Wedge impacting on pendulum 1

[mikeJW]

I have a massive wedge rising vertically and impacting on the side of a pendulum.
The pendulum consists of a hard sphere suspended from a rod and is constrained to only rotate about its’ fulcrum.
The wedge nose/slope has an angle Q to vertical.
What is the instantaneous velocity Vx (horizontal) of the pendulum for a vertical velocity of Vy on the wedge?

After a bit of thought I decided to consider the sphere to be falling vertically onto a stationary wedge and use the horizontal component of the rebound.
Vx = Vy.Sin(2Q)

What is the correct method for determining Vx?

 

[zekeman]

You would be right(if the pendulum were on a string) except for the fact that the pendulum rod restrains the vertical component of the ball.
I get
Vx=Vytan(Q)
Momentum equation

MdVy=F1dt---F1 = vertical component of impact force
mdVx=F2dt---F2 = horizontal component impact force

Energy equation
VyMdVy=VxmdVx
Substituting
VyF1dt=VxF2dt
Vx=Vy*F1/F2
But F2/F1=tan(Q) since M>>m
Vx=Vytan(Q)

 

[mikeJW]

:-D Many thanks zekeman.
I can now calculate the restoring force to return the pendulum to the wedge in a given time.
Incidently, on the second impact I will add on a horizontal velocity of Cr.Vx2 where Vx2 is the velocity when the pendulum returns to the new part of the wedge. Cr is the coefficient of restitution.
I expect the new Vx = VyTan(Q) + Cr.Vx2
Vx2 is solved from the intersection of the sineusoidal motion of the pendulum and the surface of the wedge.
Thank you for your prompt reply.

 

[mikeJW]

hello zekeman
Just returned to the problem and looked at using Vx = VyTan(Q) as you suggested.
:-( This can't be right! Vy.Tan(Q) is the vertical velocity to gently run up the wedge surface. No impact!!
Velocity due to impact must be higher than this. The sphere has to bounce off the wedge.

 

[GregLocock]

You have an impact problem not a statics problem. Simplify the target by calling it a spherical object constrained to move horizontally.

Now using your knowledge of the elastic properties and damping and mass of the wedge and the target you can work out the momentum transferred.

Some idea of the relative masses and so on might lead to useful approximations.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

I looked this over again and have to stand by that equation. The one error I made was a typo--which should be F1/F2=tan(Q), not the inverse, but did not affect the result. I would like to see somebody else either challenge or confirm this outcome. Maybe a simple experiment could do this.

I also noted that it doesn't look like there is separation of the wedge from the ball as you point out That doesn't mean there is no impact. It is due to the mass discrepancy of the ball and wedge.

As an example,consider a large mass striking a very small object (M>>m). A linear impact will result in the two bodies with approximately the same post impact velocity equal to the initial high mass velocity, thus no separation.

 

[GregLocock]

Um, OK. let me know if it works out for you. In certain circumstances (very slow impacts) it will. In fast ones it can't.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

"Um, OK. let me know if it works out for you. In certain circumstances (very slow impacts) it will. In fast ones it can't"
--------------------------------------------------------------
WHA?????

 

[IRstuff]

Usually, the only time objects travel together after a collision is in an inelastic collision.

TTFN

FAQ731-376

 

[GregLocock]

zekeman-the scenario you are suggesting only works where forces are always normal to the contacting surfaces, and the collision is elastic. As soon as it gets more complex, such as where the projectile speed is high compared with the speed of vibration in the solids (what I meant by high speed) , or where shear waves (friction) are present, or where energy is transformed into other than KE then the simple equation won't apply.

The reason is roughly that your KE equation is an inequality.

A good example of this is crash between cars. The momentum equation applies, but the KE equation provides a boundary to the observed behaviour.


Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

Greg,
Wow.My goodness,you are looking for wave equations and friction induced shear forces, etc, etc, and energy absorption modelling? How do you plan on solving this problem?? I good PHD problem??
While,you are technically correct, I believe for the problem at hand, a first cut conservative approach, the one presented, assumes an elastic collision (and therefore no crushing KE absorption). Also,assumed a normal force to the wedge surface, which implies the absence of friction, a valid assumption here.
So I am assuming that the wedge and the ball are reasonably solid and elastic. If the OP has a problem with that assumption, he could give more materials data.
I don't expect a 100% accurate answer, but I think it is a reasonable first approach and making it more complicated at this stage will serve no useful purpose and NO answers will be forthcoming.
Moreover, the OP can experiment with this and come to his own conclusion as to the validity of this estimate.

 

[mikeJW]

Reminder: I am trying to determine the initial horizontal velocity of the sphere. It is a collision. The bodies are hard and elastic. The sphere does leave the wedge, I have high speed video to prove it.
I saw the problem as a collision between objects.
Impulse forces, duration of collision are indeterminate and irrelevant.
Collision between free-bodies is trivial but add in constraints and the solution is not at all clear.

Thanks for your inputs but I am staying with my own estimate until someone provides a better equation.
I am not happy with my equation for the collision which is why I asked the question on Eng-Tips

For your info, the bodies are actually two cams. The sphere is a small light steel object and the wedge is a hard plastic surface on a large heavy steel object.
The small cam is constrained as described and does bounce off the large cam. I am trying to predict its motion after the collision. That means I need the initial launch velocity, nothing else.
The collision is elastic, it happens 10^10 times in the life of the machine without failure.

