What is the velocity of the opposing rod in a hexagon mechanics problem?

[StatusX]

Here's a mechanics problem that seems pretty straightforward, but I can't get anywhere on it. You have a system of six rods arranged in a regular hexagon, connected by hinges at each vertex, giving some non-rigid structure. Now you apply a force to one of the rods in a direction perpendicular to it, towards the center of the hexagon. This will have the effect of both adding to the total momentum of the hexagon system, as well as squashing it. The question is, if after you apply the force, the rod you were touching gets a velocity v, what is the velocity of the opposing rod? Even a hint on how to approach this would be appreciated.

 

[tiny-tim]

Hi StatusX!

Assuming the rod does not turn, how does the "front" half of the hexagon compare with the "back" half?

 

[StatusX]

I'm not sure what you're asking. Its a 2D problem, and the rod you push (call it the bottom rod) doesn't turn, it just moves upwards. Then I'm asking for the upward speed of the top rod.

 

[tiny-tim]

… the rod you push (call it the bottom rod) doesn't turn, it just moves upwards.

Yes, and so what hapopens to the other rods?

 

[StatusX]

That's part of what you need to figure out. Presumably, the bottom rod exerts some torque on them, they push the rods above them, and those push the top rod. But I'm not sure how all the forces work out.

 

[tiny-tim]

That's part of what you need to figure out. Presumably, the bottom rod exerts some torque on them, they push the rods above them, and those push the top rod. But I'm not sure how all the forces work out.

never mind the forces …

what's the geometry? ​

 

[StatusX]

Here's a picture. Assuming we apply F for a (short) time t so that the vertical velocity of the bottom rod is v, the question asks for the velocity of the top rod. To be more precise, let's assume (for now) the 6 vertices are point masses, and they're connected by massless rods. But I'm also interested in the case when the mass is uniformly distributed along the rods.

 

[Dr.D]

It looks like you can assume left-right symmetry, so you only have to analyze half the problem. Then start writing impulse and momentum relations, remembering that the reaction at each pin has two components.

 

[StatusX]

It looks like you can assume left-right symmetry, so you only have to analyze half the problem. Then start writing impulse and momentum relations, remembering that the reaction at each pin has two components.

Can you elaborate a little, maybe explain how this would work for the first joint? I know this is very simply, but I haven't done this stuff in a while and I'm getting stuck.

 

[tiny-tim]

just woke up … :zzz:

It looks like you can assume left-right symmetry …

from which the geometry tells you that there must also be front-back symmetry …

so the velocities of the front and back rods must … ?

 

[Dr.D]

I'm not quite sure what "front" and "back" refer to, but left and right refer to the two sides of the picture and the fact that they are just alike. What happens on the left must also happen on the right in exactly the opposite sense horizontally, and the same sense vertically. So if a point in the right side moves to the right and upward, the symmetric point on the left side moves to the left and upward. Thus you really only need to look at on side or the other.

 

[StatusX]

Let me be more specific with what's confusing me. For this, let's focus on a simpler system: two horizontal rods or length L connected by a frictionless hinge:

-------O-------

Now I want to know what happens if I apply an impulse I to the left end of the left rod. If there was no second rod, it would just start moving upward at speed I/m and spinning clockwise at angular velocity 6I/mL. But how does the second rod affect this? I'm not even sure what forces are present at the hinge. Can someone clear this up?

 

[rcgldr]

Now I want to know what happens if I apply an impulse I to the left end of the left rod.

Maybe this will help. For a given force, the linear reactions (f = m x a, a = f/m), are independent of any angular reactions. The upwards acceleration of the center of mass of the 2 rod system is equal to the applied force divided by the mass of the 2 rod system.

In the hexagon case, even during the period the hexagon is collapsing, the center of mass accelerates via the same rule (a = f/m).

The rest of this seems pretty complicated, somewhat similar to statics and dynamics in engineering classes.

 

[StatusX]

I know it's complicated, but the problem is I don't even know how one would do it in principle. That is, I don't care how to solve the equations, I just want to know what the equations are. Specifically, what forces act on the second rod?

 

[Phrak]

The velocities of the front and back rods are equal and opposite with respect to...

 

[StatusX]

the center of mass? Even if that is true, I don't see how it helps, because I don't know the motion of the center of mass (I'm only given the velocity of the bottom rod, not the applied force).

 

[Phrak]

the center of mass? Even if that is true, I don't see how it helps, because I don't know the motion of the center of mass (I'm only given the velocity of the bottom rod, not the applied force).

That's right, the center of mass. It helps to get all the symmetries of the problem lined-up. Maybe Tiny see's something I don't, but the problem, where the mass is at the hinge points, is easier. Otherwise, you may have to account for the angular inertia of each stick.

 

[tiny-tim]

Maybe Tiny see's something I don't …

Hi everyone!

I've been watching this thread with interest …

I think that's the best we can do on the information given …

there's no reason to believe the collision is elastic, so anther equation is needed.

