Answer:
Suppose a particle
of mass m is moving II to X-axis starting from B, where OB = b. Fig. 5(a).39.
At any time r, let the particle be at P,
where \[x=OA=BP=\upsilon
t\], \[y=b\], \[z=0\].
Components
of velocity are
\[{{\upsilon
}_{x}}=\frac{dx}{dt}=\upsilon \], \[{{\upsilon }_{y}}=\frac{dy}{dt}=0\], \[{{\upsilon
}_{z}}=0\]
As \[\overset{\to
}{\mathop{L}}\,=\overset{\to }{\mathop{r}}\,\times \overset{\to
}{\mathop{p}}\,=\overset{\to }{\mathop{r}}\,\times m\overset{\to
}{\mathop{\upsilon }}\,=m(\overset{\to }{\mathop{r}}\,\times \overset{\to
}{\mathop{\upsilon }}\,)\]
\[=m\left|
\begin{matrix}
{\hat{i}} & {\hat{j}} & {\hat{k}} \\
\upsilon t & b & 0 \\
\upsilon & 0 & 0 \\
\end{matrix}
\right|\]
\[\therefore
\] \[\overset{\to }{\mathop{L}}\,=m\hat{k}[\upsilon t\times 0-\upsilon
b]=-m\upsilon b\hat{k}\]
i.e.
angular momentum has magnitude \[m\,\upsilon \,b\] and is directed along
negative z-axis. Thus angular momentum remains constant.
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