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

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(D) The angular momentum $L$ of a particle about a point $P$ is given by the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p} = m\vec{v}$.Mathematically,$\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.The magnitude is given by $L = r p \sin 	heta$,where $	heta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.In this case,the particle is moving along the line $PC$,and point $P$ lies on this same line.Therefore,the position vector of the particle with respect to point $P$ is collinear with the velocity vector $\vec{v}$.This means the angle $	heta$ between the position vector and the velocity vector is $0^\circ$ (or $180^\circ$).Since $\sin(0^\circ) = 0$,the angular momentum $L = mvr \sin(0^\circ) = 0$.

As shown in the figure,a particle of mass $m$ moves with a velocity $v$ along the line $PC$. What is the angular momentum of this particle with respect to point $P$?

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