Question 381: Centre of Mass and Rotational Motion

Question

A particle of mass $$m$$ moves along line PC with velocity $$v$$ as shown. What is the angular momentum of the particle about P?

Accepted Answer

Angular momentum $$(L)$$
$$=$$ ( linear momentum ) $$ 	imes $$ ( perpendicular distance of the line of action of momentum from the axis of rotation)
$$ = mv 	imes r$$
$$ = mv 	imes 0$$
$$=0$$
[ Here $$r=0$$ because the particle is 
 moving through the line PQ and  
r is the perpendicular distance from line PQ of the particle ]

Answer

$$mvL$$

Answer

$$mvl$$

Answer

$$mvr$$

Answer

zero

Centre of Mass and Rotational Motion

A particle of mass $m moves along line PC with velocity v$ as shown. What is the angular momentum of the particle about P?

Choose the correct answer: