Consider a system of spring mass as shown in the figure.entire loop is free but only one nail is there at the point A... Initially the mass is at rest and then released. Assume that the spring is initially stretched and is in equilibrium in point C.

Now on the loop PQ (P=initial position of bead, Q=diametrically opposite) of the path the centre of the path is Point A. We are assuming that the loop is a massless, smooth rod. So the normal provided by the loop on the bead will pass through the axis. But the force of spring on the bead will be tangential to the path and the torque related to this force on the particle about the considered axis will be nonzero. Also no other forces are acting on the bead. So about the considered axis the angular momentum of the bead is changing. But if we analyse the entire loop+spring (massless)+bead system then there is no non-zero external torque acting on system. Since the only moving portion of the system is the bead angular momentum of the entire system is also changing. Note that the spring force is an internal force for the considered system. Where am I wrong ? Sketch of the problem

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    $\begingroup$ The path center is never at A because the path is a spiral and not an arc. $\endgroup$ – ja72 May 12 '14 at 17:18
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    $\begingroup$ This is such a radical change to the question that it should ideally be posted separately. On the other hand, you've already invested a lot of reputation in your bounty, so it's fair to keep it here. Nevertheless, you should keep the old version of the question under the current one, out of respect for the effort by the people who tried to answer that one. $\endgroup$ – Emilio Pisanty May 26 '14 at 21:45
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    $\begingroup$ Questions that have answers should not be changed to invalidate the answers. I have rolled back the question and refunded the bounty. Feel free to ask the modified question separately. $\endgroup$ – dmckee --- ex-moderator kitten May 26 '14 at 23:17
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    $\begingroup$ @Floris It looked to me as if the bounty was intended to draw attention to the edit; I didn't feel that it was fair to the OP to insist on rolling back the edit without refunding the bounty and I felt the edit needed rolling back. I presume that Dvij will either re-instate it on this question as asked or re-ask the modifier version and append the bounty to it once allowed. $\endgroup$ – dmckee --- ex-moderator kitten May 27 '14 at 3:04
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    $\begingroup$ @dmckee- AFAIK the bounty was set on the original (rolled back) question, not the completely changed one. $\endgroup$ – Floris May 27 '14 at 10:38

I think I initially misunderstood the question. I now believe there are three components here: a rigid massless rod in the shape of a spiral; a massless spring that is wound around the rod in a frictionless manner; and a bead at the end of the spring. The entire system is anchored at the point A by a single nail - thus, the system is free to rotate about this point A (it is a hinge, i.e. it fixes the position of the spiral at A, but not the direction).

Now it would be instructive to look at the forces on the system - remembering Newton's third law. You drew the following (I am simplifying the spiral to make the drawing clearer but the principle is the same): enter image description here

In reality, what you have is the following - at every point in the system there must be a force balance by Newton's third law:

enter image description here

The spring is pulling "down" on the mass $M_1$ and on the nail $A$; the reaction is a net force of $2F$ on the "spring plus rod" system, to which I am assigning a mass $M_2$.

Now what would happen in a system with finite $M_2$ is the following:

Mass $M_1$ accelerates downward with $$a_1 =\frac{F}{M_1}$$

Mass $M_2$ accelerates upward with $$a_2 =\frac{2F}{M_2}$$

A short time later, the situation will look like this (I am sorry this is a messy drawing):

enter image description here

As you can see, the mass $M_1$ is moving down, and the mass $M_2$ is moving up; the direction of the forces is changing.

Now the relative velocities of the two parts of the system: bead, and {spring + rod} after a short time $\delta t$ (before the system has had chance to change shape significantly) is given by their relative masses from the two accelerations above:

$$ \frac{V_1}{V_2} = \frac{M_2}{2M_1}$$

(and these velocities are in opposite directions). Note that this expression shows conservation of angular momentum, since

$$ 2 M_1 V_1 = M_2 V_2$$

and the distance from the center of rotation (chosen to be at A) for $M_1$ is twice the distance for $M_2$ (projected onto the horizontal axis).

Now the only thing you have to do is take the limit of the above where $M_2$ is zero, and you see that angular momentum is conserved for this system.


