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final velocity calculation
5

final velocity calculation

final velocity calculation

(OP)
Hi, I am trying to work out impact velocity, im tryng to use v^2=u+2*as

I know what my distance (s) is
I know what my inital velocity (u) is

But i dont know my acceleration
And I dont know my time

I was trying to use f=mg (i know what the mass is) then use a=f/m - but obviously end up with 9.81 as there just the same equation re arranged

any ideas, been racking my brain on this one for a few hours

cheers

RE: final velocity calculation

(OP)
no I just didnt think it was that relevent, sorry this is it:
I am trying to work out the impact force of a rock falling and hitting a surfce, dont need to worry about it bouncing or breaking up etc, just the basic force.

I was thinking KE/distance, (which i need velocity for the kinetc energy)
Another equation ive been given is mass x impact velocity - not sure baout that but either way im after the final/impact velocity.

Cheers

RE: final velocity calculation

You said you don't know your acceleration. But later you said the rock is "falling". If that is a free fall under gravity then you know your acceleration.

RE: final velocity calculation

It's pretty simple to neglect air resistance and figure the speed of the rock when it impacts. But you can't deduce an impact force from that. The rock is going to decelerate over some finite time interval and distance, which you don't know. You would have to consider deflection in the surface being impacted as well as in the rock.

Depending on what you're trying to do, it may be possible to work with the impact energy instead of actually calculating a force.

RE: final velocity calculation

a falling rock is accelerating at g, certainly. this means you know the speed of the rock as it starts to impact the ground.

but the impact force depends on the decelleration of the rock. this depends of the time of the impact event ... how long is the initial momentum reacted ? (think impuse).

there are numerous threads on impacts, very numerous, that'll come back to the same issue ... the time that the impact takes. one approach that may (or may not) help is to equate the strain energy of the surface (as it deflects, absorbing the impact) to the initial KE.

RE: final velocity calculation

impulse = change in momentum
F*t = m*v where v is velocity just before impact
As was said, you need some idea of time t duration of impact.

Ted

RE: final velocity calculation

(OP)
Thanks for the replies, I am after only after a rough answer.

Jboggs, yes the rock is simply falling, so does that mean I can assume the acceleration is 9.81?

Cheers

RE: final velocity calculation

(OP)
Unfortunately I don't know time, the rock is 32kg falling a distance of about 2.5m, if that helps at all?

Cheers

RE: final velocity calculation

sorry, student posts are not allowed

RE: final velocity calculation

(OP)
Dont worry, I am not a student. this is for a actual problem, the force i am trying to calculate is part of a much bigger equation which will lead to me buying components and actually building something, hence the reason im only after a rough answer because its not massivly significant

RE: final velocity calculation

apologies, just the very basic questions being asked. i assume you're designing the impact surface.

you can calc the impact velocity, yes?

review the numerous threads relating to impact force.

you can't determine the impact force unitll you determine the impact time.
you can assume a time, 0.01sec.
you can assume that the surface deflects.

the best method is to model the impact (or test the surface).

RE: final velocity calculation

Velocity of a falling object is v^2 = 2*g*h where initial velocity is zero. If the initial velocity is non-zero, then add it to v to calculate final velocity. You are on the correct path with your velocity calculation.

Ted

RE: final velocity calculation

The acceleration due to gravity is not an assumption, neglecting air resistance is an assumption. For a 32kg rock falling 2.5m, I'd say a very safe one.

RE: final velocity calculation

11xminurti3,

Impacts are solved by energy methods. You are working out energy the hard way.

To lift a 32kg rock 2.5m...

F=32kg×9.81m/s2 = 314N

Energy = 314N×2.5m = 785N.m

Unless you have an initial velocity of something other than zero, this is the energy you will dissipate when you hit the ground.

You still need to know your stopping distance. Use work energy to work out force and stress.

--
JHG

RE: final velocity calculation

If you have an inelastic impact, work done is force times distance to stop. For example drop rock onto soil, measure the distance penetrated. work = energy at impact. F(average force) = 0.5*m*(v^2)/d d=depth of penetration neglected change in potential energy from instant of impact to final penetration depth, W*d

Ted

RE: final velocity calculation

thining about this, i've come back to the 1st reply ... why do you want to know the impact force ? what are you going to do with it ?

are you designing the surface (and want a number to put into your calcs) ?

is this a new spec requirement that you (and your company) aren't familiar with ?

you could look at the plate ? how stiff is it ? (use Roark for displacement due to a load, hence stiffness) now it's an easy calc to compare the strain energy of the spring with the initial KE of the rock ...

