Angular momentum of a rotating sphere from a point outside the sphere

I have seen an expression for the angular momentum of a rotating sphere calculated from outside the sphere as $$L = I\omega + mvr,$$ where $v$ is the velocity of the center of mass, $m$ is the mass of the sphere, and $r$ is the distance of center of mass of the sphere from the point of calculation. My concern is if $v=0$, then $L = I \omega$, which is the angular momentum of the sphere when calculate through the center of mass. How can the angular momentum be the same when the point of calculation is changing?

• I just realized that in your question you say that $r$ is the radius of the sphere. That's not generally the formula. Instead, this formula is used when $r$ is the distance to the sphere. So it will be the radius only if the point from which you're calculating is exactly at the surface of the sphere — rather than just somewhere "outside". Maybe that will help explain some of your confusion. (Also, it's really supposed to be $\vec{v} \times \vec{r}$, so you might want to throw in a factor of $\sin\theta$ for the angle between $\vec{v}$ and $\vec{r}$.) – Mike Feb 13 '18 at 20:31

Don't confuse angular momentum and mass moment of inertia. The MMOI changes with location because of the assumption that the body rotates about that point (and hence the COM moves with $v \neq 0$).

But angular momentum depends on the position only if there is linear momentum present. This is entirely equivalent to a torque varying by position in the presence of forces, and of linear velocity varying by position in the presence of rotational velocity. The equations look the same too when considering an arbitrary point A and the center of mass point C.

$$\begin{array}{r|c} \mbox{Quantity} & \mbox{Transformation} \\ \hline \mbox{Linear Velocity} & {\boldsymbol v}_A = {\boldsymbol v}_C + {\boldsymbol r} \times {\boldsymbol \omega}\\ \hline \mbox{Angular Momentum} & {\boldsymbol L}_A = {\boldsymbol L}_C + {\boldsymbol r} \times {\boldsymbol p}\\ \hline \mbox{Torque} & {\boldsymbol \tau}_A = {\boldsymbol \tau}_C + {\boldsymbol r} \times {\boldsymbol F} \end{array}$$

Where ${\boldsymbol r}$ is the position vector of the center of mass C relative to the point of interest A.

• So in the absence, of linear momentum ${\boldsymbol p} = m\, {\boldsymbol v}_C$ the angular momentum ${\boldsymbol L}_A$ at an arbitrary point A is the same as the angular momentum at the center of mass ${\boldsymbol L}_C = \mathrm{I}_C {\boldsymbol \omega}$.

• Also, in the absence, of rotational velocity ${\boldsymbol \omega}$ linear velocity ${\boldsymbol v}_A$ at an arbitrary point A is the same as the linear velocity at the center of mass ${\boldsymbol v}_C$. (pure translation)

• Finally, in the absence, any applied force ${\boldsymbol F}$ the equipollent torque ${\boldsymbol \tau}_A$ at an arbitrary point A is the same as the equipollent torque at the center of mass ${\boldsymbol \tau}_C$. (pure torque applied)

Now to consider the mass moment of inertia about a point A not the center of mass C.

1. Assume the body rotates about point A with rotational velocity ${\boldsymbol \omega} \neq 0$ and ${\boldsymbol v}_A=0$.
2. Find the velocity of the center of mass, $\require{cancel} {\boldsymbol v}_C = \cancel{{\boldsymbol v}_A} - {\boldsymbol r} \times {\boldsymbol \omega}$.
3. Find the momentum, ${\boldsymbol p} = m\,{\boldsymbol v}_C = - m\,{\boldsymbol r} \times {\boldsymbol \omega}$
4. Find the angular momentum at the center of mass, ${\boldsymbol L}_C = \mathrm{I}_C {\boldsymbol \omega}$.
5. Find the angular momentum at point A, ${\boldsymbol L}_A = {\boldsymbol L}_C + {\boldsymbol r} \times {\boldsymbol p}$ $${\boldsymbol L}_A =\mathrm{I}_C {\boldsymbol \omega} - m\, {\boldsymbol r} \times ({\boldsymbol r} \times {\boldsymbol \omega})$$

The first part of this expression is the intrinsic angular momentum and the second part the parallel axis theorem.

