Force of impact of a pendulum.
Force of impact of a pendulum.
(OP)
I was asked to come up with a crude impact tester to test the failures of painted surfaces based on different impacts. Heres what I've got: basic pendulum with hammer type head; weighing 2.2lbs (with pendulum; 5.5ft) attached by hinges to a crossbar, with an angleometer at the pivot point. What I need to know is: using the angle from normal as a variable, how can I figure the force of impact given these criteria?
Thank you, Matt Booth, fts Technologies
Thank you, Matt Booth, fts Technologies





RE: Force of impact of a pendulum.
You need to know where your centre of gravity of your pendulum acts in terms of distance along the swinging arm from the pivot point.Then if you know its position or radius along the arm you can work out its vertical drop from say the horizontal. Having also established the combined mass of the arm and weight you can work out the potential energy of the pendulum as it reaches the vertical position just before impact with the sample. After impact the angle which the pendulum arm went through will be recorded on your device which you can use to see how much energy was left in the pendulum, just calculate the height the centre of gravity of your arm went through after impact and then calculate the potential energy = mass * gravity const* height reached.
Subtracting the potential energies before and after impact will give you the energy absorbed by the test piece.
I am not sure that it will be easy to calculate the actual breaking force that the pendulum delivers, however you will have the energy required to break the sample which you can use for comparison purposes.
regards desertfox
RE: Force of impact of a pendulum.
Maui
RE: Force of impact of a pendulum.
I would just like to point out that the Charpy impact tests
are based on conservation of energy using a pendulum.
Secondly whilst I agree that I have made no allowance for friction, heat dissipation etc perhaps you could explain how using the conservation of momentum avoids losses due to friction and heat dissipation and how the length of time the
moving pendulum is in contact with the speciman can be determined easily without resorting to some high tech measuring equipment?
regards desertfox
RE: Force of impact of a pendulum.
RE: Force of impact of a pendulum.
Thanks for your response.
I suggest you use the energy balance that I outlined earlier, should you need any further help just shout.
regards desertfox
RE: Force of impact of a pendulum.
The tester you describe is nearly the same used to measure bullet energy in a ballistics tests some years past. In use, a tape was attached to the pendulum and when the pendulum was struck, it pulled the tape through a retainer.
The length of travel was measured and energy computed. This may give you some direction to search. It also says your method has some validity.
Good luck.
Griffy
RE: Force of impact of a pendulum.
Desertfox, my reply was not intended to offend you. I am familiar with the Charpy impact tester and the theory behind its use. I also know that for tool steels such as airmelt H13, the scatter in the Charpy data is typically 50% or more of the actual test values at room temperature. For relatively low values on the Charpy scale, the test can result in significant scatter. Based upon your reply, you assumed that Destructotron was actually breaking his specimens during this test. But he did not state that he was doing this. He stated that he was studying the failures of painted surfaces based upon different impacts.
The force of the impact generated by the pendulum is equal to the time rate of change of the momentum, and one approach that could be used to accurately determine the net force imparted to the specimen is to determine the contact time at the point of collision. This can be accomplished by wiring a simple electrical circuit which is closed when contact is established between the specimen and the pendulum. In this circuit an inexpensive digital timer (similar to a digital stopwatch) can be included that turns on when the circuit is closed and turns off when it is open. This will provide an accurate value for the time of impact (obviously, an unpainted specimen will need to be run to perform this part of the test).
RE: Force of impact of a pendulum.
No offence taken and yes I did assume he was breaking the samples.
I would also point out that if the samples are held rigid and are not taken to failure then the laws of momentum conservation are invalid, ie the sample does not under go a change in momentum and Newtons laws of Restitution have to
be applied in that case. However if the pendulum bounces back from the sample recording the height the pendulum centre of gravity reaches the difference in intial and final
potential energies will estimate the energy absorbed by the sample.
regards desertfox
RE: Force of impact of a pendulum.
If calculating the impact force is as difficult as it seems, why not use the potential energy of the hammer as the basis for comparison of damage to the paint?
Easy to measure, easy to repeat.
Or drop ball bearings from a known height onto the painted surface.
Jeff
RE: Force of impact of a pendulum.
