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Can someone help clarify this equation from classical dynamics? It doesn't seem to make sense. Here's my textbook's explanation.

A particle has position vector [itex]\vec{r}[/itex] in a non-rotating, inertial reference frame (the 'un-prime' frame). Suppose we want to observe the motion of this object in some rotating reference frame (the 'prime' frame) whose origin coincides with the origin of the inertial reference frame and is rotating with a constant angular velocity [itex]\vec{\omega}[/itex] (which points along the axis of rotation) with respect to the inertial frame.

First, suppose that our particle appears stationary in the rotating reference frame. Then, in the inertial reference frame, the particle appears to rotate about the axis of rotation:

[tex]\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}.[/tex]

(This makes sense so far.)

Now suppose that the particle moves with constant velocity [itex]\vec{v'}[/itex] as observed in the rotating reference frame. Then

[tex]\frac{d\vec{r}}{dt} = \vec{v'} + \vec{\omega} \times \vec{r}.[/tex]

^ I don't understand why that is true. For instance, suppose that in the rotating reference frame, [itex]\vec{v'}[/itex] = (1,0,0) and [itex]\vec{r}[/itex] at time t=0 is (0,0,0). I would expect that, in the inertial reference frame, the particle's position vector [itex]\vec{r}[/itex] should rotate about [itex]\vec{\omega}[/itex] while increasing linearly in magnitude. However, it seems that the second equation above predicts that the change in [itex]\vec{r}[/itex] will have one component from the cross-product term and one component from [itex]\vec{v'}[/itex], which is always (1,0,0) and thus always points along the x-axis of the inertial frame. This seems inappropriate; doesn't [itex]\vec{v'}[/itex] need to be transformed to [itex]\vec{v}[/itex] in order to be added in that equation?

A particle has position vector [itex]\vec{r}[/itex] in a non-rotating, inertial reference frame (the 'un-prime' frame). Suppose we want to observe the motion of this object in some rotating reference frame (the 'prime' frame) whose origin coincides with the origin of the inertial reference frame and is rotating with a constant angular velocity [itex]\vec{\omega}[/itex] (which points along the axis of rotation) with respect to the inertial frame.

First, suppose that our particle appears stationary in the rotating reference frame. Then, in the inertial reference frame, the particle appears to rotate about the axis of rotation:

[tex]\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}.[/tex]

(This makes sense so far.)

Now suppose that the particle moves with constant velocity [itex]\vec{v'}[/itex] as observed in the rotating reference frame. Then

[tex]\frac{d\vec{r}}{dt} = \vec{v'} + \vec{\omega} \times \vec{r}.[/tex]

^ I don't understand why that is true. For instance, suppose that in the rotating reference frame, [itex]\vec{v'}[/itex] = (1,0,0) and [itex]\vec{r}[/itex] at time t=0 is (0,0,0). I would expect that, in the inertial reference frame, the particle's position vector [itex]\vec{r}[/itex] should rotate about [itex]\vec{\omega}[/itex] while increasing linearly in magnitude. However, it seems that the second equation above predicts that the change in [itex]\vec{r}[/itex] will have one component from the cross-product term and one component from [itex]\vec{v'}[/itex], which is always (1,0,0) and thus always points along the x-axis of the inertial frame. This seems inappropriate; doesn't [itex]\vec{v'}[/itex] need to be transformed to [itex]\vec{v}[/itex] in order to be added in that equation?

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