Why is the angular momentum perpendicular to | vedclass

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Question

(A) In rotational motion about a fixed axis,every particle of the rigid body moves in a circular path in a plane perpendicular to the axis of rotation

Vedclass · Accepted Answer

Because the particles move in planes perpendicular to the axis.

Vedclass · Answer

(A) In rotational motion about a fixed axis,every particle of the rigid body moves in a circular path in a plane perpendicular to the axis of rotation.Let the axis of rotation be the $z$-axis. The position vector $\vec{r}$ of a particle can be decomposed into two components: $\vec{r}_{\parallel}$ (along the $z$-axis) and $\vec{r}_{\perp}$ (in the $xy$-plane).The velocity vector $\vec{v}$ of any particle is given by $\vec{v} = \vec{\omega} 	imes \vec{r}$. Since $\vec{\omega}$ is along the $z$-axis,$\vec{v}$ lies entirely in the $xy$-plane.The angular momentum is $\vec{L} = \sum (\vec{r} 	imes \vec{p}) = \sum m(\vec{r} 	imes \vec{v})$.Substituting $\vec{r} = \vec{r}_{\parallel} + \vec{r}_{\perp}$,we get $\vec{L} = \sum m((\vec{r}_{\parallel} + \vec{r}_{\perp}) 	imes \vec{v})$.Since $\vec{r}_{\parallel}$ is parallel to the $z$-axis and $\vec{v}$ is in the $xy$-plane,their cross product $\vec{r}_{\parallel} 	imes \vec{v}$ results in a vector perpendicular to the $z$-axis.However,for a symmetric body or when summing over the entire rigid body,the components of angular momentum perpendicular to the axis of rotation cancel out due to the circular symmetry of the motion,leaving only the component along the axis of rotation $(L_z)$.Thus,the net angular momentum perpendicular to the axis,${L_ \bot }$,is zero.

Why is the angular momentum perpendicular to the axis,denoted as ${L_ \bot }$,zero in a rotational motion about a fixed axis?

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