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I'm somewhat stuck on this very simple question, but I can't convince myself of a satisfactory answer. I tried to look through the previous questions asked to avoid asking a duplicate I hope.

Any rotating object... say a rod in this case, sitting stationary in free space, when pushed at one end will cause a rotation about the center of mass. This means the pushed end and other end will move in opposite directions of each other.

The only way that the rod can transmit force down its length is through the interaction of each particle pushing/pulling on the next (tension perpendicular to the length) as it travels down the rod. Once you reach the center of mass, there is no induced rotation. How do the particles past the center of mass receive a push or pull at all if there is a point in the center which has no movement?

Additionally, how does this original force in the positive direction turn into movement in the negative direction past the center of mass? Essentially both of these questions can be answered by an explanation of how the particles inside an object behave when a torque is applied... which I cant find an answer to.

Edit: So I found this other post which asks a very similar question: Why does a rigid body rotate and not simply translate when pushed with an instantaneous force?

I feel that the corded particles explanation answer there is of the form I am imagining for this, but it seems that it doesn't explain the other side rotating backwards.

The particles in the rod near the push start to move forward, which "pull" on the neighboring particles via tension. Then at some point, this tension turns into compression, effectively pushing the particles past the center of mass downward.

Does anyone have a satisfactory internal particle explanation involving tension/compression caused by an initial force at one end which results in the final behavior of rotation of a rod? My understanding of rotational motion feels incomplete to me if I can't explain intuitively why something rotates when pushed off-center besides "if one side goes up, the other must go down".

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  • $\begingroup$ @Spiry00 I don't think there is a "simple" explanation as to why the forces behave the way they do. $\endgroup$
    – Karthik
    Apr 22 '19 at 9:12
  • $\begingroup$ A real rod isn't a 2-force member. It can transmit side loads, just as you break a stick by holding the ends and pushing down the middle. $\endgroup$ Mar 27 '20 at 3:05
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In the case you are examining, what makes the opposite half of the rod move is its inertia: indeed when you hit one end of the rod you will not only cause the whole rod to spin, but also its center of mass to move, if the rod is not holded through its center (you can try with a pencil). The center of mass of the rod will start moving (following the end you kicked) and the opposite end will follow them. In a reference frame that moves with the center of the rod you will see the rod rotating, but in the frame which was seeing the rod being still at the beginning the rod will be now also translating.

So the other half of the rod moves because it is dragged, not because the material creates a force opposite to the one impressed on the object.

Note though that if the center of the rod was jointed so that it would not translate when kicked, the whole motion would be a pure rotation (in the frame which sees the joint being still).

Now the forces acting in and on the rod will be that of the initial kick, the internal forces (which characterize the rod as being a rigid oject) and the new force impressed by the joint, stopping the translation of the center of mass. The rotation will be all the same caused by the kicked end dragging the rest of the object according the constraint acting on the system (the joint).

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  • $\begingroup$ Yes, I follow all of what you say here. The part I am curious about is how exactly you characterize the "dragging" of the rest of the rod. I know that for simplicity, we say that a single force acting on a rigid body can be turned into a force and a torque located at the center of mass. But what bugs me is how exactly the push forward at some location causes movement of each particle of the rest of the rod... "how" is it dragged? $\endgroup$
    – Splry00
    Apr 20 '19 at 7:36
  • $\begingroup$ That is due to the electromagnetic bounds between the molecules (or ions) constituting the object; this bounds are approximatively unstretchable for a rigid body. By the action-reaction principle a force applied to an endpoint of such an object is transmitted to all the points of this object. Note that the classic approximation stating that this happens instantaneously will not hold anymore as the finiteness of the speed of light is taken into account. Therefore in special relativity even the "rigid" bodies can be bent. $\endgroup$
    – AoZora
    Apr 20 '19 at 7:46
  • $\begingroup$ Ok and so as the action-reaction principle is transmitting the force to all of the points of the object, why is it that the speed of the particles closer to the ends move faster in response? And as you move closer to the center of mass, it gradually becomes slower? I realize that the outsides of rotating objects must have a greater tangential velocity in order to have the same angular velocity (due to moving in a larger arc)... but when that initial force pushes the rod, what is the reason that the push's effect "lessens" as you travel down the rod? $\endgroup$
    – Splry00
    Apr 20 '19 at 8:12
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    $\begingroup$ I answered my own question here actually. It's essentially because torque increases with distance from the COM. I really appreciate the help. $\endgroup$
    – Splry00
    Apr 20 '19 at 8:32
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The only reason that you could ever give as to why that "dragging force" pushes the rod in the opposite direction, is due to the conservation of momentum.

