Conservation of Momentum in Dynamic Engineering Systems

In summary: Thanks for the link. That does help a lot.When using energy considerations to solve problems, you often equate the initial energy with the final energy, or equate the initial energy to the work done to dissipate that initial energy.In this problem, your pile driver has an initial potential energy due to gravity and it being held above the surface (after all, it took some energy to lift that pile driver up that high, right?). When it is released, that initial PE is lost and converted progressively into kinetic energy (KE) of motion (the velocity downward). When the pile driver strikes the ground and is slowed down by deformation of the ground, the work done by the ground
  • #1
Engineeringworks23
3
0
Homework Statement
Homework relating to mechanical principles and applications
Relevant Equations
v = u+at
V(squared) = U(squared) + 2as
I = mv - mu
I = Ft
F=ma
Hi there could someone please help me with this question we have not been taught the content of this question what so ever so help would be very much appreciated. As you will be able to tell by the question it is not very difficult however I am not excellent at engineering or physics and alongside not being taught this content I am unfamiliar of what to do next. For the question I have to use the principle of conservation of energy.
 

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  • #2
Hello @Engineeringworks23, Welcome to Physics Forums.

If you are expected to use conservation of energy for this, start by considering what the source of the energy "input" is, and where it finally ends up. So what energy source is "powering" the system?
 
  • #3
Engineeringworks23 said:
Homework Statement:: Homework relating to mechanical principles and applications
Relevant Equations:: v = u+at
V(squared) = U(squared) + 2as
I = mv - mu
I = Ft
F=ma

Hi there could someone please help me with this question we have not been taught the content of this question what so ever so help would be very much appreciated. As you will be able to tell by the question it is not very difficult however I am not excellent at engineering or physics and alongside not being taught this content I am unfamiliar of what to do next. For the question I have to use the principle of conservation of energy.
Welcome to PhysicsForums.

So you should add the gravitational potential energy equation to your list of Relevant Equations. Does that sound familiar? The GPE is related to the mass of the object and how far up it is (before it starts falling)...

Also, what is the relationship between work done and force and distance (that the force is applied through)?
 
  • #4
Thanks and also thanks for responding so fast. The source of the energy input would be the pile driver, so are you saying that I need to consider it first before anything else?

gneill said:
Hello @Engineeringworks23, Welcome to Physics Forums.

If you are expected to use conservation of energy for this, start by considering what the source of the energy "input" is, and where it finally ends up. So what energy source is "powering" the system?
 
  • #5
Thanks for responding! This equation doesn’t sound familiar to me as we have not had class time since before Christmas meaning we have not yet covered it. Work done can be calculated by multiplying force and distance?
berkeman said:
Welcome to PhysicsForums.

So you should add the gravitational potential energy equation to your list of Relevant Equations. Does that sound familiar? The GPE is related to the mass of the object and how far up it is (before it starts falling)...

Also, what is the relationship between work done and force and distance (that the force is applied through)?
 
  • #6
Engineeringworks23 said:
Work done can be calculated by multiplying force and distance?
Yes, very good. In this problem, they are simplifying things by asking for an average force multiplied by the distance to stop the pile driver. In the real world the force of stopping the pile driver would not be constant, so you would need to use some calculus tools to solve the problem. But this one should be simpler than that. :smile:

When using energy considerations to solve problems, you often equate the initial energy with the final energy, or equate the initial energy to the work done to dissipate that initial energy.

In this problem, your pile driver has an initial potential energy due to gravity and it being held above the surface (after all, it took some energy to lift that pile driver up that high, right?). When it is released, that initial PE is lost and converted progressively into kinetic energy (KE) of motion (the velocity downward). When the pile driver strikes the ground and is slowed down by deformation of the ground, the work done by the ground over that deformation distance has to be equal to the energy that the pile driver had before and during the strike and deformation. Does that make sense? Energy is not created or destroyed in these types of problems -- it is "conserved" but can be converted from one type of energy to another (like the work done to dissipate that energy).

[Edit -- please see the correction for this part by @haruspex below]

So here is a link to help you understand gravitational PE:

https://en.wikipedia.org/wiki/Gravitational_energy

And as you say the work done to stop the pile driver is the (average) force multiplied by the stopping distance. Be sure to include the stopping distance in the overall height of the pile driver when calculating the initial PE of the pile driver -- it doesn't just travel from its initial height to the ground, it goes a bit below the ground, right?

Go ahead and use these hints to start writing the equations that you think apply here, and we can help you if you get stuck. BTW, it helps if you look through the LaTeX Guide link at the lower left of the Edit window so that you can post your math equations in the standard format. Thanks. :smile:
 
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  • #7
Engineeringworks23 said:
Work done can be calculated by multiplying force and distance?
In general, one has to integrate force wrt distance: ##\Delta E=\int F(x).dx##. But in this case the resistive force of the ground can be taken as constant, so can be taken outside the integral ##\Delta E=F\int .dx=F\Delta x##, so yes, just force times distance.
(Note: the question is wrong to ask for the average force; you have no way to find that if it is not constant. It should say, find the resistive force assuming it is constant.)

But it worries me that the question just says to use conservation of energy. To solve it correctly requires three steps:
  • Conservation of energy during the fall of the hammer.
  • Conservation of momentum during the fraction of a second of the impact on the pile.
  • Conservation of energy again during the descent of pile+hammer.

Have you been taught about conservation of momentum? If not, maybe you are only expected to apply conservation of energy, in which case you can do the whole thing in one step, but arriving at the wrong answer: there is a significant loss of useful energy to heat during the impact! (And don't forget that in the last step there are two masses losing GPE.)

If you have done momentum conservation then I would think you are expected to do all three steps. Maybe the "use conservation of energy" is only meant to refer to the last step... though there is still the remote possibility that they want you to use the invalid approach of energy only in order to demonstrate the difference when, later, you are instructed to use momentum conservation too.
 
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  • #8
haruspex said:
Conservation of momentum during the fraction of a second of the impact on the pile.
How can you apply conservation of momentum? According to an argument it doesn't hold because we have
  • The force of weight of the hammer
  • The normal force from the ground which increases during the impact
 
  • #9
Delta2 said:
How can you apply conservation of momentum? According to an argument it doesn't hold because we have
  • The force of weight of the hammer
  • The normal force from the ground which increases during the impact
You have to assume the duration of the impact is arbitrarily brief. The exchange of momentum is independent of the duration. For the hammer (positive down), ##\Delta (mv)=mg\Delta t-\Delta p##, and for the pile ##\Delta (MV)=\Delta p+Mg\Delta t-F_r\Delta t##. Letting ##\Delta t## tend to zero, ##\Delta (mv)=-\Delta p## and##\Delta (MV)=\Delta p##.
 
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