RE: final velocity calculation

(OP)
Thanks for the replys guys, quick question can I just do this to get a rough answer?

v^2 = 2*g*h = 7m/s

KE= 1/2*m*v2 = 784j

Force at impact = KE/distance = 318N - seems very low, I know alot of you are saying I should be working out energy etc, but does this work for a rough answer?

Cheers

RE: final velocity calculation

No, that's been the point. There have been at least a dozen similar posts from people like you attempting to calculate force without knowing the time or the deformation or the material. Your impact time or material deformation is what determines the actual force.

TTFN
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RE: final velocity calculation

You cannot make an assumption of how far the rock moves during its deceleration from hitting speed to 0 speed so you cannot get a meaningful answer to what you are asking.

If you hit a cement floor, the displacement for the decel is probably close to 0.01mm

If you hit a rubber floor, the displacement for the decel may be 10mm.

This is a variation of 1,000 - so you can see you have not defined the stopping distance enough to get any meaningful 'ballpark' answer.

You appear to be trying to figure out how many G you must design your widget to withstand if dropped from a certain distance. The answer as you can see will be anything from say 1G to 1,000G. You will not get there from here with the info you have shared.

RE: final velocity calculation

Pick two impact duration times: 0.010sec, 0.005sec

m*v/t = 32kg*7m/s/0.01sec =

m*v/t = 32kg*7m/s/0.005 =

Ted

RE: final velocity calculation

11xminurti3,

Forget the math for a minute and try this thought on. Impact is a two-way street. The moving object applies a force to the static one, and the static one applies an equal and opposite reaction force to the falling object.

If I drop an egg from 6 feet onto my new Tempurpedic mattress, it doesn't break. The soft, conforming, yet firm mattress gives the egg a good distance to decelerate, somewhere between 5 and 10 mm from my eyeball estimate of the dent it made. The force was very low.

If I drop an egg from 6 feet onto my garage floor, I have a mess to clean up. The concrete made the poor little egg decelerate to zero in almost-zero distance. The force was high.

Those are two cases in which the speed and energy of the object (egg) are the same. In the high-dollar mattress case, the resulting impact force was not enough to even break the yolk inside the egg, much less its shell. In the garage floor case, the resulting impact force was plenty to smash the egg into about a 1-meter diameter slippery mess.

There are other examples you could try with your fine bed and a garage floor to demonstrate the reaction force that the bombarded surface provides.

How high will a tennis ball bounce off of a superb mattress versus a slab of oil-stained concrete?

How much will you be spending on a new Nikon camera when you drop it on the concrete versus the top of a magnificently-engineered piece of sleeping equipment?

How hard will I be thrown into the steering wheel if I'm going 100 MPH and stop in a few feet or so when I hit a huge tree? That was a component of the impact force that moved me. If I crash into a rubber band instead, and it stretches half a mile before it brings me to a stop, I'll have virtually no force at all applied to my body. Both driving incidents were in the same car at the same speed.

In each case, you need to know the deceleration distance to determine the force. I hope I didn't come off as too sarcastic, but you have been given the same information in every prior post.

By the way, I'm not a Tempurpedic salesman. However, after my wife wore me down enough, I now own one. It's the best thing that ever happened to my back and my sleep quality. bigglasses

Best to you,

Goober Dave

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RE: final velocity calculation

OK. Use work = energy F*d = 0.5*m*v^2

F = KE/d d is the distance the rock or egg travels after it impacts the surface Increasing d decreases F, the force due to impact. d is NOT the distance the rock or egg falls to impact.

Ted

RE: final velocity calculation

Peak forces during a collision can be large, especially with rigid objects, but it actually takes work to yield or break something. If you are dealing with a structure that can in any way be treated as a spring, say a diving board, you can relate kinetic energy to structural displacement.

Say I drop a bowling ball 5 m onto a coil spring from a car, what is the peak force developed by the spring?

Bowling ball = 5kg
h= 5m
KE= mgh = 250J

Spring
k = 25000 N/m
F = kx

Spring energy storage:
E = kx2/2
x = (2E/k)1/2
x = (500/25000)1/2
x = .14m

F = .14m x 25000 N/m = 3500 N

Knowledge of the spring design, (wire dia, coil dia, steel yeild stress) will tell you if the spring will yield at the peak deflection or not.

