How Fast Must the Lid Travel to Tip Over the Entire Assembly?

[resjsu]

Hello,
See attached PDF for basic depiction of an issue I am currently working on.

If lid starts off with angle = 0 (closed) and I fling it open until it hits the "stop" (interferes with bottom causing it to stop), how fast must lid being traveling to tip over entire assembly? From this I want to see how much force I must apply to get lid traveling at specified speed.

My approach was to use conservation of angular momentum and find the lid angular momentum at the instant it reach the max opening and then insert this momentum into the momentum equation for the assembly, which would give me the velocity the assembly would experience correct?

But I cannot make the connection of how to then equate if the assembly will actually tip because in the static diagram I know the moment the weight of the assembly (counter balance) is exerting and it is in ft-lbs. But, when I calculate for angular momentum the assembly experiences it is in (lb-ft^2)/s

Any guidance at all would help.

Thank you

 

[Simon Bridge]

You need to use angular momentum and the energy stored in the motion together ... you need enough energy transferred to the vessel to get it to tip over. I'd work the problem backwards: how much energy is needed to tip the vessel? At exactly the tipping energy, the vessel will have no angular momentum about the contact point ... so what was the initial angular momentum? This is what the lid needed to impart.

You could also do it in terms of impulse ... impulse is change in momentum but allows that maybe not all the force of an impact goes to changing the moment of inertia of the impacted. Also consider if the lid sticks or bounces... and reaction forces at the hinges as it accelerates.

 

[resjsu]

Simon,
I want to keep the problem as simple as possible and would rather not consider the lid sticking or bouncing. The real world application is a heavy metal lid and base, there is not much bouncing occurring.

So I did start down this track of trying to calculate the momentum needed to tip the assembly. Please let me know if any of my steps are incorrect;
-The initial state will just be the static state. From here I can calculate the moment the center of mass of the cart exerts on the assembly, this will be the moment keeping it from tipping. From this I can calculate the force required to tip the cart. Here is what I don't understand, say I have the force needed tot ip the cart, how do I relate this to the angular momentum of the assembly? i.e. say the force is 300 pounds how does this relate to angular momentum with units (lb-ft^2)/s?

Similarly, I believe I can calculate the lids angular momentum I just don't know what to do translate this momentum to the lid.

I could post the work I've done but nothing is solid.

Thank you,
Aurelio

 

[Simon Bridge]

The lid must either bounce or stick ... if it slams into place and stays there it has stuck.
I didn't look at the attachment.

The usual force of tipping calculation is the force at the action point needed to oppose the torque from gravity.
An impact would be modeled by an impulse ... which is change in momentum. You are trying to change angular momentum.

 

[resjsu]

Simon,
Please see the attached file.

Method
First, calculate linear impulse of lid to find velocity of lid when it impacts bottom.
Next, calculate conservation of momentum to determine velocity of bottom after impact
Next, calculate linear impulse of bottom using velocity found in step 2

Where I am stuck
The linear impulse of bottom equation gives me a force calculated over time. I am assuming this force acting on the bottom will be equal tot he force the lid delivers at the point of impact;
How do I calculate the force the lid delivers at impact?
When this force is calculated how does the linear impulse equation for the bottom help me out?
How much or how fast does the angular momentum of the bottom need to change for the assembly to tip over?

I appreciate all of the help.

Thank you again
Aure