Question

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

• A. ${\overrightarrow{L}}_O$ and ${\overrightarrow{L}}_P$ do not vary with time.
• B. ${\overrightarrow{L}}_O$ varies with time while ${\overrightarrow{L}}_P$ remains constant.
• C. ${\overrightarrow{L}}_O$ remains constant while ${\overrightarrow{L}}_P$ varies with time.
• D. ${\overrightarrow{L}}_O$ and ${\overrightarrow{L}}_P$ both vary with time.

Answer 1

The correct option is C ${\overrightarrow{L}}_O$ remains constant while ${\overrightarrow{L}}_P$ varies with time.

Angular momentim about $\text{O}$ :
As we know, angular momentum $\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{V})$
Here $\overrightarrow{r}=\overrightarrow{OA}$. Since mass is undergoing circular motion,$\overrightarrow{V}$ will be in the tangential direction of a circle in the plane of circular motion.


${\overrightarrow{L}}_O=m(\overrightarrow{OA}\times \overrightarrow{V})$
Angular momentum will be in z-direction (perpendicular to the plane of circular motion).
After some time, mass is at $B$. Velocity will be in tangential direction. Hence
${\overrightarrow{L}}_O=m(\overrightarrow{OB}\times \overrightarrow{V})$ has the same magnitude and direction.
i.e ${\overrightarrow{L}}_O$ remains constant.

Angular momentum about point $\text{P}$ :
Here, $\overrightarrow{r}=\overrightarrow{PA}$
and angular momentum will be $\overrightarrow{L}=m(\overrightarrow{PA}\times \overrightarrow{V})$
$\overrightarrow{V}$ will be in tangential direction in the circular plane. Hence, the direction of angular momentum will be as shown in figure. (by right hand thumb rule).


After time $t$, when particle is at $\text{B}$, $\overrightarrow{r}=\overrightarrow{PB}$ and velocity vector will be in tangential direction.
So, angular momentum $\overrightarrow{L}=m(\overrightarrow{OB}\times \overrightarrow{V})$ and the direction of angular momentum (from right hand rule ) as shown in the figure.

Hence, magnitude of angular momentum will be same about point $\text{P}$, but direction is changing.
$\therefore$ Angular momentum about point $\text{O}$ is constant, but about point $\text{P}$ it is varying.