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Let's say we have a metal rod of consistent density sitting flat on a frictionless surface. I intuitively understand that if I push one of its ends away from me, (at a right angle to the length of the rod) it will spin on the surface.

I'm asking if anyone can explain in simple (thought experiment like) terms why the rod rotates instead of simply moving directly away from my push and not rotating. Pictures would be great! enter image description here

I've looked at What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass?, and If I push or hit an object in space will it rotate or move along a straight line? and Farcher's answer to the second question comes as close as anything to explaining the phenomenon the way I'm requesting, but I still don't entirely understand.

Like I said, I understand intuitively what will happen, but when I try to picture the individual particles of the rod I can't understand why the rod rotates instead of just moving off as a whole away from my impulse. If I cause part of the rod to move up, shouldn't all of the rod follow, just like if I grab the rod at its edge and lift it up? Obviously that's not what actually happens though. I know it must have something to do with torque, leverage, etc. but I still can't explain it.

I hope this wouldn't be considered a duplicate question since the main difference is I'm asking for a layman's explanation, not a mathematical proof. If it is the case that it can't be explained without a rigorous mathematical proof that's fine, but in that case sorry for the duplicate!

Thanks for any assistance!

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    $\begingroup$ The "explanation" is that the force is not acting on the center of mass (or at a point in line with it) and hence causes a torque, but you need to understand that "the center of mass" and "torque" are introduced exactly because of the observation that a body, when accelerated by an external force, will also undergo a rotation unless that force is acting in this particular way. In physics explanations are secondary to observations. What we do is to accept that something happens the way it happens and then we come up with a description that matches the phenomenon. $\endgroup$ – CuriousOne Jun 10 '16 at 20:11
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    $\begingroup$ @CuriousOne Anything in newtonian dynamics should be explainable with forces and the three newton laws, the rest (torque, energy conservation, angular momentum, etc, are just convenient shortcuts) $\endgroup$ – user83548 Jun 10 '16 at 20:29
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    $\begingroup$ I looked at some of the answers in your links and I am amazed how many false answers one can find to this problem. As a major hint: the acceleration is not the same for different parts of the body! Even in a freely rotating body (i.e. without any external forces) the mass elements along the instantaneous axis of rotation are not being accelerated, but every other part of it is! Please be aware that in general the instantaneous axis of rotation will change with time (often enough in a chaotic way). Any solution that doesn't talk about different accelerations is false. $\endgroup$ – CuriousOne Jun 10 '16 at 20:30
  • $\begingroup$ @brucesmitherson: And it is, just not the way many folks are trying to explain this. In an extended body not all mass elements have the same acceleration and we need to have a set of constraining equations that define the shape of the rigid body. The resulting dynamics is rich and complicated and most high school textbooks are walking a very fine line trying to teach some of the basics while shielding the students from the reality that rigid body rotation is, in general, chaotic. $\endgroup$ – CuriousOne Jun 10 '16 at 20:33
  • $\begingroup$ @CuriousOne : Your comments are again sounding very much like an answer. So why don't you submit a definitive answer instead of all these extended comments? $\endgroup$ – sammy gerbil Jun 10 '16 at 21:12
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enter image description hereIt is easier to explain in terms of torque and angular momentum, but it can also be seen with some difficulty using only internal forces. Imagine that instead of a continuous rod you have instead a bunch of particles attached by cords. If you push the mass at the edge this one will tend to move up first, but due to the cord will also try to rotate around the second particle. In addition, this second particle will also be pushed up by the first particle and move up, but then the third particle will do the same to the first two, it will make them rotate around it, as they push it up. Go in sequence until yo reach the last particle and you will have a rotating bunch. It will likely not move rigidly as it is not a rigid body. Now replace the cords by superstrong and small springs and you get something very close to a rotating rigid rod.

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  • $\begingroup$ The kicker is that not all parts of the body are accelerating the same. This, by the way, is a pretty good place to see why the particle approximation causes more harm than good when taken too far. Instead of accepting that it is an approximation of extended body dynamics, we are trying to pretend that it is natural to "synthesize" extended bodies from particles. Bad idea. $\endgroup$ – CuriousOne Jun 10 '16 at 20:56
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    $\begingroup$ but aren't bodies really made of particles? plus it is an explanation in layman's terms, I am sure you can come with a much better one (no irony intended). $\endgroup$ – user83548 Jun 10 '16 at 20:59
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    $\begingroup$ Real bodies are not made of particles. They are made of atoms and molecules. Again, a "particle" is the name of an approximation, it's not a "thing". Newton's laws do not requires particles. They do require that one distinguishes carefully between the case where a body accelerated homogeneously vs. where different parts of a body accelerate differently. $\endgroup$ – CuriousOne Jun 10 '16 at 21:23
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    $\begingroup$ Unlike the rigid rod the particles on the left will not move backwards below the initial line, because the cords can only exert tension not compression. But this changes when the cords are replaced by stiff springs. $\endgroup$ – sammy gerbil Jun 10 '16 at 21:55
  • $\begingroup$ So in this model would we say the rotation is a result of the initial impulse taking time to propagate through the whole object? $\endgroup$ – cyclingLinguist Jun 18 '16 at 15:29
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Consider the midpoint of the rod that divides the rod into a left half and a right half. We are applying a force to the right end in the upward direction. Now consider the left half. It has to (overall) go in the direction of the force applied - the rod doesn't break, right? So if we think of just the left half then there must be some net force on it in the upward direction. If you just consider the left half then the only force that can accelerate it has to come from its 'joint' with the rest of the rod - which is at the mid-point. So we must conclude that the right half exerts an upward force on the left half with the point of the application being the midpoint.

