Acceleration of falling object on impact

[sunny0302]

I have been out of school for a few years so my physics is rusty. I need help figuring something out for my job. I need to know how high I must drop an object so that it hits the ground with an impact of 1000g's. The object is a cylindar and weighs 2lbs.

I have been confusing myself. I know that the object will fall with 1g acceleration. I think need an equation that allows me to find the instantaneous acceleration at the moment of impact. I think I can assume an infinitely hard ground for now. I have found that instantaneous acceleration is delta v/delta t where t->0.

~~~~
If a=9806 m/s^2 then as t-->0, the velocity would be 9806 m/s?

So then I think I need to use another equation to back out the height

I found this equation for free-fall with no air resistance but I'm not sure it helps.
y(t)=-1/2*g*t^2 + v_o*t + y_o

I found this equation for impact vel just before impact = sqrt(2*g*h). But if I set it equal to 9806 I get h=4,903,000 meters...that seems very high.

I am also worried when the equations don't take mass into account.

~~~~~

Another option I thought of:
F=ma
F=(.907 kg*9806m/s^2)
F=8894N

And then I would need some sort of equation that let's me back height out of Newtons.

Can anyone help me? I'm going in circles.

 

[Fightfish]

If a=9806 m/s^2 then as t-->0, the velocity would be 9806 m/s?

No. For any velocity, if the object comes to rest in a time t->0, the acceleration of the object tends to infinity!
$$<a> = \frac{\Delta v}{\Delta t} = \frac{v}{\Delta t}$$
So, unless you can estimate the duration of impact, it is difficult to find out the velocity required to produce that acceleration. It is very likely that the velocity required is much less than 9806m/s, since the object is likely to stop in much less than a second.

I found this equation for impact vel just before impact = sqrt(2*g*h). But if I set it equal to 9806 I get h=4,903,000 meters...that seems very high.

I am also worried when the equations don't take mass into account.

This equation comes from conservation of energy.
$$\Delta Potential Energy = \Delta Kinetic Energy$$ (magnitude-wise)
The mass actually cancels out, so no worries there.

$$m g h = \frac{1}{2} m v^{2}$$
$$v = \sqrt{2gh}$$​

 

[sunny0302]

No. For any velocity, if the object comes to rest in a time t->0, the acceleration of the object tends to infinity!
$$<a> = \frac{\Delta v}{\Delta t} = \frac{v}{\Delta t}$$
So, unless you can estimate the duration of impact, it is difficult to find out the velocity required to produce that acceleration. It is very likely that the velocity required is much less than 9806m/s, since the object is likely to stop in much less than a second.


This equation comes from conservation of energy.
$$\Delta Potential Energy = \Delta Kinetic Energy$$ (magnitude-wise)
The mass actually cancels out, so no worries there.

$$m g h = \frac{1}{2} m v^{2}$$
$$v = \sqrt{2gh}$$​

Ok, so if I assume it stops in .01 s, the velocity changes to 98.06s.
The using the second equation I would get h=490.3 meters.

Is there any good way/equation to estimate how long it would take to stop so I can get a better height estimate?

 

[Bob S]

If an object falls from a height h, and stops in a distance x, the deceleraton a is gh/x meters per second². If h is 10 meters, and x = 1 cm, then a = 1000 g's.
Bob S

 

[sunny0302]

If an object falls from a height h, and stops in a distance x, the deceleraton a is gh/x meters per second². If h is 10 meters, and x = 1 cm, then a = 1000 g's.
Bob S

Well, that seems simple enough. Thanks!

 

[Lsos]

It seems simple, but figuring out the "x" isn't. That x is NOT something you can just approximate off the top of your head, either. You must do some research to figure this out. Also, keep in mind that the "x" will change as the drop height changes.

Anyway, if you're just going to approximate x, you might as well skip all the math and physics and this entire thread, and simply guess the height. It saves a lot of trouble and calculation, and the end result is pretty much the same...

 

[sunny0302]

It seems simple, but figuring out the "x" isn't. That x is NOT something you can just approximate off the top of your head, either. You must do some research to figure this out. Also, keep in mind that the "x" will change as the drop height changes.

Anyway, if you're just going to approximate x, you might as well skip all the math and physics and this entire thread, and simply guess the height. It saves a lot of trouble and calculation, and the end result is pretty much the same...

Assuming I know the weight, cross-sectional area and material of the object, are there any equations that can help me figure x out? Are there standard "give" estimated values for the ground. For instance, 40N of force makes the ground give .1 cm?

