question_answer

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

A)                 \[mvb\,\hat{k}\]             

B)                 \[-mvb\,\hat{k}\]

C)                 \[mvb\,\hat{i}\]              

D)                 \[mv\,\hat{i}\]

Correct Answer: B

Solution :

                    We know that, Angular momentum                 \[\overrightarrow{L}=\overrightarrow{r\,}\times \overrightarrow{p}\] in terms of component becomes                 \[\overrightarrow{L}=\left| \,\begin{matrix}    \hat{i}\,\, & \hat{j}\,\, & {\hat{k}}  \\    x\,\, & y\,\, & z  \\    {{p}_{x}} & \,\,{{p}_{y}}\,\, & {{p}_{z}}  \\ \end{matrix}\, \right|\]                 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 \[{{p}_{y}}=0\]                 \ \[\overrightarrow{L\,}=\overrightarrow{k\,}\left[ vt\times 0-b\,mv \right]=-mvb\,\hat{k}\]