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# A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in a X-Y plane with centre at O and constant angular speed $\omega$. If the angular momentum of the system calculated about O and P are denoted by ${{\overrightarrow{L}}_{0}}\text{ and }\overrightarrow{{{L}_{p}}},$ respectively, then:A.${{\overrightarrow{L}}_{0}}\text{ and }\overrightarrow{{{L}_{p}}}$ do not vary with time.B.${{\overrightarrow{L}}_{0}}\text{ }$varies with time while $\overrightarrow{{{L}_{p}}}$ remains constantC.${{\overrightarrow{L}}_{0}}$remains constant and $\overrightarrow{{{L}_{p}}}$varies with time.D.${{\overrightarrow{L}}_{0}}\text{ and }\overrightarrow{{{L}_{p}}}$both vary with time.

Last updated date: 13th Jun 2024
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Hint: The angular momentum is a vector quantity which varies with the point under consideration. Here, the angular momentum of the same body as observed from two different points the centre of rotation O and the fixed-point P are considered.
Formula Used: The angular momentum is calculated using the formula:
\begin{align} & \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p} \\ & \text{or} \\ & \overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v}) \\ \end{align}
Where, $\overrightarrow{L}$ is the angular momentum
$\overrightarrow{r}$ is the radius of the circle in which the mass undergoes circular motion
$\overrightarrow{p}$ is the linear momentum
$v$ is the linear velocity

Angular momentum is the measure of linear momentum in a circular motion. It is a vector quantity like the linear momentum.
Let us consider the given situation. An object of mass m attached to a fixed string at point P is rotating around a fixed-point O with a constant angular velocity $\omega$. We have to determine the angular momenta experienced by the body with respect to point O and P.

We can begin with the point O, the center of rotation.
$\overrightarrow{{{L}_{0}}}=\overrightarrow{r}\times \overrightarrow{p}$
The linear momentum ‘mv’ being substituted gives, $\overrightarrow{{{L}_{o}}}=m(\overrightarrow{r}\times \overrightarrow{v})$
We know that the angular velocity is a constant. Also, the radius of the rotation is a constant.
\begin{align} & \Rightarrow \overrightarrow{{{L}_{o}}}=m(\overrightarrow{OR}\times \overrightarrow{v)} \\ & \Rightarrow \left| \overline{{{L}_{o}}} \right|=mrv \\ & \\ \end{align}
The magnitude of the angular momentum at O remains constant throughout.

And, the direction of the angular momentum at O is always along the Z-direction.
Now, let us consider the angular momentum at point P due to the mass.
\begin{align} & \Rightarrow \overrightarrow{{{L}_{p}}}=m(\overrightarrow{PR}\times \overrightarrow{v}) \\ & \Rightarrow \overrightarrow{{{L}_{p}}}=m((\overrightarrow{PO}+\overrightarrow{OR})\times \overrightarrow{v}) \\ \end{align}
It is clear from the above equation that $\overrightarrow{PR}$changes its direction at every instant. As a result, the direction of the angular momentum at P keeps on varying with time.
And the magnitude is given by,
\begin{align} & \overrightarrow{{{L}_{p}}}=m\left| \overrightarrow{PR} \right|\overrightarrow{v}\widehat{k} \\ & \left| \overrightarrow{{{L}_{p}}} \right|=m\left| \overrightarrow{PR} \right|\overrightarrow{v} \\ \end{align}
Which remains constant with time.
From the above, we can conclude that $\overrightarrow{{{L}_{o}}}$remains constant both in direction and magnitude, whereas, $\overrightarrow{{{L}_{p}}}$has a constant magnitude, but the direction changes.
Therefore, $\overrightarrow{{{L}_{o}}}$ remains constant and $\overrightarrow{{{L}_{p}}}$varies with time.

The right answer is given by option C.

We can calculate the angular momentum by using the angular velocity, $\overrightarrow{\left| L \right|}=m{{\omega }^{2}}r$.