Now consider the right half. Since it applies a force on the left half in the upward direction, according to Newton's third law, the left half must apply a downward force (of equal magnitude) on the right half - with the point of the application being the midpoint. So when the right half is thought isolated, the only external forces on it are our upward force on its right end + the downward force (by the left half of the rod) on its left end. Now one can easily visualize that the right part must have some rotation.

You can go on dividing the left part into further smaller and smaller half and conclude that every right part will have to rotate in a similar manner. So one can roughly visualize that the motion has to be what it comes out to be. The overall motion is, of course, described precisely by the calculations based on Newton's laws.

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I think the confusion is that a single push at the corner doesn't "feel" like you're "doing a rotation". Here's an explanation of why it must be, by symmetry.

  • Suppose that pushing it up on the right end, as shown in your diagram, gives the rod angular velocity $\omega$ and linear velocity $v$. (We have not assumed $\omega \neq 0$, we're going to show that.)
  • By symmetry, pushing it down on the left end gives it angular velocity $\omega$ and linear velocity $-v$.
  • Then if you do both at once, by the linearity of Newton's laws, the final angular velocity is $2 \omega$ and the final linear velocity is $0$.

But doing both at once is literally taking the rod in your two hands and turning it, so $2 \omega \neq 0$, and thus $\omega \neq 0$.

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It is not clear what your problem is. You say that you understand intuitively what will happen, and that Farcher's answer comes close to explaining it for you. What exactly do you still not understand?

You ask about picturing the individual particles of the rod. Lucas and Bruce address that. But you also say

If I cause part of the rod to move up, shouldn't all of the rod follow, just like if I grab the rod at its edge and lift it up?

I think there is a misunderstanding here.

The difference when you 'lift' one end of the rod (keeping it horizontal) is that you are actually applying a torque as well as a force. This torque counter-acts the torque which appears in Farcher's explanation. If the rod is heavy (ie there is a gravity force as well as the forces you apply) you may actually have to lift with one hand and pull down with the other : you may not have the strength to apply enough torque with one hand. But some torque is still needed even without gravity. You cannot 'lift it up' only at the edge, because you cannot apply enough torque by holding only a narrow portion at the edge.

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Any rigid body is composed of particles bonded together. This bonding allows forces applied on one part to transferred to all other parts of the body. Thus when a net force is applied, all the particles will translate together at the same time. We describe this motion is a translation of the center of mass.

In addition, the bonded particles try to maintain constant distance with each other resulting in internal forces resisting any stretching of the bonds. The component of these forces perpendicular to the motion cause centripetal accelerations about the center of mass. This results in a rotating motion.

The overall effect is described with these two laws of motion

  1. The net force acting on the entire rigid body motivates the center of mass only to move linearly.
  2. The net torque about the center of mass motivates the entire rigid body to rotate about the center of mass.
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  • $\begingroup$ This is what I was sort of picturing when I was trying to think it out. Can you explain a little more (or show with a picture) how the centripetal component comes about? $\endgroup$ – cyclingLinguist Jun 12 '16 at 16:02
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Assume that the rod is made by some particles (or molecules).

enter image description here

The force $F$ is acting on one particle of the rod as you see in the figure above.

If we isolate that particle, we will have figure below (free body diagram):

enter image description here

Force $f_1$ is applied by the particle next to the first particle. As you see in the figure, there is a distance between $F$ and $f$ (although this distance is very small in fact). Thereby, the first particle will rotate counter clockwise.

If we isolate next particle:

enter image description here

There is a distance between two forces again and this particle will rotate too.

You can extend this to whole particles of the rod.


Update:

I should confess that my answer isn’t completely correct. Because this is not an answer. This is a simple explanation for helping to better understanding. When someone asks me about principles, I usually talk about it for a while to make an insight for that person. There is no answer for reason of principles (if you know what is a principle, you certainly confirm what I am saying). We cannot prove them, we can accept them or not. So, the answer to “why” for a principle is “because”.

But, if someone asks us about a principle and we say him/her “because”; we haven’t helped to him/her. I think we should talk with him/her about that principle and express some examples in real life or some thought experiments and etc. for clarification. I think we should try to help as much as we can. In current question, OP is asking about “why $\Sigma \vec F=m\vec a_G$ and $\Sigma M_G=I_G\alpha$ ($G$ is center of mass) are equations of motion of a rigid body in plane motion?” There is no answer for this question except “we have accepted them”.

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  • $\begingroup$ You can't discuss rigid body motion with a force diagram, alone. Not all particles are accelerating at the same acceleration. $\endgroup$ – CuriousOne Jun 10 '16 at 20:23
  • $\begingroup$ @CuriousOne I didn't discuss about rigid body motion. I discussed about particle motion. Furthermore, I wanted to give a simple explanation for OP that can imagine better. $\endgroup$ – lucas Jun 10 '16 at 20:29
  • $\begingroup$ Your explanation is wrong, though, because you are leaving out the essential bit. Please take a look at rigid body motion and how this is really derived. $\endgroup$ – CuriousOne Jun 10 '16 at 20:34
  • $\begingroup$ You are, by the way, completely forgetting about the force component in the rod. If there is no force in the rod, why does it matter? $\endgroup$ – CuriousOne Jun 10 '16 at 22:18
  • $\begingroup$ Sorry, when I apply force to a "particle," it does not apply force in the opposite direction to an adjacent particle. And if it did, that particle would be applying force in the original direction to the next particle. $\endgroup$ – WGroleau Jun 11 '16 at 9:40
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Yes, it rotates but it's center of mass (i.e. The full rod) also translates. The force causes ma on the center of mass, and rotation around the center of mass from torque = I x Omega. You need both equations to get the motion.

This is physics 101

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protected by Qmechanic Jun 11 '16 at 12:37

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