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## Spanish experiment

**Moderator:** Torsten

### Spanish experiment

Hi Lasse,

Just because the weight on the fixed line had the positional lead before the bricks hit the floor doesn't necessarily mean it was the one travelling the fastest at the time. Very hard to determine without some sort of accurate velocity measurement.

Cheers, James.

Just because the weight on the fixed line had the positional lead before the bricks hit the floor doesn't necessarily mean it was the one travelling the fastest at the time. Very hard to determine without some sort of accurate velocity measurement.

Cheers, James.

- Lasse Karlsson
**Posts:**4251**Joined:**Wed Jan 09, 2013 9:40 pm**Location:**There, and back again-
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### Spanish experiment

Hi James

It was the brick on the table that had the lead, didn't look at the falling weights, those wheren't the interesting part

Cheers

Lasse

It was the brick on the table that had the lead, didn't look at the falling weights, those wheren't the interesting part

Cheers

Lasse

Your friendly neighbourhood flyslinger

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### Spanish experiment

Hi Lasse,

I was talking about the two masses on the table. I should have said 'one on a fixed line' and one incorporating a spring.

Cheers, James

I was talking about the two masses on the table. I should have said 'one on a fixed line' and one incorporating a spring.

Cheers, James

- Lasse Karlsson
**Posts:**4251**Joined:**Wed Jan 09, 2013 9:40 pm**Location:**There, and back again-
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### Spanish experiment

Hi James

Ok, can you explain it to me? I'm the last guy to be posting physics stuff anyway, so I can't really get my head around this:

Two bricks side by side on a table. The get accelerated for 56 cm's, and one of them get's ahead. But the one that is ahead, isn't travelling the fastest/has the highest velocity at that point?

Cheers

Lasse

Ok, can you explain it to me? I'm the last guy to be posting physics stuff anyway, so I can't really get my head around this:

Two bricks side by side on a table. The get accelerated for 56 cm's, and one of them get's ahead. But the one that is ahead, isn't travelling the fastest/has the highest velocity at that point?

Cheers

Lasse

Your friendly neighbourhood flyslinger

http://www.karlssonflyfishing.com

***Bring Mark back!!!!!! ***

http://www.karlssonflyfishing.com

***Bring Mark back!!!!!! ***

### Spanish experiment

Hi Lasse,

I was once sat at the traffic light in my home town with a police car behind me. I accelerated off in my usual, slightly heavy right footed, way. I got to 30mph, backed off and maintained that speed. A few moments later the blue lights came on and I was pulled over. The police woman accused me of doing 50 mph in a 30 limit. I explained to her that she was the only one who was speeding, as she did 50 to catch me up, but I never exceeded 30 (and she never overtook me). Luckily she was then called away, because she wasn't buying it at all .

(that's a true story by the way).

The point of this is that if the brick that is behind only matches the speed of the one in front, it will never catch it up. In order to catch it up it must exceed the speed of the leading brick. Therefore when it actually overtakes, the speed is probably well in excess of the brick that was leading first. Judging the time/position when the speed of the lagging brick first exceeds that of the one it's chasing requires careful measurement.

Does that make sense?

Cheers, James.

I was once sat at the traffic light in my home town with a police car behind me. I accelerated off in my usual, slightly heavy right footed, way. I got to 30mph, backed off and maintained that speed. A few moments later the blue lights came on and I was pulled over. The police woman accused me of doing 50 mph in a 30 limit. I explained to her that she was the only one who was speeding, as she did 50 to catch me up, but I never exceeded 30 (and she never overtook me). Luckily she was then called away, because she wasn't buying it at all .

(that's a true story by the way).

The point of this is that if the brick that is behind only matches the speed of the one in front, it will never catch it up. In order to catch it up it must exceed the speed of the leading brick. Therefore when it actually overtakes, the speed is probably well in excess of the brick that was leading first. Judging the time/position when the speed of the lagging brick first exceeds that of the one it's chasing requires careful measurement.

Does that make sense?

Cheers, James.

- Paul Arden
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### Spanish experiment

So in order for the two masses to hit the floor at the same time, the brick with the spring would require a much higher mass. And then if the two bricks did hit the floor at the same time the brick with the spring would fly off the table out out the door.

My way of looking at this therefore, is that a rigid rod is limited purely by how quickly we can move our arms, whereas a flexible rod is limited by how much force we can apply. Speed vs Strength. Does this make any sense in engineering terms or is it bollocks?

