# Angular Momentum **URL:** \ **Category:** Factual Questions\ **Created:** [March 14, 2012, 5:17pm UTC](https://boards.straightdope.com/t/angular-momentum/615754 "2012-03-14T17:17:16Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Vorpal\_Blade](https://avatars.discourse-cdn.com/v4/letter/v/9de0a6/32.png) [@Vorpal\_Blade](https://boards.straightdope.com/u/Vorpal_Blade)\ **Post date:** [March 14, 2012, 5:17pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/1 "2012-03-14T17:17:16Z")
I throw a giant Frisbee out into space, free of (most) gravitational influence. It is spinning but part of its design includes rockets aligned tangentially which, when fired, counter the spin. In other words, the Frisbee is spinning until the rockets are fired. Once the rockets go, the Frisbee gradually stops spinning and through clever throttling eventually stops spinning. Here’s my question: where did the always-conserved momentum go? ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [March 14, 2012, 5:34pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/2 "2012-03-14T17:34:42Z")
Into the rocket exhaust? ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [March 14, 2012, 5:56pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/3 "2012-03-14T17:56:06Z")
Change in angular momentum id _torque_. You exerted torque on the frisbee and this changed its angular momentum. Angular momentum is conserved only when no torque is applied. Otherwise you couldn’t start anything spinning, or stop anything that is spinning. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [March 14, 2012, 6:11pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/4 "2012-03-14T18:11:47Z")
Torque > **[Torque](https://en.wikipedia.org/wiki/Torque)** > > In physics and mechanics, torque is the rotational analogue of linear force. It is also referred to as the moment of force (also abbreviated to moment). It describes the rate of change of angular momentum that would be imparted to an isolated body. > The concept originated with the studies by Archimedes of the usage of levers, which is reflected in his famous quote: "Give me a lever and a place to stand and I will move the Earth". Just as a linear force is a push or a pull applied to a body, a to... Angular Momentum > **[Angular momentum](https://en.wikipedia.org/wiki/Angular_momentum)** > > In physics, angular momentum (sometimes called moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important physical quantity because it is a conserved quantity – the total angular momentum of a closed system remains constant. Angular momentum has both a direction and a magnitude, and both are conserved. Bicycles and motorcycles, flying discs, rifled bullets, and gyroscopes owe their useful properties to conservation of angular momentum. Conserv Th... ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [March 14, 2012, 6:49pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/5 "2012-03-14T18:49:45Z")
> [@Vorpal\_Blade](#): > > I throw a giant Frisbee out into space, free of (most) gravitational influence. It is spinning but part of its design includes rockets aligned tangentially which, when fired, counter the spin. In other words, the Frisbee is spinning until the rockets are fired. Once the rockets go, the Frisbee gradually stops spinning and through clever throttling eventually stops spinning. > > Here’s my question: where did the always-conserved momentum go? When the frisbee stops spinning, you now have a cloud of rocket exhaust that is rotating around a common center, i.e. the center of the frisbee. That cloud of rocket exhaust, considered in aggregate, has the angular momentum that used to be in the frisbee. After a while, that exhaust gas has a large radial velocity relative to the center of the frisbee, but that’s irrelevant; you need only consider the tangential velocity of the exhaust gas cloud (relative to the center of the frisbee) in order to calculate its angular momentum, and that tangential velocity isn’t going to change over time. Draw a diagram of your frisbee of radius R, showing rockets at its edge aimed tangentially. Project the rocket plumes off into the distance, and you might be able to visualize what’s happening. ---
**Author:** ![bizerta](https://avatars.discourse-cdn.com/v4/letter/b/3be4f8/32.png) [@bizerta](https://boards.straightdope.com/u/bizerta)\ **Post date:** [March 14, 2012, 6:54pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/6 "2012-03-14T18:54:29Z")
> [@Machine\_Elf](#): > > When the frisbee stops spinning, you now have a cloud of rocket exhaust that is rotating around a common center, … NO. Once fired, the rocket exhaust continues in a straight line at a constant speed forever. ---
**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\ **Post date:** [March 14, 2012, 7:12pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/7 "2012-03-14T19:12:28Z")
> [@bizerta](#): > > NO. Once fired, the rocket exhaust continues in a straight line at a constant speed forever. Yes, for certain values of “rotating”. 🙂 A particle moving forever in a straight line carries angular momentum around any point in the universe, and for each origin that angular momentum is conserved. Recall that the size of the angular momentum vector is ||r||\*||p||_sin(theta). When the particle is far away, r is large, but sin(theta) is small to compensate. As the particle gets closer, r gets smaller, but sin(theta) gets bigger, and when the particle makes its closest approach to the origin, sin(theta) = 1 and ||L|| = ||r||_||p||. So each particle the rocket exhaust carries an angular momentum of ||r||\*||p||, where r is the radius of the frisbee, and p is the momentum of the particle. The main contribution to the norm of the angular momentum vector just shifts from the sin(theta) factor to the ||r|| factor so the whole product is conserved even as the particle moves away. And if you want to use “has angular momentum about a centre” as your definition of “is rotating about a centre”, then the exhaust is rotating. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [March 14, 2012, 8:40pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/8 "2012-03-14T20:40:11Z")
