Confusing Problem 4

[04jchatter]

Hello Everyone,

I am a product designer so my mechanical engineering is a bit rusty. I have been tasked with putting together a test program for a traffic sign. One of the tests is a simulated impact. The way this is done is by hitting the product with an (undetermined) mass using a pendulum. The arm has to be 1.25 meters long and the impact energy has to be 150 Nm.

My problem is that, that is the only information the standard gives to work with. I have estimated (as best I can) the period of the pendulum:

T=2π√L/G For which I have calculated T=2π√1.25/9.81 = 2.25

However After that I am not sure were to go, I need to prove that the energy that will impact the sign is 150 Nm.

If possible a break down of how the Pendulum period works and how to calculate pendulum energy would be excellent.

Many Thanks,

04jchatter

 

[imcjoek]

Student post in disguise...?

http://www.physicsclassroom.com/mmedia/energy/pe.cfm

 

[IRstuff]

Perhaps a description or title of the standard might be useful. What seems to be missing is the mass at the end of pendulum.

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[Dougt115]

First why care about the period? Provided the sign has to survive the impact the period is irrelevant as all of the energy will be transmitted to the sign and there is no oscillation.

Second the energy is a function of the height, Potential energy = mgh or weight*height. All you need is the center of mass of the pendulum and any added weight at the end. Use the change in height of the center of mass and the weight of your system and you have the energy.

 

[zdas04]

This is a second law problem. With the trajectory you can come up with the velocity at impact. Once you've determined the mass, the only thing missing is how long it takes for the velocity to reach zero after first impact. Once you can determine that then F=m*a.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[GregLocock]

That's a pretty funny test, you would get very different results using a 6 kg mass and a 600 kg mass.

Conservation of energy is the method, not the period of the pendulum

Cheers

Greg Locock


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[04jchatter]

@Dought115, my mistake I should have stated that the sign is supposed to be impacted and bounce back up. @GregLocock & IRstuff, I am Trying to deturmin the Mass of the weight that is what I want to know. I have the energy in the system as 150Nm but I do not have the mas of the weight.

Thank you to everyone for you help.
@imcjoek - I am a student but I am currently undertaking a work placement so this Is a real Industry problem but thank you for the Link

Kind Regards,

04jchatter

 

[LittleInch]

Am I missing something here or is there an error in describing energy as Nm. To me Nm is a torque value and energy is measured in Joules??

Hence the answer would seem to be simple - if 150Nm is actually 150Joules, then just work out for whatever mass you have what the height above the vertical position you need to generate 150J on the basis that all the potential energy is translated into kinetic energy and all is then transferred to the sign when it hits it. You might need to experiment with different masses to get the velocity right to have the best effect.

The practical difficulty will be that the pendulum will either bounce back after hitting the sign, or continue along its path (just much slower) or will get in the way of the sign bouncing back.

A diagram always helps!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[kingnero]

charpy impact test ?

 

[IRstuff]

"To me Nm is a torque value and energy is measured in Joules"

The primary units of a joule is newton*meter, i.e., force*distance

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[04jchatter]

Hello everyone,

Firstly many thanks for all your helpful advice, what you have all said has provoked me to think about the problem differently.

For those who were interested, This is the final result I got from the equations I found, the values I got seem reasonable to me to if what I have done makes sense please say.

(Also if it makes no sense please also say)

KE=1/2 x M x V²

Where: - M = Mass of bumper
V = Velocity
KE = Kinetic Energy

V= √2xGxL(1-Cos(∂max))

Where: - G = Acceleration due to Gravity (9.81m/s²)
L = Length of arm
∂ = Max Angle
So: - V= √2xGxL(1-Cos(∂max))

Therefore: V= √2x9.81x1.25(1-Cos(90max))
= 5.96m/s²

KE=1/2 x M x V²

Therefore: 150 =1/2 x M x 5.96²
Rearrange to find M: -

(1⁄2 x M x 〖5.96〗^2)/150 So: - 1/M=(1⁄2 x 〖5.96〗^2)/150=0.1184

M=1/0.1184=8.44 kg

Mass of Bumper = 8.44kg

Once again, Many thanks
04jchatter

 

[Dougt115]

Two things.

Your velocity units should be m/s

The max angle is not 90° it is 180° (1-cos180=2). (But don't use a lift of 180° as there is a chance the weight may go over the other side and hit your sign from the back.)

 

[LittleInch]

Makes no sense to me and I've looked at it a couple of times now. your v I calculate as 4.95. when you do that you get 12.23 - but see simpler way below.

[b]Always[/b] double check your calculations......

Why not use the energy balance.

Assuming a the bottom of the pendulum you have your sign and all the potential energy turns into kinetic energy, then

energy = 150J = mass x G x height at 90 degrees

So height - 1.25m at 90 degrees
Mass = 150 / 9.81 x 1.25) = 12.23kg.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[IRstuff]

I agree with LI here, except... I get 12.237 ;-)

that's just from the round off of using 9.81 instead of 9.80665 (that's just what's built-in with Mathcad)

The OP's calculator appears to be broken; the PE calculation should have resulted in 4.9514 m/s, which, when plugged into the KE equation would have resulted 12.237 kg. Note also, their first calculation has m/^2 for units of V.

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[zdas04]

The equations were kind of sloppy above. The velocity equation should be:

v=(2*g*L*(1-cos(α))^0.5

(the way the square root symbol was written above said "square root of 2 times g times L times the quantity 1-cos(α) when the real velocity is the square root of everything. It doesn't take much to do a simple unit analysis and see that the way it written gives you a "velocity" of m²/s²

If L = 1.25 m then v = 4.95 m/s

So
m = KE/v^2= 150 N*m/(4.95 m/s)^2 = 6.118 N*s^2/m = 6.118 kg

If the pendulum is raised to 179.5 degrees then the velocity becomes 7.11 m/s and the required mass drops to 3.059 kg.

I can't figure out why so many engineers have had so much trouble with this bit of high school math.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[IRstuff]

David, did you lose a factor two in that last KE calculation, .i., m = 2*KE/v^2?

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[zdas04]

Dang I'm glad that I wasn't too mean about checking work. Yes I did lose the 1/2. So at 90° it should be 12.2 kg, just like LitteInch said. At 180° (minus) it is 6.118 kg.

I really can't understand why I joined the engineers that can't do this high school math problem.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[IRstuff]

Well, misery does love company ;-)

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[racookpe1978]

12.2 kg on a lever-arm only 1.25 meters?

28 some-odd lbs swinging naturally from rest on a pivot arm only 49 inches long?

That's not even the energy of a bicycle at near-walking speed! For a road sign specification? (I am assuming it hits somewhere near the ground - not at waist-high. ) You'd put a dent in a 4x4 fence post. Maybe. )

 

[GregLocock]

Having hit the occasional road sign, perhaps they are designed to bend out the way rather than killing the driver. If not, why not?

Cheers

Greg Locock


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