1. ## Curl (physics-y)

Stewart 16.5 #37 (a)

Let $\displaystyle B$ be a rigid body rotating about the $\displaystyle z$-axis. The rotation can be described by the vector $\displaystyle {\bf w} = \omega {\bf k}$ where $\displaystyle \omega$ is the angular speed of $\displaystyle B$, that is, the tangential speed of any point $\displaystyle P$ in $\displaystyle B$ divided by the distance $\displaystyle d$ from the axis of rotation. Let $\displaystyle {\bf r} = < \! x, y, z \! >$ be the position vector of $\displaystyle P$.

By considering the angle $\displaystyle \theta$ in the figure, show that the velocity field of $\displaystyle B$ is given by $\displaystyle {\bf v} = {\bf w} \times {\bf r}$.

I'll try to produce the figure mentioned in the problem if anybody wants it, but basically $\displaystyle \theta$ is the measure of the angle between $\displaystyle {\bf r}$ and $\displaystyle {\bf k}$.

Now part of my problem is that I know jack about physics so I don't really know what a velocity field should represent. Angular speed? Tangential speed? I'm not totally clear on what the difference between these are anyway. Seriously, I know nothing about physics, I haven't taken a single class in it. But I guessed about this much:

Half the battle will be done if I show that speed $\displaystyle s$ at the point $\displaystyle P$ is $\displaystyle s = |{\bf v}| = |{\bf w} \times {\bf r}| = |{\bf w}||{\bf r}| \sin \theta = (d \cdot \omega) |{\bf r}| = (d \cdot \frac{s}{d})|{\bf r}|$ but something here is clearly confused (bah-dum-ching) since I have no guarantee that $\displaystyle |{\bf r}| = 1$.

The other half of the battle should be done by noting that the cross-product of of the two vectors will be normal to the plane formed by them thus pointing the vector in the appropriate direction. That's not exactly rigorous but I'm kind of okay with that at 1:30 AM.

2. Originally Posted by ragnar
Stewart 16.5 #37 (a)

Let $\displaystyle B$ be a rigid body rotating about the $\displaystyle z$-axis. The rotation can be described by the vector $\displaystyle {\bf w} = \omega {\bf k}$ where $\displaystyle \omega$ is the angular speed of $\displaystyle B$, that is, the tangential speed of any point $\displaystyle P$ in $\displaystyle B$ divided by the distance $\displaystyle d$ from the axis of rotation. Let $\displaystyle {\bf r} = < \! x, y, z \! >$ be the position vector of $\displaystyle P$.

By considering the angle $\displaystyle \theta$ in the figure, show that the velocity field of $\displaystyle B$ is given by $\displaystyle {\bf v} = {\bf w} \times {\bf r}$.

I'll try to produce the figure mentioned in the problem if anybody wants it, but basically $\displaystyle \theta$ is the measure of the angle between $\displaystyle {\bf r}$ and $\displaystyle {\bf k}$.

Now part of my problem is that I know jack about physics so I don't really know what a velocity field should represent. Angular speed? Tangential speed? I'm not totally clear on what the difference between these are anyway. Seriously, I know nothing about physics, I haven't taken a single class in it.
The velocity field of B is the velocity vector of each point of B.

Half the battle will be done if I show that speed $\displaystyle s$ at the point $\displaystyle P$ is $\displaystyle s = |{\bf v}| = |{\bf w} \times {\bf r}| = |{\bf w}||{\bf r}| \sin \theta = (d \cdot \omega) |{\bf r}| = (d \cdot \frac{s}{d})|{\bf r}|$ but something here is clearly confused (bah-dum-ching) since I have no guarantee that $\displaystyle |{\bf r}| = 1$.

The other half of the battle should be done by noting that the cross-product of of the two vectors will be normal to the plane formed by them thus pointing the vector in the appropriate direction. That's not exactly rigorous but I'm kind of okay with that at 1:30 AM.

3. So I'm still lost on this problem since I'm still not sure what kind of velocity we're talking about or what the difference is between the kinds of velocity. And using the only notion of velocity I understand (not the angular one), I still get the wrong result.