Mathematically the rotational velocity vector if factored out of this expression and the 3×3 matrix between the momentum and velocity is the mass moment of inertia

$$\mathrm{I}_A = \mathrm{I}_C - m \,[{\boldsymbol r} \times] [{\boldsymbol r} \times]$$

A little mathematical trick here is to convert the vector cross product into a matrix operation $a \times b = [a \times] b = \left[\matrix{0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0}\right] \pmatrix{b_x\\b_y\\b_z}$

Angular momentum depends on point from which we are calculating. On every case this is valid. Angular momentum is equal to sum of momentum about a point and momentum about center of mass. your case although you change the point the velocity of centre of mass is 0. Since there will be no momentum about the point due to rotation of centre of mass. But the momentum about center of mass is same for everypoint since the sphere is rotating itself irrespective of a point.

• so does it mean angular momentum of a rotating sphere about any point inside and outside the sphere is Iw i.e is independent of the point the point of calculation. – sachin Feb 12 '18 at 16:16
• Angular momentum will change according to point. Momentum will be same if the sphere is not rotating about the point you taken . In your case the sphere is not rotating about the point outside. It is rotating only itsel that is about centre of mass . If you give some value to 'v' in your's the momentum will change according to that point. See the parallex axis therem of moment of inertia will get a idea of what i am telling. – Sri vishnu Bharat Feb 12 '18 at 16:21

First of all, it's important to remember that the $\omega$ in your formulas is always the angular velocity measured relative to the center of mass — not whatever point you've chosen. So at the most basic level, that's how the angular momentum stays the same. But why it stays the same is a little more complicated.

It's easiest to understand if you look at a simpler case: the idealized dumbbell. Suppose we have two dense objects of mass $M/2$, separated by a distance $D$ and connected by a massless rod. Let's say their center of mass is at the origin, and object 1 is at $x=D/2$, while object 2 is at $x=-D/2$. Now suppose that object 1 is moving with velocity $v$ in the $+y$ direction, and object 2 is moving with velocity $v$ in the $-y$ direction. So we can calculate the total angular momentum: \begin{align} \vec{L} &= \vec{R}_1 \times \frac{M}{2}\vec{V}_1 + \vec{R}_2 \times \frac{M}{2}\vec{V}_2 \\ &= \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(-\frac{D}{2} \hat{x}\right) \times \left(-\frac{M}{2} v \hat{y} \right) \\ &= \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) \\ &= \frac{1}{2} D M v\, \hat{x} \times \hat{y} \\ &= \frac{1}{2} D M v\, \hat{z}. \end{align} Note how, in going from the second to third lines, the two minus signs cancel out, so each mass contributes the same amount to the total angular momentum.

But now let's suppose that we move the whole system in some direction $\vec{\delta}$. Then, $\vec{R}_1$ becomes $\vec{R}_1 + \vec{\delta}$ and $\vec{R}_2$ becomes $\vec{R}_2 + \vec{\delta}$. We can do the whole calculation again, and we'll find the these additions cancel out: \begin{align} \vec{L} &= \left(\vec{R}_1+\vec{\delta}\right) \times \frac{M}{2}\vec{V}_1 + \left(\vec{R}_2+\vec{\delta}\right) \times \frac{M}{2}\vec{V}_2 \\ &= \left(\frac{D}{2} \hat{x}+\vec{\delta}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(-\frac{D}{2} \hat{x}+\vec{\delta}\right) \times \left(-\frac{M}{2} v \hat{y} \right) \\ &= \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(-\frac{D}{2} \hat{x}\right) \times \left(-\frac{M}{2} v \hat{y} \right) \\ &\quad + \left(\vec{\delta}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(\vec{\delta}\right) \times \left(-\frac{M}{2} v \hat{y} \right) \\ &= \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(-\frac{D}{2} \hat{x}\right) \times \left(-\frac{M}{2} v \hat{y} \right) \\ &\quad + \left(\vec{\delta}\right) \times \left(\frac{M}{2} v \hat{y} \right) - \left(\vec{\delta}\right) \times \left(\frac{M}{2} v \hat{y} \right) \\ &= \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) + \left(\frac{D}{2} \hat{x}\right) \times \left(\frac{M}{2} v \hat{y} \right) \\ &= \frac{1}{2} D M v\, \hat{x} \times \hat{y} \\ &= \frac{1}{2} D M v\, \hat{z}. \end{align} Because both objects were moved by the same distance $\vec{\delta}$, but they still have opposite velocities, these contributions cancel out.

Now, in general, the total angular momentum for any object can be calculated by adding up all the contributions from each little part of the object. Wikipedia has a good discussion here, in which the equivalent cancellation looks like $\sum_i m_i \vec{v}_i = 0$.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Feb 14 '18 at 18:20

protected by Qmechanic♦Feb 13 '18 at 20:56

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