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{\bf Question}

Find the tangential and normal acceleration of a particle which
moves on the ellipse ${\bf r} = a \cos \omega t {\bf i} + b \sin
\omega t {\bf j}$.

\vspace{.25in}

{\bf Answer}

$\ds \dot {\bf r} = \omega(-a\sin\omega t{\bf i}+ b \cos \omega
t{\bf j})$

$\ds \ddot{\bf r} = -\omega^2(a\cos \omega t{\bf i} + b \sin
\omega t{\bf j}) = -\omega^2{\bf r}$

The tangent unit vector is in the direction of ${\bf v}$

Therefore \ \  $\ds  {\bf t} = \frac{(a \sin \omega t{\bf i} + b
\cos \omega t {\bf j})}{\sqrt{a^2 + b^2}}$

The normal unit vector is orthogonal to ${\bf t}$ and given from
the Serret-Frenet formulae: $\ds \frac{d{\bf t}}{ds} = +\kappa{\bf
n}$

$\ds  \frac{d{\bf t}}{ds} = \frac{dt}{ds} \frac{d{\bf t}}{dt} = -
\frac{dt}{ds}\frac{\omega}{\sqrt{a^2+b^2}}(a\cos \omega t{\bf i} +
b \sin \omega t{\bf j})$

Therefore the unit vector parallel to $\ds \frac{d{\bf t}}{ds} =
{\bf u} = \frac{{\bf r}}{\sqrt{a^2+b^2}}$

The tangential component of acceleration is
\begin{eqnarray*}  &  & {\bf
a} \cdot {\bf t} = -\omega^2 {\bf r} \cdot {\bf t} \\ & = &
-\omega^2(a\cos \omega t{\bf i} + b \sin \omega t{\bf j}) \cdot
\frac{-a\sin\omega t{\bf i}+ b \cos \omega t{\bf
j}}{\sqrt{a^2+b^2}} \\ & = & -\omega^2\frac{(b^2 - a^2)
}{\sqrt{a^2+b^2}} \end{eqnarray*}

The normal component of acceleration is

\hspace{.5in}$\ds {\bf a} \cdot {\bf n} = -\omega^2 {\bf r} \cdot
\frac{{\bf r}}{\sqrt{a^2 + b^2}}= -\omega \frac{a^2 \cos^2 \omega
t + b^2 \sin^2 \omega t}{\sqrt{a^2 + b^2}}$



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