Sum of torque

dlouth15

This is going to show that the physics doesn't change if we have a rotating arm rather than rotating the first disk.

Define a new set of rotations in terms of the old ones.

Subtract [latex]\omega_1[/latex] from the first disk's rotation bringing it to zero.

[latex]\displaystyle{\omega_{1}^{\prime}=\omega_{1}-\omega_{1}=0}[/latex]

For the second disk we need to add [latex]\omega_1[/latex] as it is a clockwise rotation.

[latex]\displaystyle{\omega_{2}^{\prime}=\omega_{2}+\omega_{1}}[/latex]

For the third disk we need to subtract [latex]\omega_1[/latex] because it is counterclockwise.

[latex]\displaystyle{ \omega_{3}^{\prime}=\omega_{3}-\omega_{1}}[/latex]

Finally we have the rotation of the arm:

[latex]\displaystyle{\omega_{arm}^{\prime}=\omega_{1} }[/latex]

Now using these new rotation rates, disk 1 is stationary and we've adjusted the values of the other rotation rates.

The work done with these new values is:

[latex]\displaystyle{W=Ft\left|\left(r_{1}\omega_{arm}^{\prime}+r_{2}\omega_{arm}^{\prime}\right)-r_{2}\omega_{2}^{\prime}\right| }[/latex]
[latex]\displaystyle{ =Ft\left|r_{1}\omega_{arm}^{\prime}+r_{2}\omega_{arm}^{\prime}-r_{2}\omega_{2}^{\prime}\right|}[/latex]

This is similar to what we found earlier for the rotating arm system.

Now we substitute back our original values.


[latex]\displaystyle{W=Ft\left|r_{1}\left(\omega_{1}\right)+r_{2}\left(\omega_{1}\right)-r_{2}\left(\omega_{2}+\omega_{1}\right)\right|}[/latex]
[latex]\displaystyle{=Ft\left|r_{1}\omega_{1}+r_{2}\omega_{1}-r_{2}\omega_{2}-r_{2}\omega_{1}\right| }[/latex]

and finally simplifying the above we get

[latex]\displaystyle{ W=Ft\left|r_{1}\omega_{1}-r_{2}\omega_{2}\right|}[/latex]

which of course is what we had before for the non-rotating arm system.

In other words changing to a system with the arm rotating and the first disk fixed has made no difference to the calculation of work done against friction. But of course we already knew this since the rotating arm system is merely the fixed arm system when viewed from a rotating frame of reference.

neufneufneuf

I'm agree with the calculation of heating, if the arm rotates or not it's ok.

In the case with the arm is fixed, w1 and w2 counterclockwise, w3 clockwise, I have a work from torque at -FRtw1-Frtw2+kFrtw2-kFrtw3 and H=FRtw1+Frtw2, the sum is at kFrtw2-kFrtw3, it is 0. F4=k|F|

In the case with the arm is rotating, when I count the energy from all torques, I find Ft(rw2-krw2-krw3-(1+k)(R+r)w1+k(R+3r)w1), I made a mistake. The sum of all energy is Ft( 2krw1-krw2-krw3) and like w1=(w2+w3)/2, the sum is 0 without sliding. If there is sliding between the grey disk and the brown disk, there is more energy from heating and the sum Ft( 2krw1-krw2-krw3) is always at 0, no ?

dlouth15

neufneufneuf wrote: »

I'm agree with the calculation of heating, if the arm rotates or not it's ok.

In the case with the arm is fixed, w1 and w2 counterclockwise, w3 clockwise, I have a work from torque at -FRtw1-Frtw2+kFrtw2-kFrtw3 and H=FRtw1+Frtw2, the sum is at kFrtw2-kFrtw3, it is 0. F4=k|F|

In the case with the arm is rotating, when I count the energy from all torques, I find Ft(rw2-krw2-krw3-(1+k)(R+r)w1+k(R+3r)w1), I made a mistake. The sum of all energy is Ft( 2krw1-krw2-krw3) and like w1=(w2+w3)/2, the sum is 0 without sliding. If there is sliding between the grey disk and the brown disk, there is more energy from heating and the sum Ft( 2krw1-krw2-krw3) is always at 0, no ?

