Force, Work, and Energy

The problem I am having is conceptual. Say that a particle of mass m starts from rest, and then moves along the path r(t) = <x(t), y(t)>. Does the forcing function always have to equal F(t) = m <x''(t), y''(t)>? (Just asking for confirmation on this part.) I'm going to proceed assuming that F(t) must equal the second derivatives, because it seems to be correct at the end of my derivation.

To calculate the work done by a force field and a path, the line integral is given by :

W= ∫ (F•T) ds

T is the unit tangent vector, r'(t)/||r'(t)||, and ds is the length, ||r'(t)|| dt. Therefore,

W= ∫ F • r'(t) dt

Because of the way I defined F(t), I can plug it in directly:

W= ∫ m <x''(t), y''(t)> • <x'(t), y'(t)> dt

Additionally, accounting for the force supplied against gravity (in this part, I am not sure whether the sign is correct. Someone PLEASE PLEASE check this):

F= m <0, g>

I can combine to obtain this:

W= ∫ m <x''(t), y''(t) +g> • <x'(t), y'(t)> dt

W= ∫ m *x'(t) *x''(t) + m *y'(t) *y''(t) + mg y'(t) dt

W = 1/2 m x'(t)^2 + 1/2 m y'(t)^2 + mg y(t) [a,b]

This seems to be in agreement with the work-energy theorem, which states that:

W = K = 1/2 mv^2

Now comes my first question. What if I have the same conditions, but instead of assuming that the input force is F = m <x''(t), y''(t)>, I assume that F = <g(x,y), h(x,y)>? If I go through the procedure starting from here:

W= ∫ F • r'(t) dt

Does this integral only tell me the work done by the field? What if I wanted to find the field that supplied the rest of the force?

My second question is, what if I'm given that the initial position of a particle is (0,0), and that the forcing function is F = m <x''(t), y''(t)>? Then do I assume that the particle always travels the path r(t) = <x(t), y(t)>? What if my path isn't time dependent?

Third, what if there is friction along this path? How do I account for this?

Fourth, my book (Giancoli, 4th ed.) states that the "work done by a varying force is":

W= ∫ F • dl = ∫ F cos(θ dl

This is drastically different from my derivation. Can someone please explain this difference? I'm leaning toward the impression that my book is completely wrong, and the author tried to cater to dummies who have no real idea how to operate mathematics.

Fifth, what if the path isn't time dependent?

Someone please help me. I'm pretty desperate, because I have a midterm on this stuff on Thursday (tomorrow). To be precise, in 36 hours.

The problem I am having is conceptual. Say that a particle of mass m starts from rest, and then moves along the path r(t) = <x(t), y(t)>. Does the forcing function always have to equal F(t) = m <x''(t), y''(t)>? (Just asking for confirmation on this part.)

Yes.

To calculate the work done by a force field and a path, the line integral is given by :

W= ∫ (F•T) ds

T is the unit tangent vector, r'(t)/||r'(t)||, and ds is the length, ||r'(t)|| dt. Therefore,


Because of the way I defined F(t), I can plug it in directly:

W= ∫ m <x''(t), y''(t)> • <x'(t), y'(t)> dt

[quote]
Additionally, accounting for the force supplied against gravity (in this part, I am not sure whether the sign is correct. Someone PLEASE PLEASE check this):

You need to define which direction you're taking gravity in with regards to the rest of the system. Then the correct sign for g should just drop out. Obviously its recommended to take g parallel to one of the axis. Normally its defined as in the direction of -y (or - z in 3d), but you need to specify.

My second question is, what if I'm given that the initial position of a particle is (0,0), and that the forcing function is F = m <x''(t), y''(t)>? Then do I assume that the particle always travels the path r(t) = <x(t), y(t)>? What if my path isn't time dependent?

Yes you assume it travels the path r(t) if the path is time dependent. If the path isn't time dependent?
Then you have to consider a path that IS time Dependant to perform the integral, but because of the independence, no matter what path you choose you'll get the same answer. (F would be a conservative force)

Third, what if there is friction along this path? How do I account for this?

If there is friction on the path, generally it will be *hidden* in the F= (term) since if force isn't constant, one of the reasons for that could be the friction.

