A particle of mass M moves with a constant ve | vedclass

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Question

(C) The angular momentum $\vec{L}$ of a particle with respect to the origin is given by $\vec{L} = \vec{r} 	imes \vec{p}$,where $\vec{p} = M\vec{v}$.

Vedclass · Accepted Answer

Remains constant

Vedclass · Answer

(C) The angular momentum $\vec{L}$ of a particle with respect to the origin is given by $\vec{L} = \vec{r} 	imes \vec{p}$,where $\vec{p} = M\vec{v}$.Since the particle moves parallel to the $X$-axis with constant velocity $\vec{v} = v\hat{i}$,its position vector is $\vec{r} = x\hat{i} + y\hat{j}$.Thus,$\vec{L} = (x\hat{i} + y\hat{j}) 	imes (Mv\hat{i}) = Mv(x(\hat{i} 	imes \hat{i}) + y(\hat{j} 	imes \hat{i}))$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get $\vec{L} = -Mvy\hat{k}$.Here,$M$,$v$,and $y$ (the perpendicular distance from the $X$-axis) are all constant.Therefore,the angular momentum $\vec{L}$ remains constant.

$A$ particle of mass $M$ moves with a constant velocity parallel to the $X$-axis. What happens to its angular momentum with respect to the origin?

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