If you displace the rod slightly from one end of it, the center of mass remains unaffected because there is no force acting on it directly. So, the conservation of momentum says that net force on the system should be zero if it has to happen. From this you could ascertain that there is a force acting on the opposite end of the rod to push it backwards.

The force that actually does this is electromagnetic (from the bonds between the atoms). But it is literally impossible to say as of now what is causing them to behave in this way, unless you use the simple Newton's laws of motion, that gives this idea of the necessity of momentum being conserved in this scenario, and that which ends up causing the non-zero torque.

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Newton’s laws; applied to translation and rotation:

Consider a system of relatively small particles. They may be atoms. They may, or may not, be bound into a rigid structure. The particles may be subject to forces from other particles within the system or to external forces coming from outside of the system. To simplify the notation, assume that all, r’s, v’s, and a’s in this discussion represent vectors. Small letters refer to individual particles, and capital letters refer to the total mass and the center of mass. The, r’s, are measured from the origin of an arbitrarily chosen inertial coordinate system (it is often convenient to do calculations in terms of the x,y, and z components of vectors). Vectors with primes are measured in a similar system that is centered on (and moving with) the center of mass. An, x, indicates a vector product and the symbol, Σ, indicates a vector sum which includes all of the particles in the system. The vector, R, which locates the center of mass is defined by the expression MR = Σmr. The derivative with respect to time give a more intuitive expression: MV = Σmv. The total momentum of the system is given by the vector sum of the momenta of all of the particles and can be expressed as the total mass times the velocity of the center of mass. Another derivative gives: MA = Σd (mv)/dt. Since the internal forces occur as equal and opposite pairs, the changes in momentum associated with these forces are also equal and opposite, and drop out of the summation. Only the component of the changes in momentum caused by the external forces remain so: MA = Σm dv/dt = Σma = Σf. (The acceleration of the center of mass is determined by the vector sum of the external forces.)

The angular momentum of the system (relative to the origin of the inertial system) is L = Σ r x mv. In the center of mass coordinates it is: Σ [(R + r’) x m(V + v’)] = Σ (R x mV) + Σ (R x mv’) + Σ (mr’ x V) + Σ (r’ x mv’) = (R x MV) + Σ(r’ x mv’) In the second (of four terms) the summation determines the velocity of the center of mass in the center of mass system (which is zero). In the third term, the summation determines the position of the center of mass in that system (also zero). Conclusion: The angular momentum of the system can be considered as a combination of two parts. One part is associated with the motion of the center of mass; the other part is associated with the rotation of the system about the center of mass.

Again, taking a derivative: dL/dt = (dR/dt x MV) + (R x M dV/dt) + Σ (dr’/dt x mv’) + Σd(r’x mv’)/dt) but dR/dt is V and dr’/dt is v’ so the vector products in the first and third term are each zero. And again, since the internal forces occur as equal and opposite pairs, any changes in angular momentum caused by the internal forces cancel out. Then dL/ dt = R x MA + Σ r’ x f. Thus the change in angular momentum associated with the motion of the center of mass depends on its acceleration which is caused by the vector sum of the external forces; and the change in the angular momentum about the center of mass is caused by the vector sum of the torques about the center of mass caused by the external forces. (i.e. The external torques acting about the center of mass cause an angular acceleration around the center of mass.)

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