Bounce Back after Impact

[Tonji]

I need help calculating the amount a rotating output shaft will bounce back after hitting a hard stop. The inertia of the rotating element is 4.85 E-04 oz-in-sec^2. This inertia is rotating at a speed of 2300 rad/sec. Between the load inertia and the hard stop is a torsion spring that has an overall spring constant of 7.3E6 oz-in/rad. The spring constant of the hard stop is 11.3E9 oz-in/rad and it appears to be structurally capable of withstanding the impact. Once the impact occurs, there is no damping or friction present in the system to limit the bounce back.

 

[zekeman]

I would say that you don't have enough info on the system as described to get a realistic answer. You need data on the energy loss at the point of impact which is precisely the question asked (paraphrased).
Your best bet is to do it empirically with test data.
It is insufficient to say that there is no loss of energy after impact. The problem is what is the loss during impact.

 

[IRstuff]

If there is no damping or friction, what prevents the inertia from simply spinning in the opposite direction after impact?

TTFN

 

[Tonji]

Thanks for the comments.

I made a mistake in my original question. We measured about 4 in-oz of friction in the system. In the past, we have analyzed impacts like this by setting the KE of the rotating element equal to the PE of the "hard stop" (usually made up of springs). This allows you to define a suitable spring constant for the "hard stop" so that all the energy can be absorbed over a specific amount of spring deflection. In most cases we use brakes to stop the rotating element just after the impact occurs. This limits the "bounce back". In this case, the rotating element is free to rebound.

 

[zekeman]

It will reverse but since the loss of energy during impact is not known you cannot determine the final rotational velocity.

 

[zekeman]

Even though I'm puzzled over your knowledge of the 4 in-oz of energy loss, if it is so, then the KE difference between preimpact and postimpact energy is then 4 in oz.That equation should yield your answer.

 

[IRstuff]

Since you stipulated no other dissipation devices other than friction, the kinetic energy of the post impact rotating element will only be degraded by the amount of frictional work consumed during the impact.

That means that the 2*compression*friction is the only significant energy loss mechanism

TTFN

 

[Cockroach]

Textbook, Mechanics of Machines, Beers & Johnson, look under dynamic applications for "impact", "coefficient of restitution" that sort of thing.

This is an extremely documented problem in first year mechanics textbooks.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[GregLocock]

zekeamn, I think he's talking about a torsional friction, not an energy loss.

Anyway, the easiest thing to do is to set up a numerical solution in Excel.



Cheers

Greg Locock