Wedge impacting on pendulum
Wedge impacting on pendulum
(OP)
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?
![[ponder] ponder](https://www.tipmaster.com/images/ponder.gif)
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?
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?
![[ponder] ponder](https://www.tipmaster.com/images/ponder.gif)
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?





RE: Wedge impacting on pendulum
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)
RE: Wedge impacting on pendulum
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.
RE: Wedge impacting on pendulum
Just returned to the problem and looked at using Vx = VyTan(Q) as you suggested.
Velocity due to impact must be higher than this. The sphere has to bounce off the wedge.
RE: Wedge impacting on pendulum
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
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.
RE: Wedge impacting on pendulum
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
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RE: Wedge impacting on pendulum
TTFN
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RE: Wedge impacting on pendulum
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
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.
RE: Wedge impacting on pendulum
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.
RE: Wedge impacting on pendulum
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
RE: Wedge impacting on pendulum
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
RE: Wedge impacting on pendulum
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
RE: Wedge impacting on pendulum
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
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.
RE: Wedge impacting on pendulum
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
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
Obviously the scale of this model is completely different to your problem.
http://greglocock.webs.com/
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: Wedge impacting on pendulum
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
RE: Wedge impacting on pendulum
Many thanks for staying with it. I am diverted onto something else for now but will be back.
GregLock we have more in common than you know.
In the 80's I was modelling 1/4 car suspension with humble computers. I wish I had todays toys. Tyres were difficult.
Every model bounced along the road like large earth movers on balloon tyres. Good luck.
RE: Wedge impacting on pendulum
The reason you are getting separation is less the impact and more because of the curvature of the cam at the point of contact. In cam dynamic, this called "separation" and occurs when the cam surface decelerates faster than the follower spring decelerates the follower.
In your case, you have a post impact behavior that is similar.
I still stand on my earlier explanation that the post impact velocity of the small sphere leaves it in virtual contact with the cam surface and your high speed camera may be showing the above separation phenomena. If you do it with a pure wedge I think you will see minimal to no separation depending on whether the cam slows down as a result of the impact.
RE: Wedge impacting on pendulum
The camera shows the motion of the sphere to be greater than the surface height of the cam (when the leaf spring has a mass on the free end). Subsequent impacts buld up!
As a golfer I am well aware of what happens when a sloping cam hits a stationary sphere. I am not very good at predicting where that sphere goes either. But the ball does not just fall off the tee-peg.
I still think it is an impact problem.
Anyone know the physics of a golf slice?
Thanks for your input so far, it's much appreciated.
RE: Wedge impacting on pendulum
Here is my take on the sphere/pendulum wedge impact.
See attached jpeg.
Fig 1 shows the relative velocity of sphere and wedge.
For convenience consider the sphere hitting the wedge not the other way round.
Fig 2 shows the freebody deflection of the sphere.
My contention is that at the instant of impact the sphere does not know it is constrained so it tries to rebound as in Fig2.
Fig3 shows the only possible direction of travel for the sphere due to constraint.
Now do I take V.CosP as the horizontal velocity or does conservation of momentum make the horizontal velocity = V?
Or is it some other velocity?
My problem in this concept is that the sphere/pendulum actually continues down at constant velocity V relative to the wedge.
The horizontal velocity to slide along the wedge is V.SinQ which is definately less than V.CosP or V so the sphere would leave the surface of the wedge.
Conservation of energy and concervation of momentum don't simply apply as there is clearly an additional horizontal velocity.
Conservation of my sanity is at risk.
Help please.
RE: Wedge impacting on pendulum
My bad,
After thinking about what I said and your comments, I took another look at my equation set,
MOMENTUM
MdVy=F1dt---F1 = vertical component of impact force
mdVx=F2dt---F2 = horizontal component impact force
ENERGY
VyMdVy=VxmdVx
Substituting
VyF1dt=VxF2dt
Vx=Vy*F1/F2
But F2/F1=tan(Q) since M>>m
Vx=Vytan(Q)
and found the error in the energy equation. I will now redo the system as follows:
MOMENTUM
M(Vyf-Vyi)=F1dt=mvf---F1 = vertical component of impact force
mvx=F2dt---F2 = horizontal component impact force
ENERGY
M(Vyf^2-Vyi^2)= M(Vyf+Vyi)(Vyf-Vyi)=mvf^2
Now dividing the two equations, the masses drop out and I get
(Vyf+Vyi)F1dt=vxF2dt
and
since Vyf is equal to Vy1 equal to V of the wedge this becomes
vx=2VF1/F2
and
vx=2Vtan(Q)
which is double the value I got before and confirms your observation of separation after impact.In fact the separation velocity is Vtan(Q) for the wedge. Of course the actual velocity would be somewhat less but is conservative.
Now I know why Tiger Woods can drive the golf ball that far!
RE: Wedge impacting on pendulum
Yes zekeman I believe you've cracked it!
It is good job my cam angles are < 45 degrees and not close to 90 degrees or some other physics would kick in.
Tigers drive and my Sunday best do hit close 80 degrees and they sound good. There's a clue there but I won't go down that road.
Thanks very much for your help.