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(B) The forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$ acting downwards. Taking the point of suspension a

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angular momentum changes in direction but not in magnitude

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(B) The forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$ acting downwards. Taking the point of suspension as the origin,the torque $\vec{	au}$ due to the gravitational force is $\vec{	au} = \vec{r} 	imes \vec{F} = \vec{r} 	imes m\vec{g}$. The magnitude of this torque is $	au = mg \ell \sin 	heta$,where $	heta$ is the angle the string makes with the vertical. This torque is always directed horizontally,perpendicular to the plane containing the string and the vertical axis. Since the rate of change of angular momentum $\frac{d\vec{L}}{dt} = \vec{	au}$,the torque causes the direction of the angular momentum vector $\vec{L}$ to change continuously as it precesses around the vertical axis. However,because the torque is always perpendicular to the angular momentum vector $\vec{L}$,the magnitude of the angular momentum remains constant. 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 $\omega \text{ rad/s}$ about the vertical axis. About the point of suspension:

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