Tangential Velocity as a Cross Product

Referring again to Fig.B.4, we can write the tangential velocity vector $ \underline{v}$ as a vector cross product of the angular-velocity vector $ \underline{\omega}$B.4.11) and the position vector $ \underline{x}$:

$\displaystyle \underline{v}\eqsp \underline{\omega}\times \underline{x} \protect$ (B.17)

To see this, let's first check its direction and then its magnitude. By the right-hand rule, $ \underline{\omega}$ points up out of the page in Fig.B.4. Crossing that with $ \underline{x}$, again by the right-hand rule, produces a tangential velocity vector $ \underline{v}$ pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since $ \underline{\omega}$ and $ \underline{x}$ are mutually orthogonal, the angle between them is $ \pi /2$, so that, by Eq.$ \,$(B.16),

$\displaystyle \left\Vert\,\underline{\omega}\times \underline{x}\,\right\Vert \...
...,\right\Vert\cdot\left\Vert\,\underline{x}\,\right\Vert \eqsp \omega R \eqsp v
$

as desired.


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Relation of Angular to Linear Momentum
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Mass Moment of Inertia as a Cross Product