Here's a general formula for work done against friction in the situation where both [latex]\omega_1[/latex] and [latex]\omega_{arm}[/latex] are allowed to be non-zero. It is based on the formula Work = Force x distance.

[latex]\displayvalue{W=Ft\left|\omega_{arm}\left(r_{1}+r_{2}\right)+r_{1}\omega_{1}-r_{2}\omega_{2}\right|}[/latex]

If we stop the arm rotating then we have [latex]\omega_{arm}=0[/latex] and

[latex]\displayvalue{W=Ft\left|\omega_{1}-r_{2}\omega_{2}\right|
}[/latex]

Which we agree is correct for the situation of the non-rotating arm.

If we fix disk 1 so [latex]\omega_1=0[/latex] and allow the arm to rotate i.e. [latex]\omega_{arm}>0[/latex] then we have

[latex]\displayvalue{W=Ft\left|\omega_{arm}\left(r_{1}+r_{2}\right)-r_{2}\omega_{2}\right|}[/latex]

Which is what you would expect if you apply work=force X distance to this situation.

Since there's a known relationship between the rotations of disks 2 and 3, we can rewrite if we wish these three equations for work in terms of [latex]\omega_3[/latex]

The relationship is [latex]\omega_3=\omega_2+\omega_{arm}[/latex], so if the arm is not rotating we simply have [latex]\omega_3=\omega_2[/latex] i.e., the disk rotates at the same rate but in the opposite direction to disk 2. If the arm is also rotating we need to add in [latex]\omega_{arm}[/latex] to account for this.

So the general formula becomes

[latex]\displayvalue{W=Ft\left|\omega_{arm}\left(r_{1}+r_{2}\right)+r_{1}\omega_{1}-\left(\omega_{3}-\omega_{arm}\right)\omega_{2}\right|}[/latex]

In the case of the non-rotating arm
[latex]\displayvalue{W=Ft\left|r_{1}\omega_{1}-\left(\omega_{3}-\omega_{arm}\right)\omega_{2}\right|
}[/latex]

In the case of the rotating arm but Disk 1 fixed
[latex]W=Ft\left|\omega_{arm}\left(r_{1}+r_{2}\right)-\omega_{2}\left(\omega_{3}-\omega_{arm}\right)\right|[/latex]

I think where you are making your mistake is that you are adding a contribution from both disk 2 and disk 3 when it should only be one of them. There's a simple mechanical linkage between the two and it doesn't involve the expenditure of work to drive one from the other.

Your intuition might be telling you otherwise but intuition can be misleading in complicated situations. In such situations you have to fall back on the mathematics.

neufneufneuf

You're right, I understood 3 previous case. Thanks

In the case:



At start I have w2 > w1 and w2 = w3. All rotationnal velocities are counterclockwise (labo frame reference). The magenta axis is fixed. The energy from heating is Frt(w2+w3) and the energy from torque is -Frt(w2+w3) the sum is at 0 but the force F6 seems to works on the arm ? How to calculate for have 0 ?

dlouth15

neufneufneuf wrote: »

You're right, I understood 3 previous case. Thanks

In the case:



At start I have w2 > w1 and w2 = w3. All rotationnal velocities are counterclockwise (labo frame reference). The magenta axis is fixed. The energy from heating is Frt(w2+w3) and the energy from torque is -Frt(w2+w3) the sum is at 0 but the force F6 seems to works on the arm ? How to calculate for have 0 ?

