Angular Momentum Variation for Circular Motion About Different Points

To determine how the angular momentum vectors $\vec{L}_O$ and $\vec{L}_P$ vary with time, we analyze the position and velocity vectors of the particle in a Cartesian coordinate system.

Let the origin of the coordinate system be at the center of the circular path $O(0, 0, 0)$, with the circular motion occurring in the $x$-$y$ plane. Point $P$ lies on the $z$-axis at a height $h$ above $O$, so its coordinates are $(0, 0, h)$.

1. Motion of the Particle

For a particle of mass $m$ executing uniform circular motion of radius $R$ in the $x$-$y$ plane with constant angular speed $\omega$, its position vector relative to $O$ at any time $t$ is given by: $\vec{r}_O = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$

The linear velocity $\vec{v}$ of the particle is the time derivative of $\vec{r}_O$: $\vec{v} = \frac{d\vec{r}_O}{dt} = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$

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2. Angular Momentum about Point $O$ ($\vec{L}_O$)

The angular momentum about the center $O$ is defined as: $\vec{L}_O = \vec{r}_O \times (m\vec{v})$

Substituting $\vec{r}_O$ and $\vec{v}$: $\vec{L}_O = m \left[ \left( R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j} \right) \times \left( -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j} \right) \right]$

Evaluating the cross product using standard unit vector relations ($\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{i} = -\hat{k}$): $\vec{L}_O = m R^2 \omega \left[ \cos^2(\omega t) + \sin^2(\omega t) \right] \hat{k} = m R^2 \omega \hat{k}$

Since $m$, $R$, $\omega$, and the direction $\hat{k}$ are all independent of time, $\vec{L}_O$ remains constant in both magnitude and direction.

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3. Angular Momentum about Point $P$ ($\vec{L}_P$)

The position vector of the particle relative to point $P(0,0,h)$ is: $\vec{r}_P = \vec{r}_O - h\hat{k} = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j} - h\hat{k}$

The angular momentum about $P$ is: $\vec{L}_P = \vec{r}_P \times (m\vec{v})$

Using the determinant form for the cross product:

$$\vec{L}_P = m \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ R\cos(\omega t) & R\sin(\omega t) & -h \\ -R\omega\sin(\omega t) & R\omega\cos(\omega t) & 0 \end{vmatrix}$$

Expanding the determinant: $\vec{L}_P = m \left[ \hat{i}\left( 0 - (-h R\omega\cos(\omega t)) \right) - \hat{j}\left( 0 - (-h R\omega\sin(\omega t)) \right) + \hat{k}\left( R^2\omega\cos^2(\omega t) + R^2\omega\sin^2(\omega t) \right) \right]$

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

While the magnitude of $\vec{L}_P$ is constant: $|\vec{L}_P| = \sqrt{(m h R \omega)^2 + (m R^2 \omega)^2}$

its direction continuously changes with time as it precesses around the $z$-axis. Therefore, $\vec{L}_P$ varies with time.

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Conclusion

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

Correct Option: C