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?

 

[mikeJW]

Hi desertfox and GregLocock
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.

 

[zekeman]

Just got a chance to follow up on your new problem.
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.

 

[mikeJW]

Thanks zekeman.
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.

 

[mikeJW]

Hi folks.
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.

 

[zekeman]

Mike,

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!

 

[mikeJW]

:-D
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.