final velocity calculation 5

[11xminurti3]

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

 

[IRstuff]

Is there some reason you're not describing the problem you are trying to solve?

TTFN
faq731-376

 

[11xminurti3]

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

 

[Jboggs]

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.

 

[JStephen]

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.

 

[rb1957]

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.

 

[hydtools]

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

 

[11xminurti3]

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

 

[11xminurti3]

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

Cheers

 

[rb1957]

sorry, student posts are not allowed

 

[11xminurti3]

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

 

[rb1957]

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).

 

[hydtools]

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

 

[1gibson]

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.

 

[drawoh]

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

 

[hydtools]

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

 

[rb1957]

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 ...

 

[11xminurti3]

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

 

[IRstuff]

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
faq731-376

 

[mikekilroy]

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.