If we have an object of mass $m$, moving on a path of constant longitude travelling at speed v, what will the force on the object be just as it passes the South Pole? We are to assume the Earth is perfectly spherical

This question involves rotating frames, the formula we were provided was $$(\ddot{\vec{r}})_S=(\ddot{\vec{r}})_{S'}+\vec\omega \times( \vec\omega \times \vec r)+2\vec\omega \times (\dot{\vec{r}})_{S'}+\dot{\vec{\omega}} \times\vec{r}$$

Where $S$ is the inertial frame and $S'$ is the rotating frame. I am not really sure how I am to go about this, I know that $F=m(\ddot{\vec{r}})$ but I'm not sure how to calculate the other terms.

  • $\begingroup$ Hint: $\vec{\omega}$ and $\vec{r}$ are parallel (or antiparallel, depending on convention) at the South Pole, so one of the terms is immediately zero. $\endgroup$ – probably_someone Jun 6 '19 at 11:02
  • $\begingroup$ Do we need to worry about the $\dot{\vec{\omega}} \times\vec{r}$ term? isnt $\vec{\omega}$ basically constant $\endgroup$ – user3184807 Jun 6 '19 at 11:16
  • $\begingroup$ Yes, that term is important, since the path is one of constant longitude, and circles of constant longitude rotate as the Earth rotates. If the object is moving quickly enough, this term might be small, but we don't know how the speed of the object compares to the tangential speed of the Earth's rotation. If the object is moving really slowly, then this term might be dominant. $\endgroup$ – probably_someone Jun 6 '19 at 12:03

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