 

[Molydood]

calculating the length/duration of the deceleration zone is tricky, it would be much easier to measure it if you have access to accelerometers, otherwise you have to think about orientation of the object under impact, the stiffness and geometry of the object, dynamics of crumple zone etc etc…. it could get pretty complex pretty quick

 

[Lsos]

Yeah, exactly. It could, and should get VERY complex VERY quickly. There's so many variables to consider, the properties of the ground, the object, etc. I think it's something that can only be approximated sufficiently with experiment. And, actually, that doesn't seem very difficult. Just take the cylinder and drop it, and measure how big of a dent it made.

Depending on the variables though, it might make no dent at all. In which case you should look at better equipment...accelerometer, high speed camera, whatever else there might be. Hell, you could probably even figure out how to use an egg to help you get the results.

Ultimately the sophistication of the measuring equipment comes down to how much your job really cares about getting accurate results. If they say that they want you to figure it out only using google and your imagination, tell them to be prepared for an error factor of a few hundred percent at best, imo.

 

[gmax137]

I have been out of school for a few years so my physics is rusty. I need help figuring something out for my job. I need to know how high I must drop an object so that it hits the ground with an impact of 1000g's.

As seen in the previous responses, this is a very complicated question. Which to me, implies that the wrong question may have been asked. You might want to discuss this with whoever posed the question, and find out from them what it really is they want to do. In other words, I doubt that what they really want to do is "hit the ground with an impact of 1000g's" since that is a pretty poorly defined request.

 

[sunny0302]

Yeah, exactly. It could, and should get VERY complex VERY quickly. There's so many variables to consider, the properties of the ground, the object, etc. I think it's something that can only be approximated sufficiently with experiment. And, actually, that doesn't seem very difficult. Just take the cylinder and drop it, and measure how big of a dent it made.

Depending on the variables though, it might make no dent at all. In which case you should look at better equipment...accelerometer, high speed camera, whatever else there might be. Hell, you could probably even figure out how to use an egg to help you get the results.

Ultimately the sophistication of the measuring equipment comes down to how much your job really cares about getting accurate results. If they say that they want you to figure it out only using google and your imagination, tell them to be prepared for an error factor of a few hundred percent at best, imo.

Thanks so much for your input. There is another way to test the shock. There is a machine but it costs $ so I think they wanted to see if they could just drop it and get about the same results. It looks like, unless they are willing to drop it many times from different heights and then measure the deformity of the ground, it can't really be done.

 

[sunny0302]

As seen in the previous responses, this is a very complicated question. Which to me, implies that the wrong question may have been asked. You might want to discuss this with whoever posed the question, and find out from them what it really is they want to do. In other words, I doubt that what they really want to do is "hit the ground with an impact of 1000g's" since that is a pretty poorly defined request.

Thank you so much for your advice. I was feeling pretty dumb struggling with this question and I'm glad to know at least that it is complicated.

 

[GT1]

If an object falls from a height h, and stops in a distance x, the deceleraton a is gh/x meters per second². If h is 10 meters, and x = 1 cm, then a = 1000 g's.
Bob S

Can you explain how did you get this relation- a=g*h/x ?

 

[Molydood]

I think you are talking about drop testing on a rig as opposed to the sort you might use for key fobs for example. Drop rigs are used for components that are fitted to other assemblies and usually at specific accelerations in each axis and fixing the test sample in place with fixtures representative of those in the assembly. Go on, stop us guessing, tell us what this is and what testing you are wanting to carry out and why, it will give you more focussed feedback, and stop the answers heading in the wrong direction.

 

[Daina-MDL]

I have been out of school for a few years so my physics is rusty. I need help figuring something out for my job. I need to know how high I must drop an object so that it hits the ground with an impact of 1000g's. The object is a cylindar and weighs 2lbs.

I have been confusing myself. I know that the object will fall with 1g acceleration. I think need an equation that allows me to find the instantaneous acceleration at the moment of impact. I think I can assume an infinitely hard ground for now. I have found that instantaneous acceleration is delta v/delta t where t->0.

~~~~
If a=9806 m/s^2 then as t-->0, the velocity would be 9806 m/s?

So then I think I need to use another equation to back out the height

I found this equation for free-fall with no air resistance but I'm not sure it helps.
y(t)=-1/2*g*t^2 + v_o*t + y_o

I found this equation for impact vel just before impact = sqrt(2*g*h). But if I set it equal to 9806 I get h=4,903,000 meters...that seems very high.

I am also worried when the equations don't take mass into account.

~~~~~

Another option I thought of:
F=ma
F=(.907 kg*9806m/s^2)
F=8894N

And then I would need some sort of equation that let's me back height out of Newtons.

Can anyone help me? I'm going in circles.

If you don't mind my asking... I am a physics student and i have never heard of impact being measured in g. (I may be wrong) I'm going to run the scenario through my professor and get back to you on that.