Fourth, my book (Giancoli, 4th ed.) states that the "work done by a varying force is":

W= ∫ F • dl = ∫ F cos(θ dl

They just expanded the dot product of work , a•b = ab cos(theta), except, here a= F and b=dl

Fifth, what if the path isn't time dependent?

Again, if a path isn't time dependent, you have to create a path that is, but you'll get the same answer regardless of the path you choose.

Yes you assume it travels the path r(t) if the path is time dependent. If the path isn't time dependent?
Then you have to consider a path that IS time Dependant to perform the integral, but because of the independence, no matter what path you choose you'll get the same answer. (F would be a conservative force)

To reinforce this concept, if I'm given that a particle is sliding down the path y=x^2 from a height of 4 to a height of 0 in a uniform gravitational field F=<0,-mg>, then it doesn't matter if I parametralize this as r(t)=<t,t^2>, even though the particle wouldn't really fall at that rate?

If there is friction on the path, generally it will be *hidden* in the F= (term) since if force isn't constant, one of the reasons for that could be the friction.

I spent an hour thinking about this. If the dot product between the force and the unit tangent vector is the magnitude of the projection of the force onto the tangent vector, then the dot product between the force and the unit normal vector is the magnitude of the projection of the force onto the normal vector.

Let's say that a particle is sliding down an irregular ramp. Since friction is F= -μFn, then would the energy lost due to friction be this?

E= -∫ μF • N ds

E= -∫ μF(t) • T'(t)/||T'(t)|| ||r'(t)|| dt

I'm not sure about this, because I've never taken vector calculus before, but I do know that doing this process over a over a surface gives the flux，or the amount of liquid that passes through the surface. My equation seems to be correct for the frictional force, but because of flux, the equation also seems to be counter-intuitive. Also, is there any way to simplify N ds? It seems really messy.

They just expanded the dot product of work , a•b = ab cos(theta), except, here a= F and b=

Could you please elaborate on this? How would you use

W= ∫ F • dl = ∫ F cos(θ dl

on a general curve, say... r(t)=<e^t, t^2>? I thought this equation only works when you're dragging a box on a flat surface and the force or angle the force applied isn't changing.


Thanks a bunch for helping out.

To reinforce this concept, if I'm given that a particle is sliding down the path y=x^2 from a height of 4 to a height of 0 in a uniform gravitational field F=<0,-mg>, then it doesn't matter if I parametralize this as r(t)=<t,t^2>, even though the particle wouldn't really fall at that rate?

Yep, thats the idea, just remember to state what values of t you're using to integrate. For r(t)=<t,t^2> from (2,4) (0,0) you should say t goes from 2 to 0. Not entirely sure a gravitational field is a good example, since g normally is a function of time, but if you're treating it as a constant, go for it.

Let's say that a particle is sliding down an irregular ramp. Since friction is F= -μFn, then would the energy lost due to friction be this?

E= -∫ μF • N ds

The energy loss due to friction is equal to the work done against it, yes.

E= -∫ μF(t) • T'(t)/||T'(t)|| ||r'(t)|| dt

I'm not sure about this, because I've never taken vector calculus before, but I do know that doing this process over a over a surface gives the flux，or the amount of liquid that passes through the surface. My equation seems to be correct for the frictional force, but because of flux, the equation also seems to be counter-intuitive. Also, is there any way to simplify N ds? It seems really messy.

N ds can be simplified, but it depends entirely on the physical system you're looking at. For example, a circle around the origin in the x-y plane, would have a normal vector either in the +z direction, or -z direction, so you could just stick a unit vector (+k or -k) instead of N, and then the equation becomes pretty easy.
There isn't a mathematical method to simplify it in general that I can think of though

Could you please elaborate on this? How would you use

W= ∫ F • dl = ∫ F cos(θ dl

if θ isn't constant? I thought this equation only works when you're dragging a box on a flat surface and the force or angle the force applied isn't changing.

If theta isn't constant, then its a horrible horrible equation. You either need to express dl as a function of theta, or express theta as a function of length. Both are doable, but its very dependent on any given situation as to how you should go about it.

What about drag forces, which are proportional to the velocity squared? It would undoubtedly take this form:

∫ bF • r'(t) dt

except this time, I have absolutely no idea how to express bF. My book doesn't cover it at all (meaning I probably don't need to know this, but I really do want to). My book mentions F=-bv, but only provides treatment for constant velocity. How would you treat both?

