A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in figure. The mass is undergoing circular motion in $x-y$ plane with centre $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by ${\overrightarrow{L}}_O$ and ${\overrightarrow{L}}_P$ respectively, then

In figure, weight$mg$ of the mass acts vertically downwards and tension T in the string acts along $K P$.

Two rectangular components of $T$ are

$T\cos \theta$ opposite to $mg$ and $T\sin \theta$ along $KO$

About $O$, torques due to $T\cos \theta$ and $m g$ cancel out. Torque due to $T\sin \theta$ is zero. Therefore, net torque about $O$ is zero.

As ${\overrightarrow{\tau }}_O=\frac{d{\overrightarrow{L}}_O}{dt}=0$,$\therefore {\overrightarrow{L}}_O$= constant

i.e., ${\overrightarrow{L}}_O$ does not vary with time.

About $P$, torque due to $T$ is zero, but torque due to $mg\ne 0$.

As ${\overrightarrow{\tau }}_P=\frac{d{\overrightarrow{L}}_P}{dt}\ne 0$

$\therefore {\overrightarrow{L}}_P$is not constant. It varies with time.