Look at it another way, in reverse. You have a giant paddlewheel floating free in space, and a rogue meteor hits a paddle and starts it spinning. The distance from center of mass about which it spins to that impact point is r, so the angular momentum imparted is that “p r sin theta”, where theta is angle between impact direction and tangent, p is ( loss of)momentum of meteor. ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [March 14, 2012, 8:46pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/9 "2012-03-14T20:46:28Z")
> [@leahcim](#): > > And if you want to use “has angular momentum about a centre” as your definition of “is rotating about a centre”, then the exhaust is rotating. I didn’t express it very well, but my claim of “a _cloud_ of rocket exhaust rotating around a common center” was about this. The plumes, all taken together, constitute a cloud of gas that is both expanding _and_ rotating about a common center. ---
**Author:** ![bizerta](https://avatars.discourse-cdn.com/v4/letter/b/3be4f8/32.png) [@bizerta](https://boards.straightdope.com/u/bizerta)\ **Post date:** [March 14, 2012, 9:10pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/10 "2012-03-14T21:10:33Z")
> [@Machine\_Elf](#): > > … The plumes, all taken together, constitute a cloud of gas that is both expanding _and_ rotating about a common center. NO, NO, NO, NO, NO. A body in motion remains in motion in a straight line unless there is an external force. Once the gas is released, there is no external force. Like the Energizer Bunny, it keeps going and going and going. The gravitational effect of the frisbee and (and the universe as a whole) is not measurable. Now the gas cloud may appear to be an ever-expanding, symmetrical (from the frisbee’s point of view) spiral, but each individual molecule is going in a straight line. ---
**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\ **Post date:** [March 14, 2012, 9:21pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/11 "2012-03-14T21:21:38Z")
> [@bizerta](#): > > Now the gas cloud may appear to be an ever-expanding, symmetrical (from the frisbee’s point of view) spiral, but each individual molecule is going in a straight line. **Machine Elf** ’s point is that these statements are not contradictory. Every particle in the cloud can be moving in a straight line, but the cloud, as a whole, can be expanding and rotating. My point was that even a single particle traveling in a straight line has an angular momentum that is conserved in the absence of external torque, even though it doesn’t “look like” it is rotating in any way. ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [March 14, 2012, 9:23pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/12 "2012-03-14T21:23:00Z")
> [@bizerta](#): > > Now the gas cloud may appear to be an ever-expanding, symmetrical (from the frisbee’s point of view) spiral, but each individual molecule is going in a straight line. There’s nothing inconsistent between that and what Machine Elf said. Suppose instead of a cloud, the frisbee fires off two ball bearings in opposite directions connected via a massless, infinitely steretchable string. Now, since the ball bearings had to be fired off at the edges of the frisbee but they have a tangental velocity, you can see that the string has an initial rotation to it (and that our ball bearings rotating around this center exactly contain the angular momentum of the frisbee). As the ball bearings get farther apart, the rotation of the string slows down, to the point where it will never quite complete a quarter of a rotation. But that’s ok, because the ball bearings are also getting farther from the center, and these two effects exactly cancel out. So yes, angular momentum is conserved. ---
**Author:** ![bizerta](https://avatars.discourse-cdn.com/v4/letter/b/3be4f8/32.png) [@bizerta](https://boards.straightdope.com/u/bizerta)\ **Post date:** [March 14, 2012, 10:46pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/13 "2012-03-14T22:46:45Z")
> [@leahcim](#): > > … My point was that even a single particle traveling in a straight line has an angular momentum … NO, NO, NO. If it’s going in a straight line, where’s the center of rotation. There isn’t one. Without rotation, it does not have any angular momentum. It has linear momentum, but not angular momentum. Once the frisbee has stopped rotating, the angular momentum has been transferred to linear momentum. The conservation of angular momentum only deals with objects that effect each other (such as a skater pulling her arms in when she spins). Once the particles leave the rocket at the tip of the frisbee, they are totally independent, free to pursue a life of religious fulfillment as they travel forth in a straight line at a constant speed. ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [March 14, 2012, 11:00pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/14 "2012-03-14T23:00:02Z")
> [@bizerta](#): > > NO, NO, NO. Once is sufficient, thanks - and please stop shouting. > [@](#): > > If it’s going in a straight line, where’s the center of rotation. You can choose any center you like. In the present case, the most convenient one is the center of the frisbee. Consider each particle of rocket exhaust to have mass m and linear absolute velocity v. The rockets are mounted at radius r from the center of the frisbee and fire the exhaust tangentially. Each particle, once it leaves the rocket nozzle, has angular momentum of m_v_r. The entire cloud is comprised of n particles of rocket exhaust, so the angular momentum of the entire cloud is n_m_v\*r. > [@](#): > > Without rotation, it does not have any angular momentum. Given that the conservation of angular momentum is a principle well established in physics, if you believe the cloud does not have any angular momentum, then I challenge you to explain where the angular momentum that was in the frisbee has gone. > [@](#): > > It has linear momentum, but not angular momentum. Once the frisbee has stopped rotating, the angular momentum has been transferred to linear momentum. No. Angular momentum in a closed system (in this case, a system consisting of the frisbee and the cloud of rocket exhaust) is conserved. ---