Honestly, I have no idea. Had a quick look through all the textbooks I own, and not one of them said anything helpful on that. Sorry

Oh wells. At least I didn't absolutely have to know it.

 By the way

.____.

Oh my god. I just had an epiphany. If F= -bv, then it makes sense to set this as a differential equation.

-b*r'(t) = m*r''(t)

0= r''(t) +b/m *r'(t)

0= r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t)

0= ∫ r''(t)*e^(b/m *t) +b/m *r'(t)*e^(b/m *t) dt

<C1,C2> = r'(t)*e^(b/m *t)

r'(t)= e^(-b/m *t) < C1, C2>

r''(t)= < C1*(-b/m)*e^(-b/m *t), C2*(-b/m)*e^(-b/m *t) >

F= ma

F(t) = < -C1*b*e^(-b/m *t), -C2*b*e^(-b/m *t) >

From here though, I run into three problems. How do I go about finding C1 and C2, which arose from my integration? Also, I understand that I ended up with the same friction function for both the x and the y directions, because I used integrating factors, but what if the particle isn't moving from rest? Finally, this forcing function implies that the force due to friction actually decreases as time goes on, or am I interpreting this wrong?

what if the particle isn't moving from rest?

delta v?

For F= -bv^2, the following:

m r''(t) = b*r'(t)^2

r''(t)/r'(t) = b/m*r'(t)

ln|r'(t)| = b/m*r(t)

r'(t) = e^(b/m*r(t))

1 = r'(t) e^(-b/m*r(t))

t+C1 = (-m/b) e^(-b/m*r(t))

(-b/m)*(t+C1) = e^(-b/m*r(t))

ln|(-b/m)*(t+C1)|= -b/m*r(t)

r(t) = (-m/b)*ln|(-b/m)*(t+C1)|

r'(t) = 1/[(-b/m)*(t+C1)]

r'(t) = m/(-bt+m*C1)

r''(t) = -bm/(-bt+m*C1)^2

However, starting from the bolded step, I could have done things differently to obtain a different answer.

ln|(-b/m)| +ln|(t+C1)|= -b/m*r(t)

r(t) = -m/b*ln|(-b/m)| -m/b*ln|t+C1|

r'(t) = (-m/b) / (t+C1)

r''(t) = (-m/b) / (t+C1)^2

I'm not sure what is going on. They both seem correct, but I'm leaning more toward the bottom one. And like before, I run into trouble dealing with the constant of integration. :/

Oh and I just noticed that the units of all three are entirely off. The only way for the units to work is if C1 and C2 have the units of velocity. Then I'm really lost, because I have to choose a C1 and C2. ;_;

Oh and I just noticed that the units of all three are entirely off. The only way for the units to work is if C1 and C2 have the units of velocity. Then I'm really lost, because I have to choose a C1 and C2. ;_;

I don't have time right now to look through the math, but thats probably a realistic conclusion to come to. When you're solving a differential equation, the most common way to solve them is with a set of boundary conditions, like assuming that your particle starts at rest, which you'd use to find the cconstants.

I went to my professor's office hours. Apparently, he was unable to help me out because, he'd never even heard of the unit normal vector, and he couldn't see how it could arise from the unit tangent vector. Also, he said he was not willing to go through the mathematical complications. I almost wanted to cry.

Anyways, he did have a point. A flat line doesn't have a unit normal vector. So I'm completely lost with this formula:

E= -∫ μF • N ds

To reconcile this, my teacher told me to find θ as a function of position, and then use the formula given in the Giancoli book:

E= -∫ μF cos(θ dl = -∫ μmg cos(θ dl

To do this, I proposed the formula:

tan(θ = -dx/dy

the gradient of a plane curve. My god. He actually didn't believe me when I told him that this is the gradient of a plane curve. He called what I was doing "unnecessary mathematical complication". "I'm sorry. I'm unable to help you. If there's anything regarding tomorrow's quiz, by all means, let me know, but you don't ever need to know this, and it'll just end up looking ugly." So he went to work on his computer, and I just continued my mathematical pursuits on his whiteboard. Plugging in, I obtained:

E = -∫ μmg cos(arctan(-dx/dy)) dl

dl = sqrt(1+(dy/dx)^2) dx

cos(arctan(-dx/dy)) = 1/sqrt(1+(dx/dy)^2)

Therefore:

E = -∫ μmg sqrt(1+(dy/dx)^2)/sqrt(1+(dx/dy)^2) dx

At this point, my professor told me that the equation couldn't be simplified further, and it was so complicated that I might as well just use the approximation technique. Nonetheless, I pursued the question a little further, and I ended up with below. It was quite simple, and either my physics teacher is a moron or he never learned calculus in the first place.

