# Can a non-rotating ball have angular momentum?

Suppose a ball is going straight along a fixed line without rotating. Now you consider a point which is not along the trajectory of the ball. If I want to consider the angular momentum about that point, as you can understand that the $$r$$ and $$v$$ vector aren't in the same direction, so $$r\times v$$ should have a value. So this object isn't rotating but has an angular momentum. Can I conclude that the ball has angular momentum without rotating ?

• Yes , that is the way it has been defined . – anna v Aug 24 '18 at 6:33
• Yes. The definition makes sense, because if a ball flew past you, you would start spinning if you caught it. – knzhou Aug 24 '18 at 6:38
• I was going to say an electron until I read the question. – Michael Aug 24 '18 at 18:54
• You should distinguish between the object on its own (frame of reference defined by its velocity vector), and the system of the static object and the observer, where line of sight is an alternate frame of reference, and you get a choice as to which frame is used for the 'angular' momentum question. Without a well defined centre of balance/gravity, you can't decide about the angular momentum. – Philip Oakley Aug 25 '18 at 21:11
• Why did you add a space before the final question mark? In English there is never a space before a comma, full stop (period), colon, semi-colon, question mark or exclamation mark. If you made this trivial edit to bump your question to the home page, such a practice is frowned upon. – Chappo Hasn't Forgotten Monica Nov 24 '18 at 1:27

Yes, it does. It may seem a bit more intuitive if you imagine a line connecting your reference point and the centre of the ball: as the ball moves, the line sweeps out an angle across the ball, so there should be some angular momentum. If you think in polar coordinates instead of cartesian, the ball is rotating.

• This means that the answer to the question's title would be "No, something which isn't rotating can't have angular momentum", right? Your "Yes, it does" would be an answer to the body of the question. – Anonymous Coward Aug 24 '18 at 14:02
• @JoseAntonioDuraOlmos I'm not sure if no triple negatives are ever not the right answer to no question. – Yakk Aug 24 '18 at 17:34
• @JoseAntonioDuraOlmos The opposite: "Yes, something which isn't rotating CAN have angular momentum." – Kyle Oman Aug 24 '18 at 18:19
• @KyleOman Can you explain the point that the ball is rotating in polar coordinates. – Nobody recognizeable Aug 25 '18 at 6:02
• @Nobodyrecognizeable draw the system at 2 different times, including radial and azimuthal unit vectors, and the velocity vector. Velocity has rotated wrt the coordinate frame. – Kyle Oman Aug 25 '18 at 8:47

I think your question comes about because you are comparing angular momentum to linear and there seems to be a difference: in the case of "normal" momentum, it's mass times velocity, it's a "real" thing... but angular, well that seems to be virtual, the object might have or not have it depending on where "you" are. And that seems strange, right?

Ok, so imagine that same ball going along that straight line, but now you are on a cart travelling at the same speed along the same line. Now what's the momentum of the ball? Zero!

This example doesn't seem so odd, but the underlying basis is the same; all of these measurements depend on what "you" are doing, or more technically, your frame of reference.

So the idea that an object can have multiple values for [MEASUREMENT X] at the same time should not be at all surprising or weird, it's kind of baked into the way these things are defined.

Yes. If you watch a vehicle going straight on road while you stand at a bus stand then the angular momentum possessed by the vehicle is $m\vec{v}\times\vec{r}$, where $m$ is mass of vehicle, $\vec{v}$ is velocity of vehicle and $\vec{r}$ is the vector starting from you and ending at the vehicle which is variable in this case.

Though I won't give you a complete answer but I would give you enough hint to think upon it.

Let a bullet moving with a speed $v$ collide with a block of comparable mass. The block will start moving. In a crude language we tend to say that the block has gained some momentum from the bullet.

Now we release a non spinning bullet towards a coin such that it hits the edge of the coin. The coin will surely spin. Thus it has acquired an angular momentum. If the bullet wouldn't have angular momentum, then how come it transfer angular momentum to the coin.

Now the thinking process is yours....

• The angular momentum of the system will still be conserved. The coin will start spinning but the bullet's trajectory will be deflected in the opposite direction too. – Mike Aug 24 '18 at 11:27

The angular momentum of an object is simply the sum of the angular momentum of its parts. If an object has angular momentum L, and N atoms, then each atom has, on average, L/N angular momentum. The atoms aren't rotating, they are revolving, and from an instantaneous point of view, they have linear motion.

If you took two balls connected with a string and spin them, they would have angular momentum. If you cut the string, the balls will fly off, and the angular momentum won't go anywhere; the balls will still have angular momentum.

Note that angular momentum is defined with respect to a particular axis. When discussing the angular momentum of an object, the axis is taken to be one that goes through the center of mass. But each object has an angular momentum with respect to every axis. The earth has an angular momentum with respect to the polar axis, and another angular momentum with respect to the sun's polar axis. For each axis, the total angular momentum of the universe is conserved.

The problem I see is that it is not clear what is meant by "rotating." An object can rotate about its center of mass, about a point away from its center of mass, or both!
So, an object can have angular momentum, even though it may not be rotating with respect to its center of mass, but moves in a circular path. Or it might be moving in a straight line but rotates about its center of mass, or both.

The answer depends on the kind of object - particle or field - you mean, and if you refer to spin- or orbital angular momentum.

Invisible or "hidden" electromagnetic field spin angular momentum is equivalent to the canonical angular momentum $$\vec{L}_{c} = \vec{r} \times q \vec{A}$$ generated by a charge $q$ at relative rest in a magnetostatic field with vector-potential $\vec{A}$ at $\vec{r}$. Example: If $q$ is located in the center of a solenoid field of flux $\Phi$, a "hidden" electromagnetic field angular momentum $L = q \Phi/ 2 \pi$ would be generated.