What Pin Position Ensures Rigid Body Rotation Upon Impact?

[TheFerruccio]

Homework Statement

Two horizontal bars connected by a frictionless pin are released and allowed to fall and impact a fixed pin. Where must the pin impact the bars to ensure rigid body rotation?

Homework Equations

Conservation of angular and linear momentum, relative velocity equation.

The Attempt at a Solution

Pin location: x
Initial system velocity: v
Left rotation: \omega_1
Right rotation: \omega_2
Left velocity: v_1
Right velocity: v_2

Coefficient of restitution was not given, so I assumed a plastic impact, whereby it impacts the pin and rotates with the surface resting against the pin (so, restitution coef = 0). I also don't think gravity matters in this problem, since the impulse force is orders of magnitude greater than the force due to gravity. Meaning: This system could happen independent of gravity, and the pin should be located in the same place.

Angular momentum about the pin is conserved. I can also use kinematics to relate v_1 and v_2 to \omega_1 and \omega_2.

So, my final equation (initial angular momentum about the pin = final angular momentum about the pin) has three unknowns: \omega_1 \omega_2 and x.

I know that, for rigid body behavior, the two \omega terms will be equal, eliminating one unknown. However, I still have two left.

My question is this: Other than angular momentum being conserved about the pin, what other equation do I use to solve for the remaining unknown terms?

 

[TheFerruccio]

Anything else I should add to this post? I have the equations written out on paper, but I think I was pretty explicit in what unknowns I needed to continue to solve for.

Basically, the rotations of both the left and the right bar must be the same, and we're asked to find where the two-bar system should impact the pin so the resultant rotation of the system is homogeneous.

 

[AlephZero]

The thing you haven't mentioned is the forces acting on the pin.

I would be inclined to do this a slightly different way. First assume the bar is rigid, apply an impulse to it at the pivot position, and find its linear and angular acceleration.

Then cut the bar into two pieces, and find the forces and moments at pin position which give the same accelerations on each piece. You are looking for the solution where the moment at the pin position is zero.

 

[TheFerruccio]

Interesting. Why would this not work using a momentum approach? Is it because there aren't enough unknowns? What would be missing in the analysis if I were to continue with a conservation of momentum approach? I have seen it solved using an impulse method, though I have seen lots of other problems solved using a momentum approach.