RE: Force of impact of a pendulum.
The units of energy are Joules
regards
desertfox
RE: Force of impact of a pendulum.
D = R-Rcos(theta)
RE: Force of impact of a pendulum.
Also, there may be standardized industry-accepted procedures already. Contact your paint vendor to find out.
http://www.EsoxRepublic.com
RE: Force of impact of a pendulum.
Maui
RE: Force of impact of a pendulum.
Its not possible to calculate the force using the energy approach.
regards desertfox
RE: Force of impact of a pendulum.
Maui
RE: Force of impact of a pendulum.
I pointed out in my first response that I thought the force
ivolved with the impact would be difficult to calculate this would be due to the small amount of time involved during the impact. I also pointed out that calculating the difference in energies was a means of comparison which also
seems to be the same thoughts as some of the other thread writers. I quote here from a text book "Mechanics For Advanced Level" by L.Bostock & S.Chandler , pub:- Nelson Thornes.
"The conservation of linear momentum is not valid in such cases for the impulse applied to the particle by the fixed surface is an external impulse; hence the momentum of the particle is changed but the momentum of the fixed object is not changed by an equal and opposite amount.
I consider the samples to be fixed objects and as such would fall into Newton's law of Restitution ie :- coefficients of restitution between 0 and 1. for calculating the speed at which the pendulum weight bounces off the sample. I think I mentioned Newton's law of Restitution in my previous post.
Maui your original post states that the energy approach is incorrect due to losses via heat, acoustic energy, friction etc perphaps you would be kind enough to explain how these
losses are avoided using your theory which I requested in an earlier post.
Regards Desertfox
RE: Force of impact of a pendulum.
losses are avoided using your theory which I requested in an earlier post."
The statement which I made in my earlier post was,"This is an inelastic collision. A conservation of energy approach will not give you the correct answer because you don't know how much of the total energy is dissipated as acoustic energy, heat, etc." In other words, by using an energy balance approach you have no way to determine the errors introduced into your calculations due to the losses that I mentioned. If these losses are small enough to be ignored, then your approach will allow the calculation of the energy absorbed by the specimen due to the impact. If they are not, then your calculations will result in errors which may lead to incorrect conclusions based upon the results of these experiments. Such energy loss mechanisms do not need to be considered in the approach which I proposed because the error in my calculations is quantifiable based solely upon the accuracy of the measurements themselves. Yours are not.
The quote above taken from,"Mechanics For Advanced Level", describes the result for the limiting case of a collision between a particle of finite mass and one which is approximated as an infinite mass. If you choose to approximate the fixtured sample as an infinite mass, then it will make the calculation of the force imparted to the sample by the pendulum even easier.
Maui
RE: Force of impact of a pendulum.
I stated:- I would also point out that if the samples are held rigid and are not taken to failure then the laws of momentum conservation are invalid, ie the sample does not under go a change in momentum and Newtons laws of Restitution have to
be applied in that case.
Your statement in the later post:- I would also point out that the laws of momentum conservation are not invalid. If that were true, then Newton's laws of motion would also be invalid.
If we are comparing what was said word for word then I said that the laws of conservation of momentum were invalid for the situation where the samples are fixed.
Further yes if we assumed the samples were fixed we could calculate the maximum force delivered to the sample assuming
that the impact was inelastic.
However your assumption of an inelastic collision is incorrect because to have an inelastic impact the pendulum and sample would have to coalesce ie there would be no seperation speed between them.
So if we have an inelastic impact in which the sample does not fail how do you measure the time of the impulse force as
your circuit would close but not open?
If the pendulum bounces off the sample then you cannot by definition have an inelastic collision. Reading the subsquent posts by Destructotron it appears to be an elastic collision which is why I stuck by the energy analysis once I realised that the samples were not being impacted to failure.
So for the record then you beleive that the conservation of linear momentum is valid in this case and that we have an inelastic collision then, explain what the velocity is of the coalesced sample and pendulum just after impact?
If you accept there is seperation between pendulum and sample after impact then what is the momentum of the sample
after this occurence as according to your theory it must under go an equal and opposite reaction therefore its velocity cannot be zero.
regards Desertfox