**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\ **Post date:** [March 14, 2012, 11:12pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/15 "2012-03-14T23:12:22Z")
> [@bizerta](#): > > NO, NO, NO. If it’s going in a straight line, where’s the center of rotation. There isn’t one. YES, YES, YES. In the absence of any force whatsoever, angular momentum is conserved using any point in the universe as the centre of rotation. Simply pick any point in the universe as your origin. Compute the quantity ||r||\*||p||\*sin(theta), and see that throughout the entire motion that quantity stays the same. If we are talking about motion in a central force field, then we have to be careful to put the origin at the centre of the force field, to eliminate any torque, but in the absence of any potential at all, we can place the origin anywhere. > [@](#): > > Without rotation, it does not have any angular momentum. It has linear momentum, but not angular momentum. Once the frisbee has stopped rotating, the angular momentum has been transferred to linear momentum. There is nothing at all inconsistent about a particle having both angular momentum and linear momentum. And in the absence of any external force there is nothing inconsistent about both being conserved. > [@](#): > > The conservation of angular momentum only deals with objects that effect each other (such as a skater pulling her arms in when she spins). This is not correct. Conservation of angular momentum about a centre occurs whenever the Lagrangian describing the system is invariant under rotations about that centre. Once the particle is free, the Lagrangian describing its motion is invariant under rotations about _any_ centre. ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [March 14, 2012, 11:30pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/16 "2012-03-14T23:30:23Z")
What if the Frisbee contacted a long strip of rubberized material on rollers. Say, something like a gigantic treadmill is space, would the answer be the same? ---
**Author:** ![kombatminipig](https://avatars.discourse-cdn.com/v4/letter/k/b5ac83/32.png) [@kombatminipig](https://boards.straightdope.com/u/kombatminipig)\ **Post date:** [March 15, 2012, 8:47am UTC](https://boards.straightdope.com/t/angular-momentum/615754/17 "2012-03-15T08:47:49Z")
This thread oddly reminds me of the failed (most likely) Russian rocket launch observed above Norway a couple of years ago: [STRANGE SPIRAL LIGHTS OVER NORWAY - YouTube](http://www.youtube.com/watch?v=Sof_QK1z4b4) ---
**Author:** ![minor7flat5](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/minor7flat5/32/258_2.png) [@minor7flat5](https://boards.straightdope.com/u/minor7flat5)\ **Post date:** [March 15, 2012, 12:31pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/18 "2012-03-15T12:31:33Z")
> [@Dr.Strangelove](#): > > Suppose instead of a cloud, the frisbee fires off two ball bearings in opposite directions connected via a massless, infinitely steretchable string. > > Now, since the ball bearings had to be fired off at the edges of the frisbee but they have a tangental velocity, you can see that the string has an initial rotation to it (and that our ball bearings rotating around this center exactly contain the angular momentum of the frisbee). > > As the ball bearings get farther apart, the rotation of the string slows down, to the point where it will never quite complete a quarter of a rotation. But that’s ok, because the ball bearings are also getting farther from the center, and these two effects exactly cancel out. I like this explanation. At first I found it confusing because the two ball bearings are supposed to go on their merry way in straight lines, and the string sounded too much like a piece of elastic (though it isn’t). Indeed, the string is simply a way of describing a line drawn center-to-center between the projectiles. Seen this way, as the two balls speed away in straight lines, tangent to the circle, the centerline defined by the two balls is still always rotating, slower and slower, as the balls go further in their perfectly straight, but parallel, paths. ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [March 15, 2012, 8:18pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/19 "2012-03-15T20:18:48Z")
> [@minor7flat5](#): > > Indeed, the string is simply a way of describing a line drawn center-to-center between the projectiles. > > Seen this way, as the two balls speed away in straight lines, tangent to the circle, the centerline defined by the two balls is still always rotating, slower and slower, as the balls go further in their perfectly straight, but parallel, paths. Exactly. I debated whether to include the string, since as you mention it slightly confuses the issue, but as you say it provides a rotating centerline. I also wanted to provide a bit of physical intuition that it is a single \*system \*we’re talking about, and that looking at one object or the other in isolation doesn’t give the full picture. ---
**Author:** ![bizerta](https://avatars.discourse-cdn.com/v4/letter/b/3be4f8/32.png) [@bizerta](https://boards.straightdope.com/u/bizerta)\ **Post date:** [March 15, 2012, 9:18pm UTC](https://boards.straightdope.com/t/angular-momentum/615754/20 "2012-03-15T21:18:14Z")
Boy, did I blow this one. I became so fixated on the fact that everything was moving in a straight line so angular momentum must be zero (which I now realize is incorrect). I’ve asked my wife for humble pie for desert tonight.