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\begin{center}
\textbf{Vector Functions and Curves}

\textit{\textbf{Applications}}
\end{center}

\textbf{Question}

What are the tangential and normal components of the Coriolis force on
an object moving with (horizontal) velocity $\underline{v}$ in the
following situations
\begin{description}
\item{(a)}
At the north pole.
\item{(b)}
At the south pole.
\item{(c)}
At the equator.
\end{description}

\textbf{Answer}

The Earth has angular velocity $\un{\Omega}$ which points due
north. If a particle is moving with horizontal velocity $v$,then the
tangential and normal components of the Coriolis force ($\un{C}$) are
related by:
\begin{eqnarray*}
\un{C}_T & = & -2\un{\Omega}_N \times \un{v}\\
\un{C}_N & = & -2\un{\Omega}_T \times \un{v}
\end{eqnarray*}

\begin{description}
\item{(a)}
At the north pole, $\un{\Omega}_T=0$ and $\un{\Omega}_N=\un{\Omega}$.
\begin{eqnarray*}
\Rightarrow \un{C}_N = \un{0}\\
\un{C}_T = -2\un{\Omega}\times\un{v}
\end{eqnarray*}
And so the Coriolis force is $90^{\circ}$ east of $\un{v}$.

\item{(b)}
At the south pole, $\un{\Omega}_T=0$ and $\un{\Omega}_N=\un{\Omega}$.
\begin{eqnarray*}
\Rightarrow \un{C}_N = \un{0}\\
\un{C}_T = -2\un{\Omega}\times\un{v}
\end{eqnarray*}
And so the Coriolis force is $90^{\circ}$ west of $\un{v}$.

\item{(c)}
At the equator, $\un{\Omega}_N=\un{0}$ and
$\un{\Omega}_T=\un{\Omega}$.
\begin{eqnarray*}
\un{C}_T & = & \un{0}\\
\un{C}_N & = & -2\un{\Omega} \times \un{v}
\end{eqnarray*}
And so the Coriolis force is vertical.

\end{description}

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