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 the $x-y$ 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 $L_O$ and $L_P$ respectively, then

Step 1: Define the coordinate system and motion parameters.

Let the center of the circle $O$ be the origin $(0,0,0)$, and let the fixed point $P$ lie on the $z$-axis at a height $h$ above $O$.

The position vector of the mass $m$ relative to $O$ at any time $t$ is given by:

$$\vec{r}_O = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}$$

where $\theta = \omega t$.

The linear momentum of the mass is:

$$\vec{p} = m\vec{v} = m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})$$

Step 2: Calculate the angular momentum about $O$.

The angular momentum about the center $O$ is defined by the cross product of the position vector and the momentum.

$$\vec{L}_O = \vec{r}_O \times \vec{p}$$ $$\vec{L}_O = (R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}) \times m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})$$ $$\vec{L}_O = m\omega R^2 \hat{k}$$

Since $m$, $\omega$, and $R$ are constant, $\vec{L}_O$ does not change with time.

Step 3: Calculate the angular momentum about $P$.

The position vector of the mass relative to $P$ includes the vertical displacement $-h\hat{k}$.

$$\vec{r}_P = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j} - h\hat{k}$$

The angular momentum about $P$ is:

$$\vec{L}_P = \vec{r}_P \times \vec{p}$$ $$\vec{L}_P = (R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j} - h\hat{k}) \times m\omega R(-\sin\theta\,\hat{i} + \cos\theta\,\hat{j})$$ $$\vec{L}_P = m\omega R \left( h\cos\theta\,\hat{i} + h\sin\theta\,\hat{j} + R\hat{k} \right)$$

Step 4: Analyze the time-dependence of both angular momenta.

Express the angular momentum about $P$ in terms of time $t$ using $\theta = \omega t$.

$$\vec{L}_P = m\omega Rh\cos(\omega t)\,\hat{i} + m\omega Rh\sin(\omega t)\,\hat{j} + m\omega R^2 \hat{k}$$

The horizontal component of $\vec{L}_P$ rotates in the $x$-$y$ plane, meaning its direction changes continuously with time.

Answer: $\vec{L}_O$ remains constant, while $\vec{L}_P$ varies with time.