I think F6 should be drawn so that it points directly away from the pink axis. I think F4 should be in the equal and opposite direction to F6.

neufneufneuf

True, F3 and F5 have the same origin in the center of axis, like F4 and F6. Forces F1 and F2 come from friction, this gives F5 and F6 on black arm and the reaction of the arm is F4 and F3. There are torques F1/F3 on red disk and F2/F4 on grey disk, these torques works at -Frt(w2+w3). But I see F6 works on black arm at Frtw1. What's wrong ?

dlouth15

neufneufneuf wrote: »

True, F3 and F5 have the same origin in the center of axis, like F4 and F6. Forces F1 and F2 come from friction, this gives F5 and F6 on black arm and the reaction of the arm is F4 and F3. There are torques F1/F3 on red disk and F2/F4 on grey disk, these torques works at -Frt(w2+w3). But I see F6 works on black arm at Frtw1. What's wrong ?

What force does F6 represent? What gives rise to it? How are you calculating it?

In general what problem are you trying to solve here?

neufneufneuf

This was my first case I study before I gave 3 next.

When w2=w3, I think there is friction between the red disk and the grey disk, no ? If yes, for me F2 cannot be alone because the disk can't move in translation, it has an axis, so F2 is apply on axis : this is F6. The arm has a reaction, this reaction is F4. F1 and F4 gives torque on disk. F6 is alone so I think it gives a torque on black arm, but I'm not sure. For me, forces F3, F4, F5, F6 are consequencies of F1 and F2, no ?

neufneufneuf

I understood, in the heating I take (w2+w3) but it is (w2-w1)+(w3-w1), the sum is at 0.

If w1 is clockwise. F6 gives -2Frtw1, the heating gives Frt(w2+w3-2w1), F1 and F2 give -Frt(w2+w3) ?

dlouth15

neufneufneuf wrote: »

This was my first case I study before I gave 3 next.

When w2=w3, I think there is friction between the red disk and the grey disk, no ? If yes, for me F2 cannot be alone because the disk can't move in translation, it has an axis, so F2 is apply on axis : this is F6. The arm has a reaction, this reaction is F4. F1 and F4 gives torque on disk. F6 is alone so I think it gives a torque on black arm, but I'm not sure. For me, forces F3, F4, F5, F6 are consequencies of F1 and F2, no ?

Whilst I don't know exactly what you are getting at, I agree that there will be friction between the two disks.

This will give rise to forces at the axes of the disks and generate a torque. If there's no input torque to the two disks maintaining the rate of rotation of the disks, then they will slow down due to friction.

The forces of friction will create reactive forces acting on the axes of the disks in equal and opposite directions. In the absence of any input torque maintaining rotation of the disks, these forces will also have an effect on the overall rotation of the system. It looks to me that F6 acting on the arm will cause a torque about the pink axis which will cause a change in the rate of rotation of the system itself.

If there's an input torque on the disks maintaining their rotation then other forces need to be drawn in to show this if you want to deal purely with linear forces. You will find that these other forces act to maintain the overall rotation rate of the system.

In general rather than trying to keep track of all the forces, you should use conservation of angular momentum. If friction slows down the disks, then that angular momentum must be preserved so the overall angular momentum of the system is constant.

neufneufneuf

I'm agree with the conservation of angular momentum. But I would like to understand like this too.


Take 2 disks in rotation around blue axis at w1 and in rotation around itself. Blue axes are fixed to the ground. There is friction between 2 disks. w2 >> w1, for example I can take w2=80 and w1=10. Like w2 >> w1, friction gives forces F1 and F2 like I drawn. All rotationnal velocities are in labo frame reference.



The heating is H=2Frt(w1+w2)=80Frt, the work from torques is T=-Frt(w2+w2)=-60Frt. If radius of disk is very small, think with radius like 0.001m and the length of arm at 1 m.

Do you understand this case ?

dlouth15

neufneufneuf wrote: »

The heating is H=2Frt(w1+w2)=80Frt, the work from torques is T=-Frt(w2+w2)=-60Frt. If radius of disk is very small, think with radius like 0.001m and the length of arm at 1 m.

Do you understand this case ?

I don't understand how you are getting -Frt(w2+w2) for heat.

neufneufneuf

For heating it's 2Frt(w1+w2) I think. But the work from torques is -Frt(w2+w2) or 2Frtw2, F1 works at -Frtw2 and F2 works at -Frtw2 too, no ?

dlouth15

neufneufneuf wrote: »

For heating it's 2Frt(w1+w2) I think.

