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{\bf Question}

A particle of mass $m$ moves along a space curve given by

$\ds {\bf r} = a \cos \omega t {\bf i} + b \sin \omega t {\bf j}.$

Find
\begin{description}
\item[(a)] the torque about the origin of the force acting upon it,
\item[(b)] the angular momentum of the particle about the origin.
\end{description}

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{\bf Answer}


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

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

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

\begin{description}
\item[(a)]

Using Newton's 2nd law: $m \ddot {\bf r} = {\bf F}$

The torque is ${\bf r} \times {\bf F} = {\bf r} \times m \ddot
{\bf r} = -m \omega^2 {\bf r} \times {\bf r} = 0$

\item[(b)]
\begin{eqnarray*} {\rm Angular\ momentum} & = &  {\bf r} \times m \dot {\bf
r}\\ & = & m \omega \left| \begin{array}{ccc} {\bf i} & {\bf j} &
{\bf k} \\ a \cos \omega t & b \sin \omega t & 0 \\ -a \sin \omega
t & b \cot \omega t & 0 \end{array} \right| \\ & = & m a b \omega
{\bf k} \end{eqnarray*}

\end{description}




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