A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $\mathrm{AB}$.…
A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $\mathrm{AB}$. If the angular momentum of the particle with respect to origin $\mathrm{O}$ is $L_A$ when it is at A and $L_B$ when it is at B, then:
$L_A=L_B$
the relationship between $L_A$ and $L_B$ depends upon the slope of the line $\mathrm{AB}$
$L_A < L_B$
$L_A > L_{B^*}$
Solution
Angular momentum \(=\) Linear momentum \(\times\) perpendicular distance between line of action of liner momentum from origin
Let d be the perpendicular distance.
\(p_A\) and \(p_B\) be the linear momentum at \(A\) and \(B\)
Angular momentum \(L_A=p_A \times d\)
\(L_B=p_B \times d\)
So, linear momentum will be equal, i.e., \(L_A=L_B\).here, \(p_A\) and \(p_B\) are equal as have equal velocity.
\(L_A=L_B\)
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