A particle of mass 1 kg is moving with a lin | vedclass

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(C) The angular momentum $L$ of a particle about the origin is given by the formula $L = |\vec{r} 	imes \vec{p}| = r p \sin 	heta$.From the geometry

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(C) The angular momentum $L$ of a particle about the origin is given by the formula $L = |\vec{r} 	imes \vec{p}| = r p \sin 	heta$.From the geometry of the motion,the perpendicular distance $d$ from the origin to the line of motion is $d = r \sin 	heta$.Therefore,the angular momentum can be expressed as $L = p \cdot d = (mv) \cdot d$.Given: mass $m = 1 \ kg$,velocity $v = 2 \ ms^{-1}$,and perpendicular distance $d = 12 \ cm = 0.12 \ m$.Substituting these values into the formula:$L = (1 \ kg) 	imes (2 \ ms^{-1}) 	imes (0.12 \ m) = 0.24 \ kg \cdot m^2s^{-1}$ (or $Js$).Thus,the angular momentum is $0.24 \ Js$.

$A$ particle of mass $1 \ kg$ is moving with a linear velocity of $2 \ ms^{-1}$ parallel to the positive $X$-axis. During this motion,its minimum distance from the origin is $12 \ cm$. The angular momentum of this particle about the origin is ....... $Js$.

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