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

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$mv\ell$

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(D) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.Mathematically,$\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v}) = m(\vec{r} 	imes \vec{v})$.The magnitude of the angular momentum is given by $L = mrv \sin(	heta)$,where $	heta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.From the geometry of the figure,the perpendicular distance from the origin $O$ to the line of motion $PC$ is $\ell$.Therefore,the magnitude of the angular momentum about $O$ is $L = m v \ell$.

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

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