A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure

Q: Why is $\mathbf{L}_O$ constant if the particle is constantly changing its position?

Even though the position vector $\mathbf{r}_O$ and velocity vector $\mathbf{v}$ are changing direction, they both always lie in the $x$-$y$ plane. Their cross product, by the right-hand rule, always points strictly along the positive $z$-axis. Since the magnitude $mvr$ is also constant, the vector $\mathbf{L}_O$ never changes.

Q: Doesn't the magnitude of $\mathbf{L}_P$ remain constant? Why does the problem say it varies?

Yes, the *magnitude* $|\mathbf{L}_P|$ is absolutely constant! However, angular momentum is a *vector*. For a vector to be constant, both its magnitude AND its direction must remain unchanged. Because $\mathbf{L}_P$ sweeps out a cone, its direction is constantly changing, meaning the vector itself varies with time.

Q: How do we know $\mathbf{L}_P$ sweeps out a cone?

The position vector $\mathbf{r}_P$ has a constant vertical component and a rotating horizontal component. When you take the cross product with the rotating velocity vector, the resulting vector $\mathbf{L}_P$ gets a constant vertical component and a rotating horizontal component. A vector with a constant vertical height and a rotating horizontal radius traces out a cone!

Q: What if the string was attached to $O$ instead of $P$?

If the string was attached to $O$, the physical tension would provide the centripetal force, but the angular momentum analysis would be exactly the same as our $\mathbf{L}_O$ calculation. It would remain perfectly constant.

Q: Is angular momentum conserved in this system?

Angular momentum about $O$ is conserved because the net torque about $O$ is zero (the tension force passes through $P$, and its horizontal component passes through $O$). However, the net torque about $P$ is NOT zero (gravity creates a torque about $P$), which is exactly why $\mathbf{L}_P$ changes with time!

Question

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 $\mathbf{L}_O$ and $\mathbf{L}_P$ respectively, then

Options:

(A) $\mathbf{L}_O$ and $\mathbf{L}_P$ do not vary with time

(B) $\mathbf{L}_O$ varies with time while $\mathbf{L}_P$ remains constant

(C) $\mathbf{L}_O$ remains constant while $\mathbf{L}_P$ varies with time

(D) $\mathbf{L}_O$ and $\mathbf{L}_P$ both vary with time

Sigma Percentile · Accepted Answer

The concept of angular momentum often feels abstract, but it becomes incredibly intuitive once you visualize it in three dimensions. In this problem, we are asked to compare the angular momentum of a particle about two different points: the center of its circular path ($O$) and a point directly above it ($P$).

Let's break down the physics and the geometry step-by-step.

### The Physical Setup: Visualizing the Motion

Imagine a small mass $m$ moving in a perfect circle of radius $R$ on a horizontal $x$-$y$ plane. The center of this circle is the origin, $O$.

Now, look directly above $O$ along the $z$-axis to a point $P$. The mass is tied to $P$ by a massless string. As the mass revolves around the circle, the string sweeps out the shape of a cone.

### The Master Equation: Angular Momentum

To determine how the angular momentum behaves, we must return to its fundamental definition. The angular momentum $\mathbf{L}$ of a particle about a specific origin is given by the cross product of its position vector $\mathbf{r}$ and its linear momentum $\mathbf{p}$:

$$ \mathbf{L} = \mathbf{r} 	imes \mathbf{p} = m(\mathbf{r} 	imes \mathbf{v}) $$

This equation tells us two critical things:
1. The magnitude of $\mathbf{L}$ depends on the distance, the speed, and the angle between them.
2. The direction of $\mathbf{L}$ is always strictly perpendicular to both the position vector $\mathbf{r}$ and the velocity vector $\mathbf{v}$, determined by the right-hand rule.

### Analyzing Angular Momentum About the Origin ($O$)

Let's first calculate the angular momentum about the center of the circle, $O$.

The position vector $\mathbf{r}_O$ points from $O$ directly to the mass. Notice that both $\mathbf{r}_O$ and the velocity vector $\mathbf{v}$ lie completely flat in the horizontal $x$-$y$ plane. Furthermore, because the motion is circular, the radius is always perpendicular to the tangent velocity.

When we take the cross product $\mathbf{r}_O 	imes \mathbf{v}$, the right-hand rule dictates that the resulting vector must point straight up along the positive $z$-axis.

$$ |\mathbf{L}_O| = m R v \sin 90^\circ = mRv $$

Since the mass, radius, and speed are all constant, the magnitude of $\mathbf{L}_O$ is constant. And because it always points straight up, its direction is also constant. Therefore, **$\mathbf{L}_O$ does not vary with time.**

### The Twist: Angular Momentum About Point $P$

Now, let's shift our reference to point $P$. This changes everything.

The new position vector $\mathbf{r}_P$ points from $P$ down to the mass. This vector is slanted; it has a downward vertical component (along the $z$-axis) and an outward horizontal component (in the $x$-$y$ plane).

When we take the cross product of this slanted $\mathbf{r}_P$ with the horizontal velocity $\mathbf{v}$, the resulting angular momentum $\mathbf{L}_P$ must be perpendicular to both. Because $\mathbf{r}_P$ is slanted, $\mathbf{L}_P$ is also tilted away from the vertical $z$-axis.

### The Sweeping Cone: Why Direction Matters

Visualize the mass traveling around the circle. As it moves, the velocity vector changes direction, and the slanted position vector $\mathbf{r}_P$ rotates with it.

Consequently, the tilted angular momentum vector $\mathbf{L}_P$ also rotates, sweeping out a cone in space!

Let's look at the math. The magnitude of $\mathbf{L}_P$ is:

$$ |\mathbf{L}_P| = m |\mathbf{r}_P| v \sin 90^\circ = m l v $$

(where $l$ is the length of the string). The magnitude is perfectly constant! However, a vector is only constant if *both* its magnitude and direction are constant. Because $\mathbf{L}_P$ is constantly rotating and changing its direction, the vector itself is changing.

### The Final Verdict

We can confidently conclude that the angular momentum about $O$ remains perfectly constant, while the angular momentum about $P$ varies with time due to its changing direction.

This elegant interplay between geometry and physics is a classic hallmark of rotational dynamics. The correct choice is **(c)**.

Select Answer:

Visualized Solution

$The Physical Setup$

$Angular Momentum Formula$

$Angular Momentum about O$

$Constancy of L_{O}$

$Angular Momentum about P$

$Direction of L_{P}$

$Variation of L_{P}$

$Final Conclusion$

The Sigma Insight: Torque and Angular Momentum

The Physical Setup

The Master Equation

Analyzing Angular Momentum About the Origin ( $O$ )

The Twist

The Sweeping Cone

The Final Verdict

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Momentum Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Momentum about O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constancy of LO​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Momentum about P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction of LP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Variation of LP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion