If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane as shown in fig. Its angular momentum with respect to origin at any time t will be

 

We know that, Angular momentum

$\stackrel{.}{\overrightarrow{\mathrm{L}}}=\overrightarrow{r}\mathrm{x}\overrightarrow{\mathrm{P}} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{component} \mathrm{becomes}$

As motion is in x-y plane (z = 0 and ${\mathrm{P}}_{\mathrm{z}} = 0), \mathrm{as} \stackrel{.}{\overrightarrow{\mathrm{L}}}=\hspace{0.17em}\hspace{0.17em}\widehat{\mathrm{k}} ({\mathrm{xp}}_{\mathrm{y}} - {\mathrm{yp}}_{\mathrm{x}}\mathrm{)}$

Here x = vt, y = b, ${\mathrm{p}}_{\mathrm{x}}= \mathrm{mv} \mathrm{and} {\mathrm{p}}_{\mathrm{y}} = 0$

$\therefore$$\overrightarrow{\mathrm{L}}=\hspace{0.17em}\hspace{0.17em}\widehat{\mathrm{k}} [\mathrm{vt}\times 0-\mathrm{b} \mathrm{mv}] = - \mathrm{mvb} \widehat{\mathrm{k}}$