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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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{\bf Question}

A particle moves under the action of the force $f(r){\bf r}.$
Prove that the angular momentum of the particle is constant.

\vspace{.25in}

{\bf Answer}

$\ds {\bf N} = f(r) {\bf r}$

${\bf N} = {\bf r} \times {\bf F} = f(r) {\bf r} \times {\bf r} =
0$

\medskip

as ${\bf N} = \dot {\bf L}$ then ${\bf N} = 0 \Rightarrow \dot
{\bf L} = 0 \Rightarrow {\bf L} = $constant.

i.e. angular momentum is constant.


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