Why do you think that? Please explain how you arrived at that expression.

neufneufneuf

w1 changes the trajectory of points on disks, so I think with big lenght of arm: 1 meter and with a very small radius 0.0001 m for example, like that the difference of trajectories from w1 don't affect points on disks or in a very small value (it's possible to calculate it). In this case w2 must be very high, 100000 for example. If r->0, trajectories are the same and depend only of w2. But H=2Frt(w1+w2) and T=-Frt(w2+w2), there is always a difference of 2Frtw1, it's possible to have r=1e-20 m with a very high w2 but the difference is 2Frtw1, no ?


With big radius, it's possible to look at the difference of trajectories due to w1:

dlouth15

I'm just looking for an explanation for how you arrived at H=2Frt(w1+w2). For example if w2 was zero and the disks were not rotating and therefore no friction, you would have H=2Frt(w1). I don't understand that.

neufneufneuf

w2 is in labo frame reference. And if w2' (arm reference) is at 0, look at the image the slope of the red arrow and the green arrow are not the same (not the horizontal arrows). I'm not sure.

With a different position it's logical no ?

dlouth15

neufneufneuf wrote: »

w2 is in labo frame reference. And if w2' (arm reference) is at 0, look at the image the slope of the red arrow and the green arrow are not the same (not the horizontal arrows). I'm not sure.

With a different position it's logical no ?

But if w2=0 (in the laboratory frame) it means that neither disk is rotating relative to each other. The combination of disks are moving about on the arms but there's never friction between them. Yet you think there is heat produced. You had H=2Frt(w1+w2), but if we stop the rotation of the disks (in the laboratory frame), then H=2Frtw1. How does this make sense?

neufneufneuf

But if w2=0 (in the laboratory frame) it means that neither disk is rotating relative to each other. The combination of disks are moving about on the arms but there's never friction between them. Yet you think there is heat produced.

No, I don't think that, if w2 =0 nothing rotates.

You had H=2Frt(w1+w2), but if we stop the rotation of the disks (in the laboratory frame), then H=2Frtw1. How does this make sense?

take the second position (vertical, it's easier), if w2=30 (labo reference) so w2' (arm reference) =40, no ? With the vertical position, it's logical that w1 don't change the heating. And sure I don't need to add in the equation. But w2 is a labo reference and for have the true w2' I need to add w1. Heating is F*(d1+d2) and d1=rw2'+Rw1 and d2=rw2'-Rw1, here w1 disseppear but for find the work from heating I need to take w2', not w2 ?

dlouth15

neufneufneuf wrote: »

No, I don't think that, if w2 =0 nothing rotates.

If w2=0 then that means that both disks have an angular velocity of zero. That means they are not rotating.

Regardless of how they move about on the arms, one non-rotating disk remains above the other non-rotating disk. There's no slipping of surfaces.

Yet your formula indicates that there's heat produced. How can this be?

Note your formula contains w2, not w'2. So we're talking about the situation where the absolute rate of rotation is zero. Not the rate of rotation relative to the arm.

neufneufneuf

With this position (blue axes are fixed to the ground):



With w2 > w1 (labo reference), you're not agree that there are forces F1 and F2 ?

dlouth15

neufneufneuf wrote: »

With this position (blue axes are fixed to the ground):



With w2 > w1 (labo reference), you're not agree that there are forces F1 and F2 ?

I agree that if w2 is > 0 (doesn't have to be > w1) there will be those forces.

