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In his answer to another post user Albertus Magnus describes the situation of a bullet hitting a rod in free space on its tip in a "purely tangential" way causing the rod to spin in a purely rotational movement. He then elaborates that this case is not realistic and should be rather treated as an external force acting on the tip of the rod. While this would remove the question, whats happening with the momentum of the bullet, this does not really make sense to me either. Any force acting on a point should cause a translation of the point and therefore apply some translational momentum to the body. Purely rotational motion seems only possible if another force at any part of the body would compensate the translational momentum, but that is not possible if a finite force is acting on only one point. Since I still don't understand this after some discussion - or rather I still beleave such process is unphysical - I decided to open a new question:

Can a force acting on a single point of a resting, freely movable body cause it to spin without causing translational movement?

If yes, how does it work? What is the meaning of a "purely tangential impulse" in this context?

If no, is there point in bringing up such an unphysical process? Does it have any meaning in any kind of mathematical thought experiment?

Also let me mention, even though this really should be obvoius, that this is no personal attack on anyone in any way.

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  • $\begingroup$ I am going to declare my previous thoughts on an impulse being capable of producing purely rotational motion in the absence of any forces of constraint to be erroneous. The fact that the rod is a rigid body will not allow it. Everything else is true. $\endgroup$ Commented Mar 8 at 22:13
  • $\begingroup$ Thank you Zaph for being persistent in your doubts, hopefully the both of us are now of a much clearer mind. $\endgroup$ Commented Mar 8 at 22:19
  • $\begingroup$ Does that include the part about the external force acting on a point causing the rigid body to spin without translating it? $\endgroup$
    – Zaph
    Commented Mar 8 at 22:21
  • $\begingroup$ Yes, it includes that and only that, everything else was precise! $\endgroup$ Commented Mar 8 at 22:23
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    $\begingroup$ I think you writing a short answer and me accepting it would be the cleanest solution. $\endgroup$
    – Zaph
    Commented Mar 8 at 22:29

1 Answer 1

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Consider that there is a system consisting of an incoming projectile (point with mass m) and a rigid rod (one dimensional segment with mass M). For the sake of simplicity let the problem be two dimensional so that the motion is entirely in the $x,y$ plane, otherwise the rod is unconstrained, i.e. it is not fastened at any point and is free to move in any fashion.

We now ask: what happens when the projectile strikes the rod imparting an impulse at some point along the rod's extension, with a particular emphasis on the conservation of momentum.

The situation is that the incoming projectile impinges against the rod imparting to it an impulse while merging with it so that the two bodies are now one. Since the system consists of the projectile and rod, we have two masses each with its own linear and angular momenta (depending upon the coordinates chosen).

Before the collision the projectile has some angular momentum according to the origin of coordinates so chosen. The rod on the other hand is stationary and has no angular momentum. After the collision, the combined rod\projectile system has two kinds of angular momentum that sum to a net total, viz. the angular momentum of the system due to rotation about its center of mass and the angular momentum associated with the translation of the center of mass itself. The principle of conservation of angular momentum demands that the angular momentum of the projectile before collision is equal to the angular momentum of the combined rod\projectile after collision.

This analysis assumes coordinate such that each mass can be characterized entirely in terms of its angular momentum. We could have chosen other coordinates where one would have had to introduce both linear and angular momentum, however, each would still be separately conserved and the resultant motion identical, i.e. the rod would move away from its initial location with both rotational and translational motion.

Do there ever arise any cases where a rigid body can experience an external force and not exhibit translation in addition to rotation. The answer is definitely not. The fact that the body is rigid will not allow for the pure rotation without translation in the absence of some force of constraint. This can be seen clearly from the standpoint that; whereas the configuration manifold of the unconstrained system is two dimensional, requiring two numbers to specify its motion, viz. $x,y$, the purely rotating system requires only one coordinate to specify its motion entirely, i.e. an angle $\theta$ will suffice. Thus, there must be an equation of constraint to eliminate one of the two previously required coordinates. 

I must gratefully acknowledge user Zaph, as their stubborn persistence, in what they knew to be right, was a guiding light for the resolution of the problem.

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  • $\begingroup$ You talk about "coordinates such that each mass can be characterized entirely in terms of its angular momentum". Could you elaborate on how that works for an extended body like the rod? If you just have its angular momentum according to some reference point, wouldn't it be unclear how much of it is in translation and how much in rotation? Also movement radially away from the reference point could not me considered this way, could it? $\endgroup$
    – Zaph
    Commented Mar 8 at 23:29
  • $\begingroup$ @Zaph No it would not be unclear, because the angular momentum consists of two terms, one for the translation of the center of mass, and one for the rotation of the rod about the center of mass. So, you can talk about the whole thing using only angular momentum. Similarly, one can consider that the incoming projectile just has angular momentum as a result of its translation. It is important to remember that an object moving in a straight line has angular momentum, if the coordinates are chosen right. $\endgroup$ Commented Mar 8 at 23:34
  • $\begingroup$ Ok so you need 2 different values of 2 differently defined terms that in sum make up the angular momentum. But still would the angular momentum part due the translation not ignore the translational motion radially towards or away from the reference point? I.e. independently of weather the bulled moves towards your reference point or away additionally to the perpendicular motion you would get the same value of the angular momentum, which depends only on the perpendicular motion and thus does not contain full information? $\endgroup$
    – Zaph
    Commented Mar 8 at 23:49
  • $\begingroup$ @Zaph Thats why I mentioned if the coordinates are chosen right. The physics doesn't depend on the coordinates, but how you discuss the problem does. $\endgroup$ Commented Mar 9 at 0:36
  • $\begingroup$ So you choose the coordinates in a way that the translational movement radially to your reference point is $0$? But this would only be true for a single point in time, no? Radial translational movements will change with time, because the angle of the radial line from the reference point to the particle with the velocity vector will change with time, when linear translation is happening. From my understanding this should be happening independently of how you choose your coordinates. $\endgroup$
    – Zaph
    Commented Mar 9 at 6:58

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