Rod Loading - Spring v Lever

[Merlin]

Yes

I shall post the graphic this evening.
Time to go to the office now.

Merlin

[Aitor]

Yes

I shall post the graphic this evening.
Time to go to the office now.

Merlin

Great! (except for the "go to the office" part)

Summarizing:
- We have two identical weights pulling two identical bricks by the force of gravity.
- One weight is connected to its brick by an inelastic string; the other weight is connected to its brick by an inelastic string which, in the middle, has a spring attached.
- Both weights hang at the same height from the ground.
- Before releasing the weights the string in both devices is taut and the spring is compresssed (no energy stored).

Then we release both weights and they reach the ground. Meanwhile the bricks move along the table. The brick connected by a spring moves a longer distance than the brick connected by a string only. Right?

[gordonjudd]

I shall post the graphic this evening.

Merlin,
I just picked up on this thread and am getting confused about parameters. Your drawing will go a long way to making this more understandable.

What were the masses of the brick and lead in your model? What was the spring constant of the spring? What is the initial deflection of the spring?

I assume the lead is allowed to drop from the same height and that release is done by releasing the hold on the brick, with the initial deflection spring force on the brick being f=m_lead*a=.49 N.

Gordy

[Aitor]

Gordy,
I see that my experiment is very difficult to visualize correctly. I hope that this drawing clears everything up.
- T is a table
- B is a brick (or whatever you want to use, I am thinking of using a crystal ball for the real experiment).
- ST is an ideal inelastic string.
- P is a pulley.
- SP is a spring.
- W is a weight (whose mass is bigger than that of B).

I assume the lead is allowed to drop from the same height and that release is done by releasing the hold on the brick...

No. As I said several times before (the last one in the summary provided in the post just previous to yours): Before releasing the weights the string in both devices is taut and the spring is compresssed (no energy stored).
So we hold the weights and release them. Holding the bricks and releasing them would mean that the spring is loaded and nobody starts a cast with the rod already loaded.

It is just as Grunde's model of brick/spring/car but instead of car using a weight that is powered by gravity, so the force it applies is always the same. That is the difference with Grunde's car, which increases the force it applies to the brick (at least that is how I understand it).
I know that a weight isn't a caster, but a car isn't a caster either and, moreover, I don't have a car to make such experiment but I don't consider difficult to get a suitable weight.

[Magnus]

Aitor

Have you made a video of this yet?

[Aitor]

Aitor

Have you made a video of this yet?

No. I first need to design and build the device. It will take some time, I am afraid.

[Aitor]

PS : James graphic is not the simulation of a cast but the study of the effect of spring stiffness on the outcome of his model.

Merlin

James' graphic was used to prove Mark "the mechanical advantage that a fexible lever has over a stiff lever during casting" as you can read here. So it looks like the simulation of a cast to me.

[gordonjudd]

No. I first need to design and build the device.

Aitor,
Determining the optimum spring constant is something Merlin should be able to work out, but finding a spring with that value might not be so easy. You can probably live with friction effects although putting the brick mass on wheels might make a difference.

Having a real world loosely wound spring would complicate this analysis since that involves the wave equation rather than the SHO equations. In the real world I don't know if you could assume a mass-less spring as is done for the SHO. The [url=C:\Users\Administrator\Downloads\Documents\GetPDFServlet.pdf] physics of a falling spring[/url] are not intuitive since we do not think in terms of distributed forces.

Gordy

[Aitor]

Gordy,
This is not the technical forum. I am just after some clues to check just by the naked eye, for those like me who have a hard time reading graphs and understanding models.

If the model says that at the end of the experiment:
The brick on the spring will cover a longer distance than the other one...
or
The brick on the string will cover a longer distance...
or
Both bricks will cover the same distance...

That is enough for me.

[Merlin]

Aitor

Here are the illustrations of the speed and distance covered by the bricks.

First, let set the parameters: the bricks have a 5 grams mass and the weights have a 10 grams mass; the spring stiffness is 0.09 N/m. The height for dropping the weights is 2 meters (I can change all these parameters to be closer to rod characteristics). World is idealize, no friction. Parameters look like toys characteristics.

Now let’s speak with colors which are linked to the graphics.





We have a blue brick and a blue weight connected by a string on one side, and a green brick connected to a red weight by a massless spring on the other side. At start, there is no energy stored in the spring. We release both weights a time = 0.

