Ultra Urgent Last Ditch

[Connolly]

Hello comrades

Me and two others are building an engineering project to be completed by tomorrow morning at 9.00 GMT.

So, its ultra urgent. We dont have time to find or work out the formula because we are way behind on tonnes of stuff.

We have a presentation tomorrow at 11.00GMT.

This is a last ditch attempt at finding someone who can help us with this. If you cant get it by 9.00GMT, or at least 10.GMT max.

The project is a double pendulum impact rig for testing helmets. So far, we have completed everything - including the pendulum rig. One very important thing still remains a problem though - the formula for calculating the actual force hitting the helmet.


Let me explain exactly what the pendulum impact rig does and how we have to use it.
Basically, its like any other impact test rig for measuring the strength of metals, except there are two pendulums, one containing the helmet with the aluminium head mould with a mass of 10kg, and the other pendulum containing the hemisphere surface for helmet impact with mass of 5kg.
During operation, the helmet and mass is raised to a certain angle, while the other pendulum containing the impact surface remains at its rest location. The helmet pendulum is released and impacts the other pendulum (hopefully causing it to break).

Now, what we need is the formula for calculating the impact force using the data taken.

The data we are to use for each operation is :

Please note - we can ONLY use this information.

1) the angle of the raised pendulum initially (helmet pendulum)
2) the max angle of the pendulum impacted (surface pendulum)
3) the angle at which the rebound of the second pendulum hits the first.
4) all the usuals including mass of both pendulums and length of pendulum arms etc.

If any of you guys could please supply us with a formula for calculating the force of impact using the information, you would be a life saver :D

Thank you very much in advance :D

[Connolly]

Ok, no problem.

We got through the presentation without the formula (just about) - complete disaster so it was. :unsure:

Ahwell - the past is the past. :P

[red team]

If its any consolation I think this is the solution:
No confirmation on this though. I'm not a physics major.


A1 = initial angle of helmet pendulum
A2 = max angle of impact pendulum
A3 = max angle of helmet pendulum, unknown
Ab = angle of impact pendulum hitting helmet pendulum
M1 = mass of helmet pendulum
M2 = mass of impact pendulum
Mt = M1 + M2
L = length of pendulums

H1 = L - L(cos(A1)), max height of helmet pendulum, (pre-impact)
H2 = L - L(cos(A2)), max height of impact pendulum, (post-impact)
H3 = L - L(cos(A3)), max height of helmet pendulum, (post-impact) unknown
x = L - L(cos(Ai)), max height of inelastic collision, (post-impact) unknown

PE1 = M1gH1 = M1g (L - Lcos(A1)), potential energy of helmet pendulum, (pre-impact)
PE2 = M2g(H2 - x) = M2g (L - Lcos(A2)), potential energy of impact pendulum, (post-impact)
PE3 = M1g(H3 - x) = M1g (L - Lcos(A3)), potential energy of helmet pendulum, (post-impact) unknown

PE1 != KE_inelastic + PE2 + PE3, energy is not conserved with inelastic collisions
sqrt(2M1(PE1)) = P + sqrt(2M1(PE3)) + sqrt(2M2(PE2)), but momentum is.

P = sqrt(2M1(PE1)) - sqrt(2M1(PE3)) - sqrt(2M2(PE2))
P = sqrt(2M1(M1gH1)) - sqrt(2M1(M1g(H3 - x))) - sqrt(2M2(M2g(H2 - x)))
Vt = P / Mt, tangential velocity

KE_inelastic = (1/2) Mt sqr(P / Mt) = Mt g x
x = sqr(P / Mt) / 2g

// Mass does not go into pendulum acc. formula which we can use to figure out time to impact
Acc2 = (-g/L) (A2), Angular Acc. for impact pendulum
0 = A2 + (1/2) Acc2 sqr(t2)
solve for t2, t2 = sqrt((-2 * A2) / Acc2)
= sqrt((-2 * A2) / ((-g/L) (A2)))

w = (-Acc2)(t2), Angular velocity at start of swing
= ((g/L) (A2))(t2)

Ab = (w)Ti + (1/2) Acc2 sqr(Ti), Ti = time to impact
solve for Ti by completing the square.

w(Ti) + (1/2) Acc3 sqr(Ti) = Ab
w + (1/2) (Acc3) (Ti) = Ab / Ti
Acc3 = 2((Ab / Ti) - w) / Ti

Acc3 = (-g/L) (A3), Angular Acc. for helmet pendulum
A3 = ((L Acc3) / -g)
H3 = L - L(cos(A3))

// Substitute back into this
P = sqrt(2M1(M1gH1)) - sqrt(2M1(M1g(H3 - x))) - sqrt(2M2(M2g(H2 - x)))
Vt = P / Mt

KE_inelastic = (1/2) Mt sqr(P / Mt) = Mt g x
x = sqr(P / Mt) / 2g, solve for the single variable: x

x = L - L(cos(As)), As angle of separation from inelastic collision
As = arccos((L - x) / L)

0 = As + (1/2) Acci sqr(Ts)
solve for Ts, Ts = sqrt((-2 * As) / Acci)
= sqrt((-2 * As) / ((-g/L) (As)))

Vf = L(w + (-g/L)(As)(Ts)), final tangential velocity before separation

Acc_av = (Vf - Vt) / Ts
Fav = Mt (Acc_av), Fav = average force at time of impact

[Connolly]

Thank you very much Red Team for that formula :D .


Ill have a look at the formula over the next couple days (have exams at the moment until this Monday coming) and if I can make heads or tales of it, give it a go on the test rig. If not, maybe you could answer any queries I might have?

To be honest, I thought the formula was going to be something pretty shortish :lol:

Thanks again for your time Read Team :D

[red team]

Note: a couple of corrections at the end of the solution which would make a bit more sense.
The pendulum doesn't entirely come to a halt when it flys apart so I think you have to take the difference between the final tangential velocity and the initial tangential velocity.

Vf = L(w + (-g/L)(As)(Ts)), final tangential velocity before separation

Acc_av = (Vf - Vt) / Ts