# Homework Help: Velocities in rotating and inertial reference frames

1. Mar 16, 2008

### gabee

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 $\vec{r}$ 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 $\vec{\omega}$ (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:

$$\frac{d\vec{r}}{dt} = \vec{\omega} \times \vec{r}.$$

(This makes sense so far.)

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

$$\frac{d\vec{r}}{dt} = \vec{v'} + \vec{\omega} \times \vec{r}.$$

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

Last edited: Mar 16, 2008
2. Mar 16, 2008

### Staff: Mentor

The first term on the RHS represents the rate of change relative to the rotating frame; the second term represents the rate of change of the rotating frame relative to the fixed frame.

No. It always points along the x-axis of the rotating frame.
Sure. If you want to express the velocity in the non-rotating reference frame you must express that velocity in terms of non-rotating coordinates.

3. Mar 16, 2008

### D H

Staff Emeritus
You are confusing the representation of a vector with the vector itself. The displacement vector between a pair of points has different coordinates when expressed in different reference frames. The displacement vector itself is an invariant. What changes upon transformation are the coordinates of the vector, not the vector itself.

While a vector represents the same thing regardless of the frame in which the coordinates are expressed, the time derivative of a vector quantity depends on the frame in which the derivative is computed. Transforming the velocity vector of some object as observed by someone fixed in the rotating frame is merely the rotating frame velocity vector expressed in inertial coordinates. This is substantially different from the velocity vector of the same object as observed by someone fixed in the inertial frame.

You can arrive at the same equation by noting that

$$\mathbf q_I = \mathbf T_{R\to I} \mathbf q_R$$

where $\mathbf q$ is some vector quantity (here assumed to be a column vector), the subscripts I and R denote the representation of the vector in inertial and rotating coordinates, and $\mathbf T_{R\to I}$ is the transformation matrix from rotating to inertial coordinates. Taking the time derivative,

$$\mathbf q'_I = \mathbf T'_{R\to I} \mathbf q_R + \mathbf T_{R\to I} \mathbf q'_R$$

The time derivative of the rotating to inertial transformation matrix is

$$\mathbf T'_{R\to I} = \mathbf T_{R\to I}\mathbf X(\mathbf \omega)$$

where $\mathbf X(\mathbf \omega)$ is the skew-symmetric cross product matrix generated from $\mathbf \omega$. Since matrix multiplication is associative,

$$\mathbf q'_I = \mathbf T_{R\to I}(\mathbf X(\mathbf \omega)\mathbf q_R + \mathbf q'_R) = \mathbf T_{R\to I}(\mathbf \omega\times\mathbf q_R + \mathbf q'_R)$$

4. Mar 16, 2008

### gabee

Got it. That's a great explanation, thank you very much!