The first weight to hit the ground is the red one. In the test conditions the landing occurs at t = 0.74 s. At that time, the blue brick has moved 1.8 m from the start and the green one slightly less than 1.4m. It is lagging behind the blue one.

At t= 0.78 s, the blue weight hits the ground, the blue brick has moved 2 meters from the start and the green one is still lagging behind at 1.63 m from its starting position. The blue brick is now travelling at a constant speed (5.1 m/s).

At t= 0.84 s, the green brick is at maximum speed (6.3 m/s) and is located at 2 meters from its starting point. The green brick is now travelling at constant speed too, since we make the assumption that bricks are on their own as they get their maximum speed.

At t =1.1 s and after moving forward 3.75 m both bricks are at the same position but the green one is overtaking the blue brick.

What happened? The spring has caught some of the potential energy of the weight and gave it all back to the brick (with our simplified assumptions). Without spring, the initial potential energy is converted: 1/3 goes into the brick and 2/3 into the weight crashing on the floor. Adding the spring changes the picture: now it is 50% for the brick and 50% for the weight. A better situation isn’t it?

Merlin.

[TrevH]

I lost my fly tying kit a while ago, so I've been looking at buying a new vice and stuff .... now I may get one of these instead :p

[TrevH]

Sorry I haven't added anything constructive, I am paying attention, but you're all way ahead of me. :upside:

Trev

[Aitor]

Great work Merlin. Thanks.

I see in the distance vs. time graph that when both bricks stop at t= 1.6
- The brick on the string (the blue one) has covered a distance of around 6.2 m
- The brick on the spring (the green one) has covered around 6.7 m.

What happened? The spring has caught some of the potential energy of the weight and gave it all back to the brick (with our simplified assumptions). Without spring, the initial potential energy is converted: 1/3 goes into the brick and 2/3 into the weight crashing on the floor. Adding the spring changes the picture: now it is 50% for the brick and 50% for the weight. A better situation isn’t it?

Definitely. The brick on the spring has got more energy from the brick than the other one.

[gordonjudd]

I see in the distance vs. time graph that when both bricks stop at t= 1.6

Aitor,
I think that 1.6 second value is the point where Merlin stopped his calculations not where the brick came to a stop in those two cases. Since there are no losses, they would continue to travel with the maximum velocity produced with the string (5.1 m/s) or the spring (6.2m/s) connections for all time after their launch.

It is the integral of the force applied over distance (applied work energy) advantage that goes to the spring connection as discussed here..

Using your m_brick=10 g and m_lead=50 g values the no stretch string connection would apply a constant acceleration .0force of f=ma =.01*8.17(.0817 N) over a distance of 2 meters to produce .16 J of work energy.

The spring connection produces a variable force depending on the deflection of the spring. Using an optimum spring constant of around .26 N/m, that spring deflection force starts out at zero reaches a maximum value of .245 N at the maximum spring deflection of .94 m and then goes back to zero. The distance that variable force is applied would be around 2.8 m. Thus the added maximum acceleration force applied over a longer distance produced more work energy and causes the KE of the spring-launched brick to be around .247 J.

For the .01 kg brick that increase in applied work energy produces increase in the maximum brick speed from around 5.7m/s to 7.0 m/s as shown below.


Good luck finding a spring that will stretch by .94 meters using your small mass values. I suspect that is an experiment that will be nearly impossible to implement for those .01 and .05 kg mass values.
Gordy

[Aitor]

I think that 1.6 second value is the point where Merlin stopped his calculations not where the brick came to a stop in those two cases. Since there are no losses, they would continue to travel with the maximum velocity produced with the string (5.1 m/s) or the spring (6.2m/s) connections for all time after their launch.

Gordy,

OK, but if we now consider the losses due to drag, and being both bodies identical, the one with the highest speed will travel the longer before stopping.

Good luck finding a spring that will stretch by .94 meters using your small mass values. I suspect that is an experiment that will be nearly impossible to implement for those .01 and .05 kg mass values.

Those aren't my values but those which Merlin has chosen (by the way, the smallest mass is only 0,005 kg)
I am thinking of implementing the experiment with marbles, a weight of some short and a commercial rubber band, so the values will be very different for sure.
I understand that the spring constant has a big effect in the results. Could it be that by varying that value the highest speed would be reached by the marble with the rigid connection, just the opposite of what the model has shown now?