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Imagine two reference frames that share same origin. One is static (S frame) and the other one is rotating (S' frame) with $\mathbf \Omega = \Omega_0 \mathbf k$. Imagine we throw a ball in an straigth line in the inertial frame. The equation describing the ball in the $S$ frame is: $$ m\frac{d^2\mathbf r}{dt^2} =0 $$ The equation describing the particle in the $S'$ frame is: $$ m\frac{d^2 \mathbf r'}{dt^2} = \mathbf F + 2m\dot{\mathbf r}' \times \mathbf \Omega +m (\mathbf \Omega \times \mathbf r') \times \mathbf \Omega $$ The force is zero, because the particle does not suffer an acceleration in the inertial frame: $\mathbf F = 0$. We will assume that the centrifugal force is zero, for simplicity. $$ m\frac{d^2 \mathbf r'}{dt^2} = 2m\dot{\mathbf r}' \times \mathbf \Omega $$ The we obtain: $$ \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix} = \begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ \Omega_0 \end{pmatrix} = \begin{pmatrix} \Omega_0 \dot{y} \\ -\Omega_0 \dot{x} \\ 0 \end{pmatrix} $$

Solving the Equation yields (ignoring the z coordinate): $$ x(t) = A(1 - \cos(\Omega_0 t)) + B\sin(\Omega_0 t) + C \\ y(t) = B(1 - \cos(\Omega_0 t)) + A\sin(\Omega_0 t) + D $$ This vector $\mathbf r_{S'}^{S'} = (x(t), y(t))$ are the position of the particle relative to $S'$ (denoted by the subscript) in the $S'$ coordinates (denoted by the superscript).

My question is, how would you express this vector fully in the $S$ frame ($r_{S}^{S}$). There is no translating because the origins are the same (hence $\mathbf r_{S'}^{S'} = \mathbf r_{S}^{S'}$), but the I do not know what is the rotation needed to go from $S'$ to $S$. Is this reasoning even right?

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    $\begingroup$ 1) why do you ignore centrifugal force? (And don’t call it centripetal) $\endgroup$
    – JEB
    Oct 24, 2023 at 16:51
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    $\begingroup$ 2) have you considered transforming an inertial path in S over to S’? $\endgroup$
    – JEB
    Oct 24, 2023 at 16:53
  • $\begingroup$ You might also want to try some limits that are easier to understand to build some intuition. Like, what does a stationary particle look like? What does a particle moving directly to/away from the center look like, in the limit that its velocity is slow compared to the rotational speed? What about when it moves fast compared to the rotational speed? $\endgroup$
    – Andrew
    Oct 24, 2023 at 17:03
  • $\begingroup$ @JEB I ignore the centrifugal force because I assume $\Omega_0$ is small, so: $\Omega_0^2 \approx 0$ $\endgroup$ Oct 24, 2023 at 18:18
  • $\begingroup$ In the small omega limit, just solve for a particle in a uniform magnetic field. $\endgroup$
    – JEB
    Oct 24, 2023 at 20:28

1 Answer 1

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In this answer I go to a wider perspective; I will discuss the type of case where a rotating coordinate system is used. (And where it is actually useful to apply a rotating coordinate system.)

There is in Grenoble, France, a facility where a large scale setup is used for experiments in fluid dynamics of a rotating system. This facility has a 13 meter diameter rotating platform, this platform is in effect a dish filled with water.

In accordance with the rotation the surface of the water is concave. (Comparison with something that we know from daily life: when you have stirred liquid in a cup the surface of the swirling fluid is concave.)

Let the angular velocity of the platform be called $\Omega$. (Capital letter 'Omega'.) The floor of the rotating platform is concave, the shape matches the shape of the surface of the water at the operating angular velocity $\Omega$.

If the platform would not be rotating then any water poured onto the platform would flow down the slope, gathering at the lowest point. The slope of the floor is redirecting gravity. At the operating angular velocity $\Omega$ the slope of the floor is providing the required centripetal force to make water residing on the platform circumnavigate the central axis at angular velocity $\Omega$.

When using a rotating coordinate system two terms are added to the equation of motion: a centrifugal term and a coriolis term.

In the case of a physically rotating platform (such as that facility in Grenoble), we have the following in the equation of motion: the centripetal term and the centrifugal term drop away against each other. They are both proportional to the angular velocity $\Omega$ and they are opposite in direction.

That type of situation is the use case for a rotating coordinate system. You get a simplification because out of three terms two drop away against each other.


All instances where it is actually useful to apply a rotating coordinate system have that element in common: the objects that you are interested in are in physical contact with the rotating system, such that they are subject to a centripetal force proportional to the angular velocity of the rotating system.


In your question you describe the case of the motion of an object that is not in any physical contact with some rotating system. In that case using a rotating coordinate system is a pointless exercise.

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  • $\begingroup$ Yeah, I know it might be easily done in cartesian coordinates. But I just wanted to derive the path of an ball following an straight line as seen form the rotating frame, for practicing and understanding the equations. $\endgroup$ Oct 25, 2023 at 22:29

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