Question

If the total angular momentum of a system of particles is zero, are all the particles at

rest? Explain.

Answer 1

Answer: No.

Step-by-step explanation:

Firstly, I want to make the argument that objects moving in a straight line can have angular momentum (you'll see why later).

Suppose we have a particle located at
`\vec{r}=-y\hat{j}`, where r describes the position vector for any particle located at any point directly below the origin. Now suppose its momentum is described by
`\vec{p}=p\hat{j}`, where p is a positive scalar quantity (in other words, the particle has a momentum that drives its motion upward). Angular momentum is a vector quantity and is described by
`\vec{L} = \vec{r} * \vec{p}`. Since both vectors point in the
`\hat{j}` direction, their cross products will give us 0 (you should convince yourself of this property if you don't believe it). Now suppose we shift this object some distance d to the right, such that its new position vector is described by
`\vec{r}=d\hat{i}-y\hat{j}`. The
`\hat{j}` components still cancel, but
`\hat{i} * \hat{j} = \hat{k}`, so our cross product for angular momentum yields
`\vec{L}=\vec{r} * \vec{p}=d\hat{i} * p\hat{j}=dp\hat{k}`. So despite having the same motion (momentum), the particle has an angular momentum. This tells us two important characteristics about angular momentum:

1. Particles moving in a straight line can have non-zero angular momentum.
2. Angular momentum depends on the choice of origin (notice how we shifted the object a distance d to the right, which you can say means we shifted the origin a distance d to the left)

Please see attachment for an image reference if you cannot picture/diagram this (ignore the gravitational force drawn on the particle).

Now, lets take a system of particles. Suppose we have two particles on the x-axis with mass m, a distance d from the origin on either side, and velocities such that one particle moves to the right and the other moves to the left. For a system, angular momentum is defined as the following:

`\vec{L}=\sum_i \vec{r_i} * \vec{p_i}=\sum_i m_i(\vec{r_i} * \vec{v_i})`

Note: mass is a scalar quantity, so the cross product is not defined for it.

For this system, we can write out the angular momentum as the following:

`\vec{L}=m(d\hat{i} * v\hat{i}) + m(-d\hat{i} * -v\hat{i})`

However, we established above that the cross product of two identical vectors yields a result of 0. So we get that the angular momentum is 0, and yet both particles move, just that one moves to the right while the other moves to the left. In other words, the total angular momentum of the system is zero while the particles are not at rest.

If this is confusing, think about it in a more mathematical way. Angular momentum is a vector quantity, and thus follows the rules of vector algebra. In other words,

`\vec{L}=\vec{L_1}}+\vec{L_2}}+\vec{L_3}}+...+\vec{L_n}}`

This is known as superposition. Suppose we constructed a system of two particles where their cross products didn't evaluate to 0. We can still use the principle of superposition to create a scenario where the angular momentum is 0.

Suppose we have a system of 2 particles where
`\vec{L_1} = -\vec{L_2}`. The total angular momentum would be defined as

`\vec{L}=\vec{L_1}+\vec{L_2}=\vec{L_1}-\vec{L_1}=0`.

Now it doesn't matter if the particles move in a straight line, a circular path, or any other type of motion. If their individual angular momentums are equal in magnitude and opposite in direction, the angular momentum of the system is zero without the particles needing to be necessarily at rest.