The way I drew the spiral, it looked like a semicircle. In reality, one property of a spiral is that there has to be a radial component to the slope - so the force on mass $M_1$ is really not just tangential (as I drew it in the initial drawing), but it has both a tangential and a radial (inward) component. This shows that there is a force available for accelerating the mass inwards from the moment the system is released. The other aspects of the above description are unchanged - although it's slightly trickier to convince yourself of the balance of forces on the light rod when it has arbitrary shape. But the same argument holds as before: when the mass of rod+spring tends to zero, it will move instantaneously in a direction that opposes tangential motion by $M_1$, thus conserving angular momentum.

one more thought

I just re-read where you state "since the only thing moving is the bead". This could be read as meaning that the rod is fixed - in other words, that the point A is fixed such that the rod doesn't rotate about A. In that case there will be a torque at A which imparts angular momentum to the rod-spring-bead system on the one hand, and an equal and opposite angular momentum on the rest of the universe - in other words, with this kind of coupling you cannot consider the system "isolated".

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    $\begingroup$ how do you make your nice looking figures? $\endgroup$ – chris May 25 '14 at 0:40
  • $\begingroup$ @chris I draw them in PowerPoint then take a screen shot (on a Mac) $\endgroup$ – Floris May 25 '14 at 0:42

Well lets hope I understood it correctly.

First of all since this is a horizontal loop and your bead is attached to the spring it would and should not touch the loop at all! (We are just solving the problem ideally, real life is tough) So lets throw out the Normals from our view.

Now you haven't mentioned whether you have your loop fixed to a table or absolutely free in space! So I'll consider two cases.

Case 1: Your loop is absolutely free in space If your loop is free in space and then you compress the spring and release, there is nothing to stop the loop from rotating in other direction accordingly to balance angular momentum. Keep in mind if you fix the loop with just one nail either at P or Q, even then the loop will rotate!

Case 2: Your loop has been fixed to a table by two nails at ends Well in this case your loop will not rotate to balance the angular momentum, but check out the normals from the nails. When the spring pushes the bead forward it alsi pushes the nail at point P tangentially back, which would be met by an equivalent normal which would then provide you your non-zero torque!

I would suggest you to experiment with it, for the first case compress the spring and release the system on the floor, even slight rotation of the whole system in opp direction would be proof of case 1. Case 2 can be tested as described.

  • $\begingroup$ Basically your hope haven't come to reality $\endgroup$ – Dvij Mankad Apr 19 '15 at 4:45

In the ideal system you are imagining (mass less spring of fixed shape, friction free, ...) there are two possibilities:

Either the nail in A fixes the orientation of the spring, which sounds closest to what you initially describe. Than it will simply mediate torque from your system to what ever it is pinned to, resolving your overall angular momentum conservation.

Or the whole spiral plus mass may rotate ideally around that nail such that no angular momentum may be passed on. But then your spring, beeing mass less, will absorb all force in the tangential direction (still no mass, no angular momentum) and the massive bead will move radially inwards (also without angular momentum). (This is the idealized limit of the more real life situation with finite mass of the spring as explained by Floris.)


Angular momentum is only conserved in systems that have rotational symmetry. Assuming a fixed loop, if you consider as your system the bead and the spring, it is not symmetrical, and therefore, the angular momentum is not conserved. But if you consider the whole universe as your system (and, for a simplified universe, we consider the bead, the spring, the loop, and the Earth), you do have symmetry: if you rotate the Earth, the system is the same. So, when you release the marble, it will rotate one way, and the Earth will rotate so slightly the other way.

If you still have doubts, let me put a simplified example with linear momentum. If you have a ball on top of a tilted plane and let it go, it will gather momentum as it goes down (it is not conserved). But if your tilted plane is on wheels, you will see that it is pushed backwards: the total momentum (ball plus plane) is conserved.

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    $\begingroup$ "Angular momentum is only conserved in systems that have rotational symmetry"? I am sorry, but that is just wrong Angular momentum is conserved in any closed system - as long as there is no net external torque, angular momentum is preserved regardless of the geometry/symmetry of the system. $\endgroup$ – Floris May 12 '14 at 21:15
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    $\begingroup$ @Floris : the angular momentum is conserved iff the lagrangian has rotational symmetry (Noether's theorem). The special case of a closed system with no external torque, this is true because our universe has rotational symmetry. $\endgroup$ – Davidmh May 12 '14 at 21:20
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    $\begingroup$ A Langrangian being symmetric has nothing to do with the system being symmetric. I think in this instance we should be able to consider the "spring plus mass" as an isolated system within which angular momentum is conserved, without invoking the rest of the universe. Which can be seen trivially by realizing that the massless spring will become a straight line after infinitesimal time. As for your statement "...our universe has rotational symmetry.": that is patently false as stated; did you leave some words out? What is the axis of rotation? $\endgroup$ – Floris May 12 '14 at 21:30

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