hat{x}}$, $\underline{\hat{y}}$ and $\underline{\hat{z}}$ those of the IR.
The rate of change of $\underline{A}$ in the RF is
$$\frac{\mathrm{d}\underline{A} }{\mathrm{d} t}_R = \underline{v_R}=\frac{\mathrm{d}A_i }{\mathrm{d} t}\underline{\hat{i}}+ \frac{\mathrm{d}A_j }{\mathrm{d} t} \underline{\hat{j}}+ \frac{\mathrm{d}A_k }{\mathrm{d} t}\underline{\hat{k}}$$
The rate of change of $\underline{A}$ in the IF is a bit more subtle, since $\underline{\hat{i}}$, $\underline{\hat{j}}$ and $\underline{\hat{k}}$ are not fixed in the IF and therefore their time derivatives are non-zero.
$$\frac{\mathrm{d}\underline{A} }{\mathrm{d} t}_I = \underline{v_I}= \frac{\mathrm{d} }{\mathrm{d} t} \left ( A_i\underline{\hat{i}}+A_j\underline{\hat{j}}+A_k\underline{\hat{k}} \right )$$
Using the product rule,
$$\underline{v_I}= \frac{\mathrm{d}A_i }{\mathrm{d} t}\underline{\hat{i}}+ \frac{\mathrm{d}\underline{\hat{i}}}{\mathrm{d} t}A_i+ \frac{\mathrm{d}A_j }{\mathrm{d} t} \underline{\hat{j}}+\frac{\mathrm{d}\underline{\hat{j}}}{\mathrm{d} t}A_j+ \frac{\mathrm{d}A_k }{\mathrm{d} t}\underline{\hat{k}}+\frac{\mathrm{d}\underline{\hat{k}}}{\mathrm{d} t}A_k$$
Notice that the first, third and fifth terms on the RHS are simply the expression of $\underline{v_R}$. Therefore, we can write
$$\underline{v_I}= \underline{v_R} + \frac{\mathrm{d}\underline{\hat{i}}}{\mathrm{d} t}A_i+\frac{\mathrm{d}\underline{\hat{j}}}{\mathrm{d} t}A_j+\frac{\mathrm{d}\underline{\hat{k}}}{\mathrm{d} t}A_k = \underline{v_R}+A_i\underline{v_{\underline{\hat{i}}}}+A_j\underline{v_{\underline{\hat{j}}}}+A_k\underline{v_{\underline{\hat{k}}}}$$
Since $\underline{\hat{i}}$, $\underline{\hat{j}}$ and $\underline{\hat{k}}$ are effectively fixed vectors in the RF, we can use the reasoning in part (1) to write that
$$\underline{v_{\underline{\hat{i}}}} = \underline{\omega} \times \underline{\hat{i}}$$
$$\underline{v_{\underline{\hat{j}}}} = \underline{\omega} \times \underline{\hat{j}}$$
$$\underline{v_{\underline{\hat{k}}}} = \underline{\omega} \times \underline{\hat{k}}$$
Therefore, substituting these values into the expression for $\underline{v_I}$,
$$\underline{v_I}=  \underline{v_R}+A_i\left ( \underline{\omega} \times \underline{\hat{i}} \right ) +A_j \left ( \underline{\omega} \times \underline{\hat{j}} \right ) +A_k \left ( \underline{\omega} \times \underline{\hat{k}} \right )$$
Rearranging,
$$\underline{v_I}=  \underline{v_R}+ \underline{\omega} \times \left ( A_i \underline{\hat{i}} + A_j \underline{\hat{j}} + A_k \underline{\hat{k}} \right )$$
Note that the expression in the parenthesis is simply $\underline A$.
$$\underline{v_I}=  \underline{v_R}+ \underline{\omega} \times \underline{r}$$
where we have switched back from using $\underline{A}$ to $\underline{r}$ (this is just notation). In other words, the velocity in the IF is equal to the sum of the velocity *in the RF and the velocity of the RF.