Cheers, Paul

My way of looking at this therefore, is that a rigid rod is limited purely by how quickly we can move our arms, whereas a flexible rod is limited by how much force we can apply. Speed vs Strength. Does this make any sense in engineering terms or is it bollocks?

Cheers, Paul

- Lasse Karlsson
**Posts:**4251**Joined:**Wed Jan 09, 2013 9:40 pm**Location:**There, and back again-
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### Spanish experiment

Hi James

That makes good sense, don't trust your speedometer

It still means that for a period of time you where travelling faster than the police car. Same as with the bricks, without the spring, it's your heavy foot, and it goes of faster in the beginning, the spring attached brick then has to speed up further to overtake, just like the policecar, but the first brick isn't maintaining speed, but slowing down, so it doesn't need to get over 50 to overtake

Cheers

Lasse

That makes good sense, don't trust your speedometer

It still means that for a period of time you where travelling faster than the police car. Same as with the bricks, without the spring, it's your heavy foot, and it goes of faster in the beginning, the spring attached brick then has to speed up further to overtake, just like the policecar, but the first brick isn't maintaining speed, but slowing down, so it doesn't need to get over 50 to overtake

Cheers

Lasse

Your friendly neighbourhood flyslinger

http://www.karlssonflyfishing.com

***Bring Mark back!!!!!! ***

http://www.karlssonflyfishing.com

***Bring Mark back!!!!!! ***

- Paul Arden
- Site Admin
**Posts:**14265**Joined:**Thu Jan 03, 2013 11:20 am**Location:**Belum Rainforest-
**Contact:**

### Spanish experiment

Here you just bribe them 100 ringgit - problem solved.

### Spanish experiment

I've been thinking about it in terms of momentum. The momentum of both systems is the same at the beginning of the experiment. Imagine we threw the two systems through the air with the same force, I think they would both travel at the same velocity, measured at some fixed point in the system. The difference is that in all likelihood the system with the spring would be oscillating whereas the other would not be able to. But fundamentally they would both arrive at a fixed point at the same time if you measure it at that fixed point within each system.

I think the same happens across the table, if both the table and the drop are sufficiently long. One system oscillates and one doesn't but fundamentally they both travel at the same velocity.

If this is accepted, the bricks and the weights of each system must be travelling at different velocities most of the time (although there are times when they are equal - like the broken clock that is right twice a day).

So the momentum of the spring system matches the momentum of the other system, if the weight of the spring system is moving faster than the other weight, the brick on the spring must be traveling slower so the momentum of the system remains the same as the other system.

If the weight on the spring hits the floor at this time, the brick is traveling slower. The brick on the string will travel further.

We have taken more momentum out of this system because we have stopped the weight when it was travelling faster than the weight on the string, leaving less momentum in the remaining part of the system, the brick.

So that the spring system wins in the competition, the weight, the brick, the spring and the length of the drop need to be tuned so that the weight on the spring meets the floor when it is travelling slower than the weight on the string.

This way we take less momentum out of the system with the spring. The brick on the spring is moving fastest and wins in the speed/distance competition.

The energy we put into each system is the same. The energy we take out of each system is not the same.

Paul mentioned about the trade off between speed and strength. I think this is an issue too, but I don't think we've got to far as considering this yet. But I am looking forward to that when we get there

My layman's idea for a Friday afternoon. I doesn't seem to work exactly because in Merlin and James's graphs the velocities of the bricks and weights are not quite in the phases I expected but it seems to be quite a small timing difference.

Cheers

Trev

I think the same happens across the table, if both the table and the drop are sufficiently long. One system oscillates and one doesn't but fundamentally they both travel at the same velocity.

If this is accepted, the bricks and the weights of each system must be travelling at different velocities most of the time (although there are times when they are equal - like the broken clock that is right twice a day).

So the momentum of the spring system matches the momentum of the other system, if the weight of the spring system is moving faster than the other weight, the brick on the spring must be traveling slower so the momentum of the system remains the same as the other system.

If the weight on the spring hits the floor at this time, the brick is traveling slower. The brick on the string will travel further.

We have taken more momentum out of this system because we have stopped the weight when it was travelling faster than the weight on the string, leaving less momentum in the remaining part of the system, the brick.

So that the spring system wins in the competition, the weight, the brick, the spring and the length of the drop need to be tuned so that the weight on the spring meets the floor when it is travelling slower than the weight on the string.

