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Q.
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular moment with respect to the origin
NTA AbhyasNTA Abhyas 2022
Solution:
Angular momentum of the particle about origin O will be given by
$\overset{ \rightarrow }{\text{L}} = \overset{ \rightarrow }{\text{r}} \times \overset{ \rightarrow }{\text{p}} = \text{m} \left(\overset{ \rightarrow }{\text{r}} \times \overset{ \rightarrow }{\text{V}}\right)$
$\text{or} \left|\overset{ \rightarrow }{\text{L}}\right| = \text{L} = \text{mrV sin} \theta $
$= \text{mV} \left(\text{r sin} \theta \right) = \text{mVh}$
Now, $\text{m,V}$ and $\text{h}$ all are constant
Therefore, angular momentum of particle about origin will remain constant. The direction of $\overset{ \rightarrow }{\text{r}} \times \overset{ \rightarrow }{\text{v}}$ also remains the same (negative z direction)