We could simplify the problem by transferring to a frame in which forces of F/2 are applied on each side, but we still seem to need another equation.

 

[StatusX]

I don't think another equation is needed. Surely you can build such a hexagonal system and apply this force, and something will happen. What could be missing? And what collision are you talking about?

 

[tiny-tim]

I don't think another equation is needed. Surely you can build such a hexagonal system and apply this force, and something will happen. What could be missing? And what collision are you talking about?

The collision between the bottom rod and the two it's connected to …

if energy is lost (and it usually is), then don't we need to know how much?

 

[StatusX]

Why would you call that a collision? The rods are always touching. In any case, assume everything is ideal: no friction in the hinges, or energy transferred as heat to the rods. Do you see some problem with making these assumptions?

 

[tiny-tim]

Why would you call that a collision? The rods are always touching.

So are the balls in "Newton's cradle" (that "executive toy").

In any case, assume everything is ideal: no friction in the hinges, or energy transferred as heat to the rods.

Not even any noise?

Well, if energy is conserved, then it's just a geometry problem …

go to that frame I mentioned earlier, so the centre of mass doesn't move, and apply conservation of energy

 

[StatusX]

?


I don't know what that toy or noise has to do with anything, and I don't see a way to solve it with conservation of energy. Do you want to sketch what you have in mind?

 

[berkeman]

Would it help to first look at the problem when the top rod is fixed? Calculate the motion and velocities of the side rods as the hexagon collapses. Then do it again, with only the 5 masses (the 5 rods) opposing the motion induced by the force on the bottom rod... It looks solvable, just a number of equations (translation of 6 rods and rotation of 4).

 

[StatusX]

But again, I don't see what equations I should be using. If I knew all the forces on each rod, then it would be easy. But I don't. For example, what are all the forces on the lower left rod? Does the upper left rod apply a force to it, and in what direction? If you could even explain the simpler two rod system I metioned earlier, it would really help me.

 

[berkeman]

But again, I don't see what equations I should be using. If I knew all the forces on each rod, then it would be easy. But I don't. For example, what are all the forces on the lower left rod? Does the upper left rod apply a force to it, and in what direction? If you could even explain the simpler two rod system I metioned earlier, it would really help me.

Your two rod example doesn't have the symmetry advantage that a 3-rod or 5-rod or the 6-rod system has. So for a 3-rod system (the bottom 3), as you pust the rod up with a force in the middle, it follows a straight line up with some acceleration, and does not rotate. That moves the two hinge points up in straight lines, and that causes vertical translation and rotation outward on the 2 rods. You should be able to calculate that motion, so maybe that's a good place to start.

Then see if you can add in the rest of the rods, with the hinges constraining the motions of the ends of the rods. Hopefully you'll be able to write enough equations for the unknowns (I'm not sure that's possible, but as said before, it's a real physical system that could be built, and it would behave the same way every time you did the experiment...).

 

[Phrak]

i've got a plan, StatusX. Since I don't know the answer, and Tiny and berkeman are so confindent, let's team up .

I have a suspicion, from what berkeman said, that the top rod will stay stationary. How this squares with Tiny's kinematic idea is beyond me. (More hints Tiny!) Is it stationary?

 

[rcgldr]

This one is beyond me, but here's a suggestion.

The velocity of the bottom rod is known V_b.
The velocity of the top rod is unknown, but call it V_t

For an assumed set of values, V_b and V_t, can you then calcuate the instantaneous velocites of the hinge points (and therefore the other rods)?

Assuming you can establish a set of equation correlating movment of hinge points and other rods to V_b and V_t, can you then use these relationships to help determine V_t for a given V_b?

 

[Dr.D]

This problem has two degrees of freedom, and I have formulated the equations of motion using the Lagrangian method. The formulation is all laid out in the attached Maple worksheet. The result is a pair of coupled differential equations with no obvious closed form solution.

It would not be difficult to do a numerical simulation, but that seems to be going beyond the spirit of the original question. I have no idea how the original question was intended to be answered, but I would certainly like to hear the answer from the one who originally posed the problem.

 

[StatusX]

I can't attach the solution because its in a secure pdf (it's out of the book "A Guide to Physics Problems Part 1"), but the way they do it is similar to the way Dr. D is approaching it, using the Lagrangian method. In fact, they use the same variables as him. The key is to note that the momentum conjugate to theta is initially zero, which allows one to get a relation betweeen the initial values of \frac{dy}{dt} and \frac{d\theta}{dt}, which one can then use to find that the initial velocity of the top rod is 1/10 that of the bottom rod.

There's two problems I have. First, why is the momentum conjugate to theta equal to zero? They seem to use the argument that the initial angular momentum is zero, but I don't see a relation between the angular momentum and the momentum conjugate to theta. For example, one is conserved and one isn't. Second, although it would help if I understood the answer to the last question, I'd really like to know how to do this problem directly, using force balancing. This is why I didn't mention this solution earlier.