A mass M hangs on a massless rod of length l | vedclass

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(D) The angular momentum is defined as $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.For point $A$ (the center of the circular path):

Vedclass · Accepted Answer

$L_A$ is constant,both in magnitude and direction.

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(D) The angular momentum is defined as $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.For point $A$ (the center of the circular path): The position vector $\vec{r}_A$ is in the plane of the circle,and the velocity vector $\vec{v}$ is tangential to the circle. The cross product $\vec{r}_A 	imes \vec{v}$ always points in the positive $z$-direction (perpendicular to the plane of the circle). Since the speed $v$ and radius $r$ are constant,the magnitude $L_A = mvr$ is constant. Thus,$L_A$ is constant in both magnitude and direction.For point $B$ (a point on the axis of rotation above $A$): The position vector $\vec{r}_B$ from $B$ to $M$ changes its direction as the mass $M$ rotates. Therefore,the angular momentum $\vec{L}_B = \vec{r}_B 	imes \vec{p}$ will have a constant magnitude,but its direction will change continuously as it precesses around the $z$-axis.

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