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(A) The angular momentum of the bob about the point of suspension $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.As th

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Angular momentum changes in direction but not in magnitude

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(A) The angular momentum of the bob about the point of suspension $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = m(\vec{r} 	imes \vec{v})$.As the bob moves in a horizontal circle,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ change their directions continuously.Specifically,the vector $\vec{L}$ is always perpendicular to the plane containing $\vec{r}$ and $\vec{v}$. As the bob revolves,this plane rotates about the vertical axis,causing the direction of $\vec{L}$ to change continuously.However,the magnitude of the angular momentum $L = mvr \sin 	heta$ remains constant because the speed $v$,the distance $r$ from the suspension point,and the angle $	heta$ between the string and the vertical axis remain constant during the uniform circular motion.Therefore,the angular momentum changes in direction but not in magnitude.

$A$ bob of mass $m$ attached to an inextensible string of length $l$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $\omega \ rad/s$ about the vertical axis. About the point of suspension:

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