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{\bf Question}

A particle with position vector ${\bf r}$ relative to an origin O,
rotates with angular velocity {\boldmath {$\omega$}} about an axis
through O.  Show that the magnitude of the centripetal
acceleration of the particle, {\boldmath {$\omega$}} $\times($
{\boldmath {$\omega$}} $\times \, {\bf r})$, is $\omega^2d$, where
$d$ is the perpendicular distance of the particle from the axis of
rotation.


\vspace{.25in}

{\bf Answer}

${ } $

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\put(1.2,1){\line(0,-1){.2}}

\put(0.7,1){\makebox(0,0){$\mbox{\boldmath $\omega$}$}}

\put(1.5,1.2){\makebox(0,0){$\bf d$}}

\put(1.75,0.5){\makebox(0,0){${\bf r}$}}


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Now ${\bf r} = \lambda {\mbox{\boldmath $\omega$}} + {\bf d}$
where $\mbox{\boldmath $\omega$}, {\bf d}$ are perpendicular.
$\Rightarrow \mbox{\boldmath $\omega$} \cdot {\bf d} = 0$

Therefore ${\bf r} \cdot {{\mbox{\boldmath $\omega$}}} = \lambda
\omega^2 + {\bf d} \cdot {\mbox{\boldmath $\omega$}} = \lambda
\omega^2$

Therefore $\ds {\bf r} = \frac{1}{\omega^2}({{\bf r} \cdot
{\mbox{\boldmath $\omega$}}) \mbox{\boldmath $\omega$} + {\bf d}}$

\begin{eqnarray*} {\rm Centripetal\ acceleration} & = & {{\mbox{\boldmath
$\omega$}}} \times ( \mbox{\boldmath $\omega$} \times {\bf r}) \\
& = & \mbox{\boldmath $\omega$} \times \left(\mbox{\boldmath
$\omega$} \times \left(\frac{1}{\omega^2}({\bf r} \cdot
\mbox{\boldmath $\omega$}) \mbox{\boldmath $\omega$} + {\bf
d}\right)\right) \\ & = & \mbox{\boldmath $\omega$} \times
(\mbox{\boldmath $\omega$} \times {\bf d}) \hspace{.3in} {\rm as\
} \mbox{\boldmath $\omega$} \times \mbox{\boldmath $\omega$} = 0
\\ & = & (\mbox{\boldmath $\omega$} \cdot {\bf d})\mbox{\boldmath $\omega$}
 - {\mbox{\boldmath $\omega$} \cdot
\mbox{\boldmath $\omega$} {\bf d}} \hspace{.3in}{\rm as\ }
\mbox{\boldmath $\omega$} \cdot {\bf d} = 0 \\ & = & -\omega^2
{\bf d} \hspace{.3in} {\rm as\ required}
\end{eqnarray*}




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