# Angular momentum of linear motion of sphere without a particle approximation

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 ($$I = d$$).

• Do you mean angular momentum, or moment of intertia? It's a bit confusing what formula you're actually using, because you set some things to one. But angular momentum would not lead to $d^2$. Sep 13, 2020 at 19:07
• @Bernhard Oops, corrected. Sep 13, 2020 at 19:15

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.

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).

• But is this valid if r is not constant for all matter? You just gave me the particle approximation again. If it is exact, then I am missing the reason. Sep 13, 2020 at 19:13
• Yes. it is constant. For it to change, you will need torque. There's no forces, so no torque, so no change in angular momentum. Sep 13, 2020 at 19:15
• Wrong word. What I mean is that different pieces of the sphere will be a different distance away from the axis of rotation. Sep 13, 2020 at 19:17
• I made an edit on the answer. Sep 13, 2020 at 19:35
• I guess that your answer is some kind of volume integral (though I'm not so sure of your notation - what does $k$ mean?). I hoped for something like the parallel axis theorem to exist for angular momentum instead of moment of inertia, but barring such a theorem, I'll probably have to try working through the integration. Sep 13, 2020 at 19:53

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

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

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

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.

• You seem to have ignored the sphere-not-particle part, which is the entire point of this question. I edited the title to make the focus clearer. Sep 14, 2020 at 6:35
• The sphere part is included in $\mathcal{I}_A \vec{\omega}$. WHy do you think I have ignored it? Sep 15, 2020 at 21:43
• I did not wish to consider the sphere's rotation, but the differences between its angular momentum and the angular momentum of an equal point mass at its center of mass. The answer by Wolphram jonny shows that I based the entire question on an incorrect assumption. I'll just delete it to prevent further confusion (as it provides no value to the site). Sep 17, 2020 at 14:18