A particle of mass m moves along a straight l | vedclass

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Question

(B) The angular momentum $L$ of a particle about an origin $O$ is given by the cross product of its position vector $\vec{r}$ and its linear momentum

Vedclass · Accepted Answer

$L_A = L_B$

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(B) The angular momentum $L$ of a particle about an origin $O$ is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$,i.e.,$\vec{L} = \vec{r} 	imes \vec{p}$.The magnitude of angular momentum is given by $L = p \cdot d$,where $p$ is the magnitude of linear momentum and $d$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.Since the particle is moving along a straight line $AB$,the perpendicular distance $d$ from the origin $O$ to the line $AB$ is the same for all points on the line.Assuming the particle moves with a constant velocity $v$,its linear momentum $p = mv$ remains constant throughout the motion.Therefore,at any point on the line $AB$,the angular momentum $L = mv \cdot d$ is constant.Thus,$L_A = L_B$.

$A$ particle of mass $m$ moves along a straight line $AB$ in the $XY$ plane. Let $L_A$ and $L_B$ be the angular momentum of the particle about the origin $O$ when it is at points $A$ and $B$ respectively. Then:

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