A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass

Q: Why does the choice of origin change the angular momentum?

Angular momentum is not an absolute property of a moving object; it's defined *relative* to a specific point. Since $\vec{L} = \vec{r} \times \vec{p}$, changing the origin changes the position vector $\vec{r}$, which in turn changes the cross product.

Q: Isn't angular momentum always conserved in circular motion?

Angular momentum is only conserved if the net external torque about that specific point is zero. Here, gravity and the tension in the rod exert torques. About point A, the net torque is zero, so $\vec{L}_A$ is constant. About point B, gravity exerts a torque, causing $\vec{L}_B$ to precess!

Q: How did we know $\vec{L}_B$ sweeps out a cone?

We found that $\vec{L}_B = mRv \hat{k} + mhv \hat{r}$. The vertical component ($mRv \hat{k}$) is constant, but the horizontal component ($mhv \hat{r}$) points radially outward and rotates with the mass. A vector with a constant vertical part and a rotating horizontal part traces a cone.

Q: Why is the velocity vector purely tangential?

In steady circular motion, the speed is constant, and the particle is constrained to a circular path. The velocity vector is always tangent to the path, meaning it has no radial or vertical components.

Q: Does the magnitude of $\vec{L}_B$ change as it rotates?

No! The magnitude is $|\vec{L}_B| = \sqrt{(mRv)^2 + (mhv)^2}$. Since $m, R, v,$ and $h$ are all constants, the length of the vector $\vec{L}_B$ remains perfectly constant, even though it's pointing in different directions.

Question

A mass $M$ hangs on a massless rod of length $l$ which rotates at a constant angular frequency. The mass $M$ moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity $\omega$. The angular momentum of $M$ about point $A$ is $L_A$ which lies in the positive z-direction and the angular momentum of $M$ about $B$ is $L_B$. The correct statement for this system is

Options:

(A) $L_A$ and $L_B$ are both constant in magnitude and direction

(B) $L_B$ is constant in direction with varying magnitude

(C) $L_B$ is constant, both in magnitude and direction

(D) $L_A$ is constant, both in magnitude and direction

Sigma Percentile · Accepted Answer

### The Setup: A Whirling Mass

Imagine you are standing in a grand ballroom, watching a dancer twirl a weight on a string. The mass $M$ is moving in a perfect, steady horizontal circle.

The rod of length $l$ connects the mass to point $B$ on the vertical axis, while point $A$ lies directly at the center of the circular path.

Our goal is to understand the angular momentum of this mass from two different perspectives: point $A$ and point $B$.

### The Master Equation

Angular momentum isn't just a random formula; it's a measure of the "rotational oomph" of an object relative to a specific origin.

Mathematically, it is defined as the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$.

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

This cross product means that the direction of $\vec{L}$ is always perpendicular to both the position vector and the velocity vector.

### Perspective 1: Angular Momentum about Point A

Let's place our origin at point $A$, the center of the circle. The position vector $\vec{r}_A$ points radially outward from $A$ to the mass $M$.

The velocity vector $\vec{v}$ is tangential to the circular path. Because the path is horizontal, both $\vec{r}_A$ and $\vec{v}$ lie entirely in the horizontal $xy$-plane.

When we apply the right-hand rule to $\vec{r}_A 	imes \vec{v}$, our thumb points straight up along the positive $z$-axis.

$$ \vec{L}_A = mRv \hat{k} $$

Since the mass moves with a steady speed $v$ at a constant radius $R$, the magnitude $mRv$ is perfectly constant. Furthermore, because it always points straight up, its direction never changes. **Thus, $\vec{L}_A$ is constant in both magnitude and direction.**

### Perspective 2: Angular Momentum about Point B

Now, let's shift our perspective to point $B$, which is located at a height $h$ above point $A$. The position vector $\vec{r}_B$ now points diagonally downwards from $B$ to $M$.

We can break $\vec{r}_B$ into two components: a horizontal radial component $\vec{r}_A$ and a vertical component $-h\hat{k}$.

$$ \vec{r}_B = \vec{r}_A - h\hat{k} $$

When we take the cross product with the tangential velocity $\vec{v}$, the math gets incredibly interesting.

$$ \vec{L}_B = m(\vec{r}_A - h\hat{k}) 	imes \vec{v} $$

$$ \vec{L}_B = m(\vec{r}_A 	imes \vec{v}) - m(h\hat{k} 	imes \vec{v}) $$

The first term is just our old friend $\vec{L}_A$, pointing straight up. But the second term creates a new horizontal component that points radially outward!

### The Sweeping Cone

Because the mass is moving in a circle, that outward-pointing horizontal component rotates along with the mass.

Imagine a vector with a fixed vertical height but a horizontal part that spins around like the hand of a clock. The tip of this vector traces out a perfect circle in the air.

In other words, the angular momentum vector $\vec{L}_B$ sweeps out a cone!

Its length (magnitude) is fixed by the constant values of $m, R, v,$ and $h$. However, because it is constantly spinning, its direction is continuously changing. **Thus, $\vec{L}_B$ is constant in magnitude but varies in direction.**

### The Final Verdict

By carefully analyzing the cross products, we've uncovered the beautiful geometry of this system.

$\vec{L}_A$ stands perfectly still, a silent sentinel on the $z$-axis. Meanwhile, $\vec{L}_B$ performs a continuous conical dance.

This perfectly matches option (d): $L_A$ is constant, both in magnitude and direction.

Select Answer:

Visualized Solution

$Steady Circular Motion$

$Angular Momentum$

$Position Vector from A$

$Angular Momentum about A$

$Position Vector from B$

$Angular Momentum about B$

$Dynamics of L_{B}$

$Conclusion$

The Sigma Insight: Torque and Angular Momentum

The Setup

The Master Equation

Perspective 1

Perspective 2

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; } Steady Circular Motion

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position Vector from A

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position Vector from B

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

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

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