Calculate height of a projectile in rotating spaceship reference frame In a rotating reference frame, or in a rotating spaceship, the apparent gravity felt by an object is the centrifugal force. Any moving object also experiences a Coriolis force. When I want to do calculations with a projectile (an object thrown across the circle in such a rotating frame), how do I take into account that the Coriolis acceleration constantly changes? I have been doing some calculations using the projectile equation and changing the acceleration variable to the Coriolis and centrifugal acceleration. But obviously, this is not correct. Lastly, how can I derive the maximum height of such projectile in this rotating frame of reference, given launch angle, angular velocity, and the radius of the circle?
Any advice would be appreciated. Thank you.
 A: The efficient way to calculate that is to use the non-rotating coordinate system.
Given is:
1 Launch velocity with respect to the rotating frame
2 Angle with respect to the circular perimeter of the rotating frame
3 Angular velocity of rotating frame with respect to non-rotating frame

The velocity with respect to the non-rotating frame is the vector sum of:
1 The velocity with respect to the rotating frame
2 The instantaneous velocity of the point of launch with respect to the non-rotating frame
The motion of the projectile with respect to the non-rotating frame is along a straight line. So you can, for example, calculate when that straightline motion will intersect the perimeter again.

By contrast, if you would insist on doing the calculation exclusively in the rotating frame then I think numerical analysis is the only way. The motion with respect to the rotating frame is not some nice function, such as a parabola, or (semi)circle, or hyperbola, etc. The simplest algorithm for plotting an arbitrary trajectory is Euler's method.
Clearly, numerical analysis is an option only if you can set up a computer to do the calculation.
A: 
The position vector $~\mathbf r~$ is
$$\mathbf r=\mathbf R+\mathbf R',$$
where
$$\mathbf R= \left[ \begin {array}{c} -\rho\,\sin \left( \theta \right) 
\\ -\rho\,\cos \left( \theta \right) 
\\ 0\end {array} \right] 
$$
$$\mathbf R'=\left[ \begin {array}{c} X\\ {\frac {\sin \left( 
\alpha \right) X}{\cos \left( \alpha \right) }}-\frac 12\,{\frac {g{X}^{2}}
{{v}^{2} \left( \cos \left( \alpha \right)  \right) ^{2}}}
\\0\end {array} \right]
 $$
so that
$$\mathbf r=\left[ \begin {array}{c} -\rho\,\sin \left( \theta \right) +X
\\ -\rho\,\cos \left( \theta \right) +X\tan \left( 
\alpha \right) -\frac 12\,{\frac {g{X}^{2}}{{v}^{2} \left( \cos \left( 
\alpha \right)  \right) ^{2}}}\\ 0\end {array}
 \right]
$$
you have one generalized coordinate which is $~X$ you can now obtain the equation of motion with :
$$T=\frac m2\mathbf{\dot{r}}\cdot \mathbf{\dot{r}} \\
U=-m\,g\,\mathbf r_y$$
and the external forces due the rotation
$$\mathbf F_s=\left[ \begin {array}{c} m \left( -{\omega}^{2}r_{{x}}-2\,\omega\,v_{
{y}} \right) \\ m \left( -{\omega}^{2}r_{{y}}+2\,
\omega\,v_{{x}} \right) \\ 0\end {array} \right]  
$$
with Euler- Lagrange you obtain differential equation
$$\ddot X(\tau)=f(X(\tau)~,\dot X(\tau)~,\omega~,v~,\alpha,\rho)$$

*

*$\alpha~$ launch angle

*$v~$ launch velocity

*$\omega~$ angular velocity

*$~\rho~$ circular radius

with the solution $~X(\tau~)$
Numerical simulation
I stop the simulation if  $~x'(\tau)\,\ge 2\,\rho\sin(\theta)~$

