Register to reply

Recoil differential equations

by moemag
Tags: differential, equations, recoil
Share this thread:
moemag
#1
Jan16-13, 06:24 PM
P: 3
I am trying to figure out what I am doing. Forgive me, I’m not very sharp when it comes to differential equations… or even a useful working knowledge of calculus. I understand the concepts but if I can’t figure out how to type it into my TI-89 or get Wolfram alpha to do it… I’m kind of stuck and would appreciate some assistance.

So to start:

I have a system that has a combined momentum of 11.38 kg -meters/second. At least that is what I am estimating it to be. I know the mass of the moving part, and I know how fast it is going at it's peak.

I built a test rig to hold the system, which contains a small explosive force, setting the system into motion. The system is butted against a load cell. After firing the peak force the load cell registered was 300 newtons.

How do I get from momentum… to 300 newton’s force?
I’m guessing it has to do with the spring rate of the load cell… which is very high. I believe the load cell only moves .127mm in its entire stroke and is good for. (http://www.transducertechniques.com/lpo-load-cell.aspx ... its LPO-5K)
I keep thinking it’s somewhere in the relationship 1/2 m*v^2=1/2 k*x^2… but I’m pretty sure that somehow I have to figure out how long it takes the spring of the load cell to de-accelerate the mass, thus producing a force… I just don’t know how to do that/express that mathematically.

-my first post... this seems like where I would put this question, forgive me if it belongs somewhere else.
Phys.Org News Partner Engineering news on Phys.org
An innovative system anticipates driver fatigue in the vehicle to prevent accidents
Squink personal factory aims to make circuit prototyping easy
Catching grease to cut grill pollution
engnr_arsalan
#2
Jan17-13, 08:21 AM
P: 41
u know force.. F=Kx..find "X" displacement of srping..coz k of spring is given by manufacturer..
put "X" in the eq u gave find "v" velocity..now multiply it with mass and get momentum
now at position one at time t=0 the velocity is v=max thus mometum is max..
at state two..t=T2 velocity is v2=0 thus momentum is zero..coz all the kinetic energy has been converted into spring's potential energy..
force is rate of change of momentum i.e
F=dM/dT=M1-M2/T1-T2
u have m1= max what ever the value u calculate and m2=0,u have time t1=0 and T2 can be calculated coz force is known..
but this an approx solution..
Jupiter6
#3
Jan19-13, 12:51 AM
P: 128
Quote Quote by moemag View Post
How do I get from momentum… to 300 newton’s force?
As engnr_arsalan answered: impulse, but you need to time your experiment, not simply rely on peak force. Can you chart your force vs. time?

A second sensor such as an accelerometer (acceleration) or video(position) would really be needed to confirm results. KE is not something you can rely on. You can rely on it's derivative (MV) but you need a way to time things.

moemag
#4
Feb4-13, 04:33 PM
P: 3
Recoil differential equations

Quote Quote by Jupiter6 View Post
As engnr_arsalan answered: impulse, but you need to time your experiment, not simply rely on peak force. Can you chart your force vs. time?

A second sensor such as an accelerometer (acceleration) or video(position) would really be needed to confirm results. KE is not something you can rely on. You can rely on it's derivative (MV) but you need a way to time things.
Thanks!


Register to reply

Related Discussions
Writing Linear Differential Equations as Matrix Differential equations Calculus & Beyond Homework 1
United States Elementary Differential Equations - 1st Order Differential Equations Calculus & Beyond Homework 1
A good book on Differential Equations and Partial Differential Equations? Science & Math Textbooks 1
Minimal Surfaces, Differential Geometry, and Partial Differential Equations General Math 0
Partial Differential Equations, Applied Nonlinear Equations, and Classical Mechanics Science & Math Textbooks 0