This way we take less momentum out of the system with the spring. The brick on the spring is moving fastest and wins in the speed/distance competition.

The energy we put into each system is the same. The energy we take out of each system is not the same.

Paul mentioned about the trade off between speed and strength. I think this is an issue too, but I don't think we've got to far as considering this yet. But I am looking forward to that when we get there

My layman's idea for a Friday afternoon. I doesn't seem to work exactly because in Merlin and James's graphs the velocities of the bricks and weights are not quite in the phases I expected but it seems to be quite a small timing difference.

Cheers

Trev

- gordonjudd
**Posts:**1382**Joined:**Sat Jan 19, 2013 11:36 pm**Location:**Southern California

### Spanish experiment

Lasse,Ok, can you explain it to me? I'm the last guy to be posting physics stuff anyway, so I can't really get my head around this:

Two bricks side by side on a table. The get accelerated for 56 cm's, and one of them get's ahead. But the one that is ahead, isn't travelling the fastest/has the highest velocity at that point?

This result is confusing is because we are looking at the difference in the instantaneous vs the average velocities of the brick. We are used to situations where the velocity of an object has a constant value and thus over a fixed length of time the object with the faster velocity will cover more distance, i.e. d=v*t.

However at a more fundamental level the velocity of an object will depend on the integral of its acceleration curve and its distance will depend on the integral of its velocity curve. Those integration can produce some non-intuitive results for non-constant or non-linear acceleration vs time curves. That is what is happening with the acceleration force applied to the brick using the spring or in Jame's "racing cars" example. Unless the cop that stops you was a physics major, good luck in explaining the integration of acceleration vs time curves to get out of a speeding ticket.

To get a better understanding of what is going on and why even the lead mass has different velocities when it hits the ground take a look at is free-body diagram using Aitor's mass values of .05 kg for the lead mass and .01 kg for the brick.

You can see there are two forces acting on it as it falls. The acceleration force of gravity (9.81 m/s.^2) is pulling it down but there is an opposing force from the acceleration force on the brick that is pulling up on it as well. The net acceleration force on the falling lead mass will thus depend on the opposing acceleration force on the brick.

For the string, that opposing force will be constant since the brick will have the same acceleration as the falling lead mass. You can determine what that acceleration value will be by looking at the relative masses of the two objects, i.e.,

fg_lead=-m_lead*g;%acceleration force of gravity on the lead mass (directed in the negative y direction)

fa_brick=m_brick*g*m_lead/(m_lead+m_brick);%acceleration force on brick will be pulling on the lead in the positive direction

f_net_est=fa_lead+fa_brick;%net acceleration force on the lead mass

For Aitor's mass values that calculation shows the net acceleration of the lead mass will be -8.17 m/s. instead of the -=9.81 m/s.^2 rate it would have had without the brick.

With the spring the acceleration force on the brick will be zero at the instant the lead mass starts to fall, will reach some maximum value when the spring is stretched its maximum amount, and then will return to zero when the deflection in the spring goes to zero. You (or more likely Merlin ) can determine the stretching in the spring with the use of a second order differential equation that is related to what goes on in a simple harmonic oscillator.

I have given up on trying to explain that because of the way a SHO operates, the maximum spring force is twice as large as the force the constant force in the string for a spring value that is properly matched to get a loaded SHO frequency that is matched to the acceleration time.

**But if you try to understand why the mass overshoots its equilibrium value when it oscillates in a SHO you can start to see why it also produces that force doubling effect.**

Until you do the calculation for yourself (such as James went to the trouble to do) then you will have to take it on faith that the comparative force on the the brick with the string and the spring works out to look like this for a 2 meter drop and a spring value of .295 N/m using Aitor's mass values. Time is running from right to left in this plot (since the lead mass starts at a 2 meter height) so the break in the slope in spring force curve corresponds to when the lead mass hits the ground.

The work on the brick is related to the area under those curves, so I think you can see the area under the green curve would be larger than the corresponding red area for the constant string force. More applied work means more energy applied to the brick and thus it will also have a higher launch velocity when its crosses the 2 meter finish line and sails off the table.

Since the retarding force on the lead mass from the spring is also larger than it is for the string the spring driven case will also slow down the fall of the lead mass more. Thus the lead mass will dissipate less energy when it hits the ground since its impact velocity will be smaller with the spring than it will be with the string.

Gordy