Angular momentum of linear motion of sphere without a particle approximation

Question

This seems like a really simple case that I haven't found covered anywhere else: Without loss of generality (scaling), we have a uniform full sphere of radius $1$ and mass $1$ moving at a speed of $1$ perpendicular to an axis of distance $d$ from its center. What is the sphere's angular momentum? I only found the trivial solution for a particle approximation ($I = d$).

Answer 1

For a system of particles $\vec L=M \vec R \times \vec V+\sum{m_i\vec r_i \times \vec v_i}$, where the first term are position and velocity of the center of mass, and in the second term the variables are relative to the center of mass. Because the object is not rotating, the second term is zero ($\vec v_i=0$ for all $i$). Thus, you can use the particle approximation.

Answer 2

Angular momentum $\mathbf L = \mathbf r \times \mathbf p$. For this case, |$\mathbf p| = m|\mathbf v| = 1$ and |$\mathbf r$|= d. If $\mathbf r$ and $\mathbf v$ are orthogonal, |$\mathbf L| = d$

Edit: As for linear motion $\mathbf p_k$ is the same for all small elements of the sphere, there is a sum of $\mathbf L_k = \mathbf r_k \times \mathbf p_k = \frac{1}{k}\mathbf r_k\times \mathbf p$.

$\frac{1}{k}\mathbf r_k = \mathbf r$ (the vector to the center of mass).

Answer 3

Here is the general 3D case, which you can use to find your specific answer.

A mass $m$ with center of mass at A is moving with velocity $\vec{v}_{A}$ and is located at $\vec{r}_{BA}$ relative to a reference point B. The body is rotating with $\vec{\omega}$, and for simplicity it has a scalar mass moment of inertia about the rotation axis of $I_{A}$.

Then linear momentum is

$$ \vec{p} = m \vec{v}_A \tag{1}$$

and angular momentum about B

$$ \vec{L}_{B} = I_{A} \vec{\omega} + \vec{r}_{BA} \times \vec{p} \tag{2}$$

where $\times$ is the vector cross product.

It is not clear if in your case the body is rotating or not. For the real general case, the mass moment of inertia term is a 3×3 matrix $\mathcal{I}_{A}$ which converts angular velocity vector to angular momentum at the center of mass

$$ \vec{L}_{B} = \mathcal{I}_{A} \vec{\omega} + \vec{r}_{BA} \times \vec{p} \tag{2}$$

The above is only valid for an inertial point B.