If a particle of mass m is moving with consta | vedclass

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Question

(b) We know that, Angular momentum
$\overrightarrow L = \overrightarrow {r\,} 	imes \overrightarrow p $ in terms of omponent becomes
$\overrightarr

Vedclass · Accepted Answer

$ - mvb\, \,\hat k$

Vedclass · Answer

(b) We know that, Angular momentum
$\overrightarrow L = \overrightarrow {r\,} 	imes \overrightarrow p $ in terms of omponent becomes
$\overrightarrow L  = \left| {\,\begin{array}{*{20}{c}}
  {\hat i\,\,}&{\hat j\,\,}&{\hat k} \ 
  {x\,\,}&{y\,\,}&z \ 
  {{p_x}}&{\,\,{p_y}\,\,}&{{p_z}} 
\end{array}\,} ight|$
As motion is in x-y plane ($z = 0 $ and ${P_z} = 0$), so $\overrightarrow {L\,} = \overrightarrow {k\,} (x{p_y} - y{p_x})$
Here $x = vt, y = b, $${p_x} = m\,v$ and ${p_y} = 0$
$	herefore \overrightarrow {L\,} = \overrightarrow {k\,} \left[ {vt 	imes 0 - b\,mv} ight] = - mvb\,\hat k$

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

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