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Referring again to Fig.B.4, we can write the
tangential velocity vector
as a vector cross product of
the angular-velocity vector
(§B.4.11) and the position
vector
:
![$\displaystyle \underline{v}\eqsp \underline{\omega}\times \underline{x} \protect$](img2852.png) |
(B.17) |
To see this, let's first check its direction and then its magnitude.
By the right-hand rule,
points up out of the page in
Fig.B.4. Crossing that with
, again by the right-hand
rule, produces a tangential velocity vector
pointing as shown
in the figure. So, the direction is correct. Now, the magnitude:
Since
and
are mutually orthogonal, the angle between them
is
, so that, by Eq.(B.16),
as desired.
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