JEE MAIN - Physics (2002 - No. 68)
A particle of mass $$m$$ moves along line PC with velocity $$v$$ as shown. What is the angular momentum of the particle about P?
$$mvL$$
$$mvl$$
$$mvr$$
zero
Explanation
Angular momentum $$(L)$$
$$=$$ ( linear momentum ) $$ \times $$ ( perpendicular distance of the line of action of momentum from the axis of rotation)
$$ = mv \times r$$
$$ = mv \times 0$$
$$=0$$
[ Here $$r=0$$ because the particle is moving through the line PQ and r is the perpendicular distance from line PQ of the particle ]
$$=$$ ( linear momentum ) $$ \times $$ ( perpendicular distance of the line of action of momentum from the axis of rotation)
$$ = mv \times r$$
$$ = mv \times 0$$
$$=0$$
[ Here $$r=0$$ because the particle is moving through the line PQ and r is the perpendicular distance from line PQ of the particle ]
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