Having some trouble with acceleration in polar coordinates So, I solved a question about acceleration in polar coordinates, but most people in my class (Classical Physics, first year at university studying Physics) disagree with my answer. So the question is about a playground roundabout with radius 3m rotating at a speed of 10m/s. A person on the roundabout throws a snowball towards the centre of the roundabout at a speed of 20m/s. What acceleration will the snowball experience that will cause it to miss the centre?
So, I started with the formula for acceleration in polar coordinates:
$$\vec{a} = \hat{r}(\ddot{r} - r\dot{\theta}^2) + \hat{\theta}(r\ddot{\theta} + 2\dot{r}\dot{\theta}).$$
I let the centre of the roundabout be the origin, and interpreted the problem like this: $r$ is the distance from the snowball to the centre of the roundabout, $\dot{r}$ is then the one dimensional velocity of the snowball, and $\ddot{r}$ would be the rate of change of $\dot{r}$. $\dot{\theta}$ is the angular velocity of the roundabout, and $\ddot{\theta}$ the angular acceleration.
So, from the problem statement, I picked out the following values:
$\ddot{r} = 0, \dot{r} = -20 \text{ ms}^{-1}, r = (3 - 20t)m, \ddot{\theta} = 0, \dot{\theta} = 3.33 \text{ rad s}^{-1}$
And then I just plugged these values into the formula above. Then I get an answer where the component of $\vec{a}$ in the $\hat{r}$ direction depends on $t$, and there is a constant component in the $\hat{\theta}$ direction.
However, most people in my class didn't use this formula, and are arguing that there should only be a corilios acceleration, e.g. the theta component in this case, as $\ddot{\theta} = 0.$ That doesn't make sense to me, as I was under the impression that the acceleration formula should always work in polar coordinates. It certainly looks like that to me when the first line in the derivation is as general as just $\vec{r} = r\hat{r}$, and then we take first and second derivatives of $\vec{r}$
So, what I want to know is who is right? Am I right that we can simply apply this formula, or are my classmates right, and there is only a corilios acceleration? Also, are the values I picked out for $\ddot{r}, \dot{r}, r, \ddot{\theta}, \dot{\theta}$ correct?
Any help would be much appreciated!
 A: Ignoring z motion in the following.
Reference frame:"lab"-- the one where roundabout is rotating. Right handed, origin at roundabout center.
The trajectory is a straight line. There is no acceleration. The reason the ball misses the center is because of its initial conditions being such-there was always an initial tangential($\hat{\theta}$) velocity.
Reference frame:"rotating"-- the one where roundabout is at rest. Coincides with lab at $t=0$
At $\mathbf{t=0}$
The object has only radial velocity($-\hat{r}$). In theory it should hit the center. The only reason it won't is if something accelerated it tangentially. This come from the pseudo-forces.
The object does experience acceleration:


*

*Coriolis: $\propto -\vec{\omega} \times \vec{v} $. Here, since $\hat{v}=-\hat{r}$, the acceleration is exactly what we want: along $\hat{\theta}$.

*Centrifugal: $\propto -\vec{\omega} \times (\vec{\omega} \times \vec{r}) $. Here, since $\hat{v}=-\hat{r}$, the acceleration is along $\hat{r}$. Won't affect hitting the center.


At $\mathbf{t\gt0}$


*

*The object is starting to move tangentially. At the same time its radial velocity is being decreased by the centrifugal force. Also Coriolis force from tangential motion is also along centrifugal. All in all the object moves as if it was going forth while curving in the direction of rotation.(see fig. 1 below.)

*Eventually the object outright turns back and seems to be escaping the roundabout(see fig. 2 below).

*now the direction of Coriolis force switches....


All in all, the object moves in an ever increasing spiral. Note that the acceleration keeps changing in time.  
Conclusion
So whose frame should we consider?.Depends on the observer-if its the person on the roundabout, its the rotating frame.  The final acceleration must, of course include gravity.
The stated values for  $\ddot{r}, \dot{r}, r, \ddot{\theta}, \dot{\theta}$ seem correct for the lab frame.   
In rotating frame
$\ddot{r} \ne 0, \dot{r} = -20 \text{ ms}^{-1}, r \ne(3 - 20t)m, \ddot{\theta} \ne 0, \dot{\theta} = 0 \text{ rad s}^{-1}$


fig 2: Trajectory initially. (The blue curve is the trajectory seen in rotating frame. Orange is where the person would be in the lab frame.X and y axes are x and y postions in meters)

fig 2: Trajectory after some time. (The blue curve is the trajectory seen in rotating frame. Orange is where the person would be in the lab frame.)


As far as the formula you have stated, its applicable to inertial
  frames only. In particular, for rotating frames, use ( $'$ denotes
  rotating frame)
$$ m\vec{a}'=\vec{F}'
 -m\frac{d\vec{\omega}}{dt}  \times \vec{r}'
 -2 m \vec{\omega} \times m \vec{v}'
 -m\vec{\omega} \times (\vec{\omega} \times \vec{r}') $$ with $$ \vec{F}'= \hat{r}’(\ddot{r}’ - r'\dot{\theta}’^2) +
 \hat{\theta}’(r'\ddot{\theta}’ + 2\dot{r}’\dot{\theta}’)\\
 \vec{v}'=\dot{r}'\hat{r}'+r'\dot{\theta}'\hat{\theta}' 
$$
  At $t=0$, with no applied force
$$
\vec{\omega}=\omega\hat{z}'\\ \vec{r}'=R \vec{\hat{r}}'\\ \vec{v}=-v
 \vec{\hat{r}}'\\ \vec{F}'=0\\\ $$   we get $$ \vec{a}'=2 m \omega v
 \hat{\theta}'+m \omega^{2}R\hat{r}' 
$$

A: Perhaps this defeats the point of the exercise, but it seems to me that if you want to know what the snowball does then why not just calculate the motion of the snowball, which has nothing at all to do with the motion of the roundabout. It just flies in a parabola, whose horizontal part is a straight line relative to the ground. If the initial velocity is towards the centre of the roundabout then it will hit the centre of the roundabout.
Even if you are intended to carry out a calculation in some other frame, it surely helps to know what the answer is when calculated the easiest way.
A: The initial radial velocity is the cause of veer-off in a rotating frame.
In circumferential dynamic equilibrium
Coreolis acceleration balances the angular acceleration
$$ \alpha_r = -2 \omega \dot {r}_{radial} $$ 
and is the de-tracking component. 
