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 circle): The po

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 circle): The position vector $\vec{r}_A$ is in the plane of the circle,and the velocity vector $\vec{v}$ is tangent to the circle. The cross product $\vec{r}_A 	imes \vec{v}$ always points along the axis of rotation (the $z$-axis). Since the speed $v$ and radius $r$ are constant,the magnitude $L_A = mvr$ is constant,and the direction is fixed along the $+z$-axis. Thus,$L_A$ is constant.For point $B$ (the pivot point): The position vector $\vec{r}_B$ from $B$ to $M$ changes its direction as the mass rotates. Although the magnitude of the angular momentum $L_B$ remains constant,its direction changes continuously as the rod sweeps out a cone. Therefore,$L_B$ is not constant in direction.Comparing the options,the statement that $L_A$ is constant in both magnitude and direction is correct.

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