Find your physics book and work the kinematics back and forth. I would dig for my copy of Shigley & Mischke, Machine design as well. You need it to be transparent to you.

RE: final velocity calculation

Quote (11xminurti3)

v^2 = 2*g*h = 7m/s
please tell your boss you're not up for this task and need someone else to do it. Before anyone gets hurt.

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

RE: final velocity calculation

Walterke, good catch. We all missed it.

Ted

RE: final velocity calculation

(OP)
Walterke, please explane what is wrong that equation.

I appreciate the advice, the rock is falling on a solid surface, the significance of the answer I am after is very minor, a best geuss would do to be honest but I did want something a little better than that

RE: final velocity calculation

"Force at impact = KE/distance = 318N" ... what on earth distance did you use to derive this (i suspect the 2.5m height) ?

i ask you, did the impact take place over this distance ? (of course not)

have you looked at the othe threads about impacts ?

have even lokked at wiki for impacts ?

have you done anything more than post here ?

sorry (but not really) ... we've tried to guide you but you haven't been able to follow.

i 2nd walter's suggestion, or hit the books, research this yourself, and then you'll be able to follow our leads.

RE: final velocity calculation

Nevermind 11xminurti3. The calculation is correct. Just the way you're writing it down is a bit unorthodox imho.

However, if it takes you 'a few hours' to figure out the acceleration of a falling object I really don't think you should be the one doing these calculations.

What kind of educational background do you have?

P.S.: I really don't mean to be rude, it's just the the next part of your calculation will be a LOT more complicated then this is, and if you don't have the proper knowledge on the subject, I'm worried about the results. As an engineer myself I make enough errors as is. Just because you know (or know where to find) the formulas doesn't mean you know how to solve this problem, as you may interpret them incorrectly.

Perhaps explain why you need this a bit more, and we might be able to help you.

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

RE: final velocity calculation

" if it takes you 'a few hours' to figure out the acceleration of a falling object..."

Not just that, but the fact that the fundamental problem is covered in high school physics. The only question should have been, "what deceleration time or distance do I use?"

TTFN
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RE: final velocity calculation

Walterke, your first comment is correct. v^2 = velocity squared which is not 7m/s
v^2 = 49m^2/sec^2

Ted

RE: final velocity calculation

I think it might be safe to assume that since 11xminurti3 has a 32kg rock that it probaly is NOT going to fall on an mattress. Even if it did, it would probably squash the mattress like the proverbial egg. That's a pretty hefty rock, guys. It's also really not falling very far.

To put it bluntly, it will go "splat." The question is not going to be "how high will this bounce," but "how big a hole is this going to make?" Of course, depending on the material the rock is comprised of*, the rock might break into pieces upon impact (similar to an egg hitting a cement floor.)

For a ROUGH estimate, assume a displacement on the order of 0.01 mm (assuming it's a hard surface.) If you're hitting sand or soft soil and it really will dig a hole, you might want to do a second calculation with a larger displacement.

*correct grammar would insist that I say "of which the rock was comprised." Bah, humbug.

RE: final velocity calculation

If you only want the barest indication of what will happen when it hits, I will tell you this. I would not get my toes under the rock, with our without steel toe boots.

RE: final velocity calculation

I dont know if this is a positive contribution to this thread, but with my background in underground mining, I do know a thing or two about falling rocks..... some alot more than 32kg. Plus the original question is very similiar indeed to a query that was put to me recently by a structural engineer who was getting heavily involved in an area he should have stayed away from.

Is this 32kg rock dropping straight from the back of a tunnel , or is it rolling down the side of a mountain?? Obviously all theoretical calculations are highly dependent on the answer to this question. Then I have to point out that the geometry of all falling rocks will vary between something approximating a perfect sphere, to something that approximates a perfect cube. Depending on what I was designing, I would want to know what the estimated surface area of the contact zone is. Even all this is very theoretical and I would have to design for a point load contact over a very small area.

Penetration of an underengineered system can easily be imagined, but perhaps the areal extent of the penetration would not be critical. I dont know if the OP is out of his area of expertise, but many posters here have, with the best of intentions, made a number of assumptions that might not be valid. Just IMHO of course.

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