E = -∫ μmg (dy/dx) dx = -μmg y(x)

Note that this came exactly from the formula he told me to use. I did absolutely nothing wrong in between. After I got to this point, some fifth-year student walked into my professor's office to request an appointment about his graduate school plans, and while the fifth-year student said he would wait until the following Monday to ask, my professor insisted that he stayed. The fifth-year student responded, "You look busy. Aren't you helping him out right now?" My professor responds, "No. He's just doing a bunch of ugly mathematics, and I think we're done here, because I can't help him out further." At this point he kicked me out.

This is why, unfortunately, my questions must be pursued here. I really do wish I didn't have to be turning to online help, but it has become my only alternative. None of the people I know have taken vector calculus, and used it in combination with physics.

Getting back on topic, the formula found after simplification just doesn't look correct. I'm not sure why. Maybe because the new formula looks so damn simple, and my teacher told me that all I would end up getting is mathematical garbage. -_-"

But in all seriousness, it doesn't look correct. The reason it doesn't look correct is that the equation that was derived became exactly the same as the potential energy. What if I started and ended at the same height, but the object rolled down some hill and then back up? Then the equation wouldn't even make sense.

One way I might be able to fix this is by finding the local minima of the function and then splitting up the integral into two pieces, but I have no way of proving that this actually provides the correct frictional force. Omfg, I feel so lost. I really do want to cry, but I have no tears to shed. ;_;

 February 17

After studying this some more, I found that on a quarter of a circle, the total friction is:

E= -μmg*cos(π/4)*r*sqrt(2)

The reason is that at π/4, the curve can be split into two parts. The upper eighth of the circle has a lower amount of normal force. The bottom eighth of the circle has a greater amount of normal force. When each point along the arc is added up, they cancel out and reduce to the normal force at π/2. A argument that I devised is below:

 Spoiler

Consider the two triangles. If each of the smaller triangles are 30-60-90s and the outer larger triangle is 45-45-90, then it is easy to show that if a particle travels along the hypotenuses of the two smaller triangles, it will experience the same amount of friction as traveling the hypotenuse as the larger triangle.

To simplify things, each of the 30-60-90 triangles have sides 1, sqrt(3), and 2. Therefore, the outer triangle has two sides 1+sqrt(3) and hypotenuse sqrt(2)*(1+sqrt(3)).

Friction is defined to be μmg*cos(θ, so a particle traveling along the inside path will lose the following amount of heat in total:

2*μmg*cos(π/3) + 2*μmg*cos(π/6)

2*μmg*(1/2) + 2*μmg*(sqrt(3)/2)

μmg*(1+sqrt(3))

The amount of friction lost to the outer path is:

sqrt(2)*(1+sqrt(3))*μmg*cos(π/4)

sqrt(2)*(1+sqrt(3))*μmg*(1/sqrt(2))

μmg*(1+sqrt(3))

Lo and behold, they're the same. If these are the same, then I can generalize this to a circle, because a circle can be constructed by drawing a tangent to all the possible subdivisions that form 90 degree angles. Therefore, the heat lost due to friction is μmg, multiplied by the angle between its endpoints, or cos(π/4), multiplied by the length of the upper arc:

E= -μmg*cos(π/4)*r*sqrt(2)

E= -μmg*r

My new derivation seems correct.

 Spoiler

Assuming these two blue lines are completely the same, only symmetric about the midpoint of the arc, then the magnitudes of the normal forces will add up to be the same.

Also, This seems to be in agreement with the formula I derived earlier in the day (see my previous post):

E = -∫ μmg (dy/dx) dx = -μmg y(x)

If this process could be applied to more general curves, then it vastly simplifies things. My attempt to do this is to subdivide the curve along the inflexion points and the local minima and maxima.

 Spoiler

Is there a way to show that the total friction above or below each of these curves is the following?

E= μmg*cos(angle between the endpoints)*(length of the upper line)

If this is exact, I'll be darned happy.