On what basis though are you calculating heat? What I mean is that, for example, work is calculated on the simple basis of work = force x distance. What is the basic physical principle you are using to calculate heat? What is producing the heat?

neufneufneuf

For me heating come from F*d, with d=d1+d2=rw2'+Rw1+rw2'-Rw1=2rw2' note that w2' is in arm reference, R is the lenght of arm, d1=rw2'+Rw1 come from the green disk and d2=rw2'-Rw1 come from the red disk, so like w2'=w2+w1 I write H=2Frt(w2+w1) but maybe it's not a "reality" just for have the good value of rotationnal velocity, for me heating is calculated with rotationnal velocities in ar frame reference, it's false ?

dlouth15

neufneufneuf wrote: »

For me heating come from F*d

But this is the same as the calculation for work. In general (i.e. referring to any specific system), why do you need to have a different formula for heat?

neufneufneuf

It's not a different formula. I use F*d, but for find 'd' I need to find the real velocity. w2 and w1 are labo reference. Look at last image (vertical position) where there is the origin of F2, the velocity is Rw1+rw2' with w2' the rotationnal velocity in arm reference, the point moves at right. The point origin of F1, move at -Rw1+rw2'. The sum of velocities is 2rw2'. Like w2'=w2+w1, the work if 2Frt(w1+w2). If w2=30 counterclockise and w1=10 clockwise, w2'=40. You're not agree with that ?

dlouth15

neufneufneuf wrote: »

It's not a different formula. I use F*d, but for find 'd' I need to find the real velocity.

So although f*d is used for both work and heat produced, the distance d is different - there's a different d for the calculation of heat than the d for the calculation of heat.

In general what, in your understanding, is the essential difference between these two distances? For the moment I'm not interested in this particular system but rather the general principle involved.

neufneufneuf

I can explain heating with formulas. For torque it's a little difficult to explain, each time I use Frtw, with 'w' in the labo reference, it's ok. In the case with 3 disks, I used these reasoning and it's ok, the sum is at 0 because there is a torque on arm. Here, one torque on arm is FR another if -FR, they cancel themselves. You're not agree that the work of a force is Frtw with w in the labo reference ?

dlouth15

neufneufneuf wrote: »

I can explain heating with formulas. For torque it's a little difficult to explain, each time I use Frtw, with 'w' in the labo reference, it's ok. In the case with 3 disks, I used these reasoning and it's ok, the sum is at 0 because there is a torque on arm. Here, one torque on arm is FR another if -FR, they cancel themselves. You're not agree that the work of a force is Frtw with w in the labo reference ?

You seem to be using one value for d for heating and another value of d for work. I'm trying to understand why you are doing this. Not for this system but for any system. I'm trying to understand the general concepts you are using.

So I suppose the general question is this. For any given system involving friction, how do you go about determining the a) work done and b) the heat produced?

I have an answer to this question but I need to hear your answer. We need to agree on the general principles before proceeding with further examples.

neufneufneuf

Yes, I would like to have general formulas. For the works from forces (torques), maybe I need to take in account the direction of forces ? In this case, works from F1 is +Frt(w1-w2') and the work from F2 is +Frt(-w1-w2'), in this case work from torques is -2Frtw2', the sum is 0.

neufneufneuf

Maybe I think it's possible to find in space a point that don't move because w1 is clockwise and w2 counterclockwise. Take this simple example:



Blue axis is fixed to the ground. I give the force F1. I need energy for that but if the point don't move or move less because w1 and w2 are in the contrary direction, I need less energy. F1 decreases kinetic energy to the red disk and F2 increases the kinetic energy of the arm. But all the mass of the red disk can be in the center (red axis). If 'r' is near 0, the kinetic energy of red disk is near 0 but the arm receives a torque from F2.

with the vertical position:



Edit: if r=0 I think inertia is at 0 and F1 will cancel very quickly w2. With the velocity of origin of F1 is 0: F1 works at Frtw2=Ft(dw1+rw1) and F2 works at Fdtw1. For have a velocity at 0 I need to have w1(d+r)-rw2'=0, so it's w2'=(w1(d+r)/r

dlouth15

neufneufneuf wrote: »

Yes, I would like to have general formulas. For the works from forces (torques), maybe I need to take in account the direction of forces ? In this case, works from F1 is +Frt(w1-w2') and the work from F2 is +Frt(-w1-w2'), in this case work from torques is -2Frtw2', the sum is 0.

You should be able to state the general principles involved in reaching your conclusion.

