From "Centrifugal force", Wikipedia:



By applying the transformation above from the stationary to the rotating frame three times the absolute acceleration of the particle can be written as: \begin{aligned}{\boldsymbol {a}}&={\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}={\frac {\operatorname {d} }{\operatorname {d} t}}{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}={\frac {\operatorname {d} }{\operatorname {d} t}}\left(\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times {\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times \left(\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+2{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})\ .\end{aligned}

Here I have seen that the last term, generally $\mathbf{\omega} {\times}\left(\mathbf{\omega} {\times} \mathbf{r} \right)$. The other terms nullify out possibly.

Question: Why do the rotating frame problems require usage of non-inertial frames instead of ground frames for general calculations?

  • $\begingroup$ "nullify out", like, when $\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=\mathbf{0}$ and $\frac{\mathrm{d}\mathbf{\omega}}{\mathrm{d}t}=\mathbf{0}$? $\endgroup$
    – Nat
    May 23, 2018 at 7:33
  • $\begingroup$ I'm not sure it's true that rotating frame problems require usage of non-inertial frames. In general I would advise students to avoid working in non-inertial frames because they can be unintuitive and it's easy to make mistakes. $\endgroup$ May 23, 2018 at 7:39
  • $\begingroup$ Get some friends together and find a children's playground with a roundabout - the larger the better. (Ignore any "children only" signs as this is in the interests of science. ) Get up to speed and then feel the forces: try some simple experiments like swinging a pendulum. That will give you some intuition to make sense of the mathematics. $\endgroup$ May 23, 2018 at 8:15

2 Answers 2


Why do the rotating frame problems require usage of non-inertial frames instead of ground frames for general calculations?

They don't. Using a rotating frame can sometimes makes a problem easier to solve than other choices of a frame of reference, particularly if the non-inertial effects can be ignored or are built-in. With regard to the latter, the fictitious centrifugal force is "built-in" when using a ground frame. The gravitational acceleration g is the vector sum of the acceleration due to gravity and the centrifugal acceleration.

With regard to the former, the Coriolis effect is typically ignored in introductory physics problems that calculate how far a cannonball flies. Assuming gravitational acceleration is a constant vector and ignoring aerodynamic drag and the Coriolis effect results in a nice simple model, parabolic flight, that students can use to solve problems. Ignoring the Coriolis effect is consonant with those other simplifying assumptions.

Those assumptions are not consonant with a cannonball (or some other ballistic projectile) that rises far above the Earth's atmosphere only to come back to Earth on another continent. That however is a problem for students of Global Thermonuclear War rather than students of introductory physics.

  • $\begingroup$ Whether rotation effects are "built in" and whether the ground frame is used are independent from each other. In geophysics it is customary to give the actually measured gravitational acceleration, which, as you point out, is smaller than the true grav. acceleration, due to the Earth's rotation. In the case of meteorology: in the equations of meteorological models the centrifugal term is omitted. If the centrifugal term would not be omitted then the centrifugal effect would be introduced twice: first by using the measured gravitational acceleration, and once again with the centrifugal term. $\endgroup$
    – Cleonis
    May 23, 2018 at 21:39

Chiming in with the answer provided by David Hammen:
Depending on the thing you want to compute you make an assessment as to which is more practical.

Some examples:
Let's say you want to calculate ballistic trajectories, in the simplified case of an Earth with no atmosphere. You have a computer, and you have the skill to program calculations.

Then it is practical to use the non-rotating coordinate system that is co-moving with the Earth's center. You find the starting velocity and direction of the projectile, and using that you find numerically how long its flight lasts. During the flight the Earth has turned underneath the flying projectile, so the last step of that calculation is to factor in how much the Earth has turned during the flight. Advantage of using the inertial coordinate system: the trajectory of the flight is a Kepler orbit; very straightforward to compute.

Now let's say you want a more accurate simulation, and you do want to take air friction into account. The Earth's air mass is co-moving with the Earth, so it has a velocity relative to the non-rotating coordinate system that you have to factor in.

Another example of a case where all the input data are given relative to the coordinate system that is co-rotating with the Earth: Meteorology. By using the coordinate system that is co-rotating with the Earth all of the input data go straight in. So that would be a case where it is more practical to use the co-rotating coordinate system.

General remark:
As emphasized by John Rennie: the basis of understanding motion is to think in terms of the motion with respect to the inertial coordinate system.

To use laws of motion there is only one choice of reference frame: the equivalence class of inertial coordinate systems. There are no exceptions to that; the 'omega' in the expression of the transformation is the angular velocity of the rotating coordinate system with respect to the equivalence class of inertial coordinate systems.
That is: the equations for the rotating coordinate system work because they reference the inertial coordinate system.


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