February 22

I met with my TA and he told me that my process seems true, and that he had never thought of friction in the way I described. However, I did make one error, and that was using Giancoli's E= ∫ F cos(θ dl. My TA told me that this formula is designed for horizontal forces, and the force I wanted was vertical. Therefore, it'd be more logical to substitute sin(θ in the place of cos(θ. Working it out:

E = - ∫ μmg sin(arctan(-dx/dy)) dl

dl = sqrt(1+(dy/dx)^2) dx

sin(arctan(-dx/dy)) = (-dx/dy) /sqrt(1+(dx/dy)^2)

Therefore:

E = -∫ μmg (-dx/dy) sqrt(1+(dy/dx)^2)/sqrt(1+(dx/dy)^2) dx

E = -∫ μmg (-dx/dy) (dy/dx) dx = ∫ μmg dx

E = μmgx

What this means is that the shape of the curve doesn't matter. The sliding friction is purely dependent on the horizontal displacement, which makes sense based on the above. Therefore, my textbook is wrong, and I am happy. Lol. Being happy when my textbook is wrong... is wrong. ._.

Simple question. Okay, not so simple question.

Both momentum and energy must be conserved. Momentum is mv, and kinetic energy is 1/2 mv^2. Just by looking at this, it's not possible that both of them can be conserved, except at very specific velocities. If both quantities are conserved, how come in a collision, momentum is considered the "more important" quantity? Why can kinetic energy just be assumed to disappear into heat? What mathematical basis is there for such an assumption?

Simple question. Okay, not so simple question.

Both momentum and energy must be conserved. Momentum is mv, and kinetic energy is 1/2 mv^2. Just by looking at this, it's not possible that both of them can be conserved, except at very specific velocities. If both quantities are conserved, how come in a collision, momentum is considered the "more important" quantity? Why can kinetic energy just be assumed to disappear into heat? What mathematical basis is there for such an assumption?

You take newton's law that states that F = dp/dt (derivative of momentum on time), after you consider your system of particles entirely. You say there are no external forces (say a collision on vaccum) and clearly dp/dt = 0, so momentum doesn't change.

Also, regarding your problem with two equations and 1 variable (v). You have to see that you are considering multiple particles so you have that the TOTAL momentum and energy is conserved. So you have {v1, v2, v3, ...} with their components on all directions of space (3 equations for momentum) plus the conservation of mechanical energy.

Maybe I should have provided an example.

A car of mass 1 kg, initially moving at 2 m/s, collides into another car of mass 1 kg, initially at rest. After the collision, they stick together, so since momentum is conserved, both cars have to move together at a velocity of 1 m/s. Now let's examine their energies. Before the collision: 1/2 (1 kg) (2 m/s)^2 = 2 J. After the collision: 1/2 (2 kg) (1 m/s)^2 = 1 J. Half the energy just "disappeared"... assumed to disappear into heat, sound, whatever other by-product.

If we do this the other way, and assume that all the energy is conserved, then the final velocities of the cars have to be sqrt(2) m/s, but it's just not... or at least, it's not in most sources. Momentum somehow takes always priority in these calculations.

What is the mathematical basis for this assumption?

Inelastic collision assumes maximum loss of cinetic energy (deformation, heat, etc) since they glue together.

When you decide that they stick together you lose one degree of liberty so you don't need the energy equation anymore, the system is already defined.

Perfect elastic collision assumes conservation of energy.

Inelastic collision assumes maximum loss of cinetic energy (deformation, heat, etc) since they glue together.

When you decide that they stick together you lose one degree of liberty so you don't need the energy equation anymore, the system is already defined.

Perfect elastic collision assumes conservation of energy.

>:[

>:[

What do you want? An hypothetical example is full of assumptions. If you want a true example get two objects and make them collide and take note of their speeds before and after...
Reality is between the two extreme cases of perfect elastic and perfect inelastic. The world PERFECT should have already hinted at that.

Also pineapple at cinetic/kinetic, with 3 different languages to write this I never know which one it is.



For a more mathematical base: With the momentum equation you still have one degree of freedom in your system. So you decide how much of the kinetic energy is conservated, either all (perfect elastic) or minimum (perfect inelastic) or anything inbetween. That decision can be arbitrary or based on experiments (if you are simulating a real system and you have data that shows how "elastic" the collision is).