You said earlier that you use force x distance for heat calculations. For work calculations it is also force x distance. You should be able to explain in general terms why you are getting different answers when you are employing the same general formula.

I could easily just say your calculation of work and your calculation of energy is wrong and here is what it should be. But that would not explain why you are wrong. In order to do this, I must get an understanding of what principles you are using to arrive at your calculations.

If [work against friction] = [force due to friction] x [distance] and [energy produced] = [force due to friction] x [distance], then the normal conclusion would be that [work against friction] = [heat produced], i.e. the quantities are the same. Yet you are getting different answers. You must either be defining the force differently or the distance differently for work and heat. But I don't know why. You should be able to answer that in general terms.

neufneufneuf

You should be able to state the general principles involved in reaching your conclusion.

I understood at least the method for find the sum at 0 in each case. The main problem is I think in relative arm reference and after I change for the labo reference. It's a real source of error.

dlouth15

neufneufneuf wrote: »

I understood at least the method for find the sum at 0 in each case. The main problem is I think in relative arm reference and after I change for the labo reference. It's a real source of error.

You are still talking about a specific system here.

neufneufneuf

You're right, just a question about this case (no friction, just a single force):



- length of arm: 'd'
- radius of disk 'r'
- w1 and w2 velocities are in labo frame reference.
- w2′ the rotationnal velocity of the disk in the arm frame reference.

I set w2′ for have the velocity at the origin of F1 at 0 m/s, w1(d+r)−rw2′=0, so w2′=w1(d+r)/r.

An external force F1 is applied a very short period of time, like w1(d+r)−rw2′=0, the velocity =0, the work needed by the force F1 is 0. I define F the value of F1 or F2. F2 appears on arm due to F1. The work from F2 on arm is Fdtw1.

I have 2 choices:

1/ The work from F1 on disk is −Frtw2′=−Ftw1(d+r).

2/ The work from F1 on disk is −Frtw2=−Fdtw1

The case 2/ gives the sum of energy at 0 but for me the force is applied to the angular velocity w2′ (all others cases I studied say it's w2' not w2). Why here it is on w2 and not on w2′ ?

dlouth15

I have to get on with some other things so I won't be responding to this thread for a while after this post.

For the examples involving friction and work, it is important to understand the general principles. That is why I was trying to get you to state what they were before proceeding. If you have in your head different principles, then you will always get different answers.

The two general principles in questions involving friction as I see them are:

1. Work = Force x distance. There's two aspect to this.

The first is that the distance here is the relative motion of the surfaces involved. So you need to identify the surfaces involved in friction and calculate their relative motion. So if surface 1 is moving at [latex]v_1[/latex] in the positive [latex]x[/latex]-direction and surface 2 is moving at [latex]v_2[/latex] in the positive [latex]x[/latex]-direction, then their relative velocity of surface 2 with respect to surface 1 is [latex]v_1-v_2[/latex]. In time [latex]\Delta t[/latex], the surface 2 will have moved [latex]\Delta t\left(v_{2}-v_{1}\right)[/latex] with respect to surface 1.

The second aspect is that the force of friction always opposes motion. So we say that the work done by friction is [latex]W_{friction}=-\Delta t\left|v_{2}-v_{1}\right|
[/latex]. It is a negative figure because the force is in the opposite direction to the motion.

2. The other principle is that the energy given off as heat [latex]E_{heat}=-W_{friction}[/latex]. So [latex]E_{heat}+W_{friction}=0[/latex]. There's never a separate calculation for heat. Once we've calculated the work, we know the heat.

So applying these principles to your earlier example:



1. What is the relative motion? Well I can see straight away that the distance the surfaces travel relative to one another in time [latex]\Delta t[/latex] is

[latex]\displayvalue{2\omega_{2}r\Delta t}[/latex]

It doesn't matter how the disks move around the arms because it is the relative motion of the surfaces that count.

So the work done by friction is

[latex]\displayvalue{W_{friction}=-2Fr\omega_2\Delta t}[/latex]

The second principle tells us that [latex]