3) Now consider a vector with a non-zero acceleration $\underline{a_R}$ in the RF. Its acceleration in the IF is
$$\underline{a_I}=\frac{\mathrm{d}\underline{v_I} }{\mathrm{d} t}_I=\frac{\mathrm{d}}{\mathrm{d} t}_I \left ( \underline{v_R}+ \underline{\omega} \times \underline{r} \right ) =\frac{\mathrm{d}}{\mathrm{d} t}_I \left ( \underline{v_R}\right ) + \frac{\mathrm{d}}{\mathrm{d} t}_I \left (\underline{\omega} \times \underline{r} \right ) $$
Note that $\underline{v_R}$ is a vector moving at a constant velocity in the RF. Using the reasoning from part (2), we can write
$$\frac{\mathrm{d}}{\mathrm{d} t}_I \left ( \underline{v_R}\right ) = \frac{\mathrm{d}}{\mathrm{d} t}_R \left ( \underline{v_R}\right ) +\underline{\omega} \times \underline{r} = \underline{a_R} + \underline{\omega} \times \underline{v_R}$$
The other therm on the RHS becomes, using the product rule for the cross product,
$$\frac{\mathrm{d}}{\mathrm{d} t}_I \left (\underline{\omega} \times \underline{r} \right ) = \underline{\omega} \times \frac{\mathrm{d}\underline{r}}{\mathrm{d} t}_I + \frac{\mathrm{d}\underline{\omega}}{\mathrm{d} t}_I\times \underline{r}$$
Therefore, the expression for the acceleration in the IF becomes
$$\underline{a_I}=\underline{a_R} + \underline{\omega} \times \underline{v_R} +  \underline{\omega} \times \frac{\mathrm{d}\underline{r}}{\mathrm{d} t}_I + \frac{\mathrm{d}\underline{\omega}}{\mathrm{d} t}_I\times \underline{r}$$
We can now replace
$$\frac{\mathrm{d}\underline{r}}{\mathrm{d} t}_I = \underline{v_I}$$
and
$$\frac{\mathrm{d}\underline{\omega}}{\mathrm{d} t}_I = \underline{\dot{\omega}}$$
to give
$$\underline{a_I}=\underline{a_R} + \underline{\omega} \times \underline{v_R} +  \underline{\omega} \times \underline{v_I} + \underline{\dot{\omega}} \times \underline{r}$$
Finally, we substitute, from part 2,
$$\underline{v_I} = \underline{v_R} + \underline{\omega}\times \underline{r}$$
to give
$$\underline{a_I}=\underline{a_R} + \underline{\omega} \times \underline{v_R} +  \underline{\omega} \times \underline{v_R} + \underline{\omega} \times \left ( \underline{\omega}\times \underline{r} \right )  + \underline{\dot{\omega}} \times \underline{r}$$
As you can see, there are two $\underline{\omega} \times \underline{v_R}$ terms. This is where the factor of 2 comes from. Grouping terms together,
$$\underline{a_I}=\underline{a_R} + 2 \underline{\omega} \times \underline{v_R} + \underline{\omega} \times \left ( \underline{\omega}\times \underline{r} \right )  + \underline{\dot{\omega}} \times \underline{r}$$
The second term on the RHS is indeed the Coriolis acceleration and it has a factor of 2 in front of it. Just multiply all of the terms by the mass $m$ in the above expression to get a formula for the net force. Therefore, the Coriolis force is
$$\underline{F}= - 2m \underline{\omega} \times \underline{v_R}$$
(the minus is there just because we have rearranged the equation in terms of the acceleration in the rotating frame).
Sorry I couldn't give a more intuitive or logical explaination - I am not aware of one, so I thought that if you saw the proper mathematical derivation you'd be satifised to "accept" that there just has to be a factor of 2 in there.
