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Qmechanic
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Assume, there are 2 people staying next to each other on a rotating disk, s.t. there is a straight line from one to the other going through the center of the disk. One throws a ball to the other.

I want to discribe the motion of the ball relative to the person, who threw it:

Let $\vec{r_p}$ be the position of the person and $\vec{r_b}$ be the position of the ball, relative to the rotation axis.

Then, the position of the ball, relative to the person $\vec{r}^p_{b} = \vec{r_b} - \vec{r_p}$. Therefore, the relative velocity is $$ \frac{d}{dt}\vec{r}^p_{b} = \frac{d}{dt}\vec{r_b} - \frac{d}{dt}\vec{r_p} = \frac{d}{dt}\vec{r_b} - \vec{\omega} \times \vec{r_p} $$ and the acceleration $$ \frac{d^2}{dt^2}\vec{r}^p_{b} = \frac{d^2}{dt^2}\vec{r_b} - \frac{d}{dt}\vec{\omega} \times \vec{r_p} - \vec{\omega} \times (\vec{\omega} \times \vec{r}_p) $$

Now, the question is, why am I losing the coriolis term here (i.e. where and why did it go wrong)? Thanks!

Assume, there are 2 people staying next to each other on a rotating disk, s.t. there is a straight line from one to the other going through the center of the disk. One throws a ball to the other.

I want to discribe the motion of the ball relative to the person, who threw it:

Let $\vec{r_p}$ be the position of the person and $\vec{r_b}$ be the position of the ball, relative to the rotation axis.

Then, the position of the ball, relative to the person $\vec{r}^p_{b} = \vec{r_b} - \vec{r_p}$. Therefore, the relative velocity is $$ \frac{d}{dt}\vec{r}^p_{b} = \frac{d}{dt}\vec{r_b} - \frac{d}{dt}\vec{r_p} = \frac{d}{dt}\vec{r_b} - \vec{\omega} \times \vec{r_p} $$ and the acceleration $$ \frac{d^2}{dt^2}\vec{r}^p_{b} = \frac{d^2}{dt^2}\vec{r_b} - \frac{d}{dt}\vec{\omega} \times \vec{r_p} - \vec{\omega} \times (\vec{\omega} \times \vec{r}_p) $$

Now, the question is, why am I losing the coriolis term here (i.e. where and why did it go wrong)? Thanks!

Assume, there are 2 people staying next to each other on a rotating disk, s.t. there is a straight line from one to the other going through the center of the disk. One throws a ball to the other.

I want to discribe the motion of the ball relative to the person, who threw it:

Let $\vec{r_p}$ be the position of the person and $\vec{r_b}$ be the position of the ball, relative to the rotation axis.

Then, the position of the ball, relative to the person $\vec{r}^p_{b} = \vec{r_b} - \vec{r_p}$. Therefore, the relative velocity is $$ \frac{d}{dt}\vec{r}^p_{b} = \frac{d}{dt}\vec{r_b} - \frac{d}{dt}\vec{r_p} = \frac{d}{dt}\vec{r_b} - \vec{\omega} \times \vec{r_p} $$ and the acceleration $$ \frac{d^2}{dt^2}\vec{r}^p_{b} = \frac{d^2}{dt^2}\vec{r_b} - \frac{d}{dt}\vec{\omega} \times \vec{r_p} - \vec{\omega} \times (\vec{\omega} \times \vec{r}_p) $$

Now, the question is, why am I losing the coriolis term here (i.e. where and why did it go wrong)?

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Ilia
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The motion of a ball thrown in a rotating frame of reference

Assume, there are 2 people staying next to each other on a rotating disk, s.t. there is a straight line from one to the other going through the center of the disk. One throws a ball to the other.

I want to discribe the motion of the ball relative to the person, who threw it:

Let $\vec{r_p}$ be the position of the person and $\vec{r_b}$ be the position of the ball, relative to the rotation axis.

Then, the position of the ball, relative to the person $\vec{r}^p_{b} = \vec{r_b} - \vec{r_p}$. Therefore, the relative velocity is $$ \frac{d}{dt}\vec{r}^p_{b} = \frac{d}{dt}\vec{r_b} - \frac{d}{dt}\vec{r_p} = \frac{d}{dt}\vec{r_b} - \vec{\omega} \times \vec{r_p} $$ and the acceleration $$ \frac{d^2}{dt^2}\vec{r}^p_{b} = \frac{d^2}{dt^2}\vec{r_b} - \frac{d}{dt}\vec{\omega} \times \vec{r_p} - \vec{\omega} \times (\vec{\omega} \times \vec{r}_p) $$

Now, the question is, why am I losing the coriolis term here (i.e. where and why did it go wrong)? Thanks!