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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.enter image description here

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2 Answers 2

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enter image description here

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 $~x'(\tau)\,\ge 2\,\rho\sin(\theta)~$

enter image description here

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    $\begingroup$ @StevenOh I can give you just the simulation code containing the differential equations, not how you create those equations. $\endgroup$
    – Eli
    Apr 18, 2021 at 15:35
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    $\begingroup$ see drive.google.com/file/d/1SZfeNUlww4FzU14RTUWoCBRlCjXMp2R2/… $\endgroup$
    – Eli
    Apr 18, 2021 at 19:39
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    $\begingroup$ @StevenOh I looked again at your problem and find out that my solution wasn't quite correct, the vector R must be independent of $~\Omega$ , see also the modification on google cloud. sorry for that $\endgroup$
    – Eli
    Apr 19, 2021 at 15:23
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    $\begingroup$ did you got the last version of my file?, please put your comment in my Maple file, so I know what is your problem $\endgroup$
    – Eli
    Apr 22, 2021 at 12:52
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    $\begingroup$ you got second order differential equation for $~\ddot{X}(\tau)~$ ( (Maple ode) thus the numerical solution give you $~X(\tau)~$ the blue line is then the x,y components of $\vec R'$ $\endgroup$
    – Eli
    Apr 27, 2021 at 16:43
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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.

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