 

[desertfox]

Hi mikeJW

If you know the mass, centre of gravity of the pendulum and the angular displacement of the pendulum after impact you can calculate its velocity.
By knowing the centre of gravity and the angular displacement you can calculate the vertical shift in the centre of gravity of the pendulum as it swings this then can be related to its potential energy of mgh assuming no loss of energy then

mgh= 1/2 * m V^2.

Alternatively it is very hard to calculate what the velocity might be without a lot more information like moment of inertia of the pendulum, friction in pendulum bearings, coefficient of restitution etc.
I agree the pendulum will seperate from the wedge momentarily at least and may bounce several times although the naked eye might not see it.
One of the problems I see with zekemans post is that the assumption that the force components in the x and y direction are the same, I cannot see in your earlier posts any information that would lead to that conclusion.

desertfox

 

[mikeJW]

:-D Ahay now we may getting somewhere.
I like the energy balance because I have not disclosed the real situation. It gets worse!
The small sphere is actually a bend on the end of a leaf spring. The bend is an arc of a circle. The spring is anchored at the opposite end.
The wedge is a cam surface in hard plastic.
The vertical velocity of the wedge at impact is > 1metre/S.
The cam slope < 45 degrees.
The leaf spring natural frequency is > 120Hz
I expect the spring to be back in contact in < 1mS.
It is important that I get a good estimate of when/where the spring returns to the cam for different speeds and wedge angles.
Once I am able to estimate the initial 'collision' velocity at the end of the spring I can deal with flight duration and subsequent collisions and rebounds.

I can probably determine the change in kinetic energy from the incremental change in potential energy in the leaf spring.
(The leaf spring is pre-loaded against a stop to maintain its position ahead of the wedge arriving.)
Will return to the problem later.

Sorry about the lack of information earlier but I under-estimated the difficulty.
Previously the sphere was a small mass on a leaf spring. The motion was measured by high speed video.
We never tried predicting the motion ahead of having prototype parts!

mikeJW

 

[desertfox]

Hi Mike,
Now you are describing a completely different scenario from the initial one I thought you had on reading your original post. If I understand your last post correctly the leaf spring is pre-loaded up against a stop, the stop I presume is independent of the wedge?
So now when you operate the cam, the wedge rises and presumably lifts the spring off its stop and then falls back again.
Is the wedge motion rotary? What details have you got of the leaf spring ie dimensions, spring rate etc.
Finally, I think it would be helpful if you could upload a drawing or sketch of your arrangement.
Looking at earlier posts which talked about momentum equations, which assumed that no energy was destroyed, would now be negated because work done or energy is used up in compressing the spring.
Regards
desertfox

 

[GregLocock]

I'm not sure that an analytical solution is your best bet in such a complex non linear case. There again the following might be a case of if the only tool you have is a hammer, then everything looks like a nail, but a simulation such as LS DYNA, ADAMS, or Working Model, would do this without breaking a sweat.

If the natural frequency of the spring is 120 hz, and the contact time is only 1ms, then predicting what is going on will be complicated by the dynamic response of the spring -it'll have higher order modes excited by the impact.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mikeJW]

The problem expands!
I am delighted that you guys are trying to help but I can't be completely clear on what is going on, so I have attached an equivalent (well a rough equivalent).
The file 'CatchCam.jpg' has been uploaded.
The notched cam travels right to left, pulled by a heavy mechanism. The leaf spring has to drop into the notch and arrest the cam motion. It is like a limit stop.
The plastic cam breaks free from the mechanism and is held.
Sometime later the mechanism returns to re-attach to the cam and the leaf-spring is rotated to release the cam and allow left to right motion.

These are small parts. The double cam length is under 10mm and the leaf-spring is under 50mm.

I have the leaf-spring predicted frequencies for 1st and 3rd order modes from 'Mechanica' but to our knowledge none of the fancy solid modellers can deal with the dynamics of impact. So bear with me why I try to get a handle on it the old fashioned way.

 

[GregLocock]

Oddly enough at work I am currently simulating the forces on a wheel and suspension as they drive (or crash) into and out of a large pothole. That bears quite a strong resemblance to your problem.

Handcalcs will get you within a factor of 2-10. They are worth doing, but not definitive.

The simulation programs I suggested will analyse that sort of problem as well as you can define the physical properties.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[GregLocock]

Just for grins here's a quick model showing that sort of problem. The leafspring is modelled as a series of rigid, massy, links joined by torsional springs, so is capable of resonant behaviour in its own right, and is preloaded against the stop. That's how we model leafsprings for trucks in the real world. The friction and elastic properties of the contact between the green disc and the cam are definable.

Obviously the scale of this model is completely different to your problem.

http://greglocock.webs.com/

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[desertfox]

Hi mikejw

Thanks for the diagram, just thinking in my lunch break you could draw acceleration diagrams for the spring rising up the profile from which you should be able to obtain forces acting on the spring, knowing the spring rate will then tell you how much the spring deflects, just a thought, I'll try to think some more and let you know if anything else springs to mind.
You could I suppose just work out the load from the deflection on the spring from the geometry you have and ignore the dynamic effect.
If you give details of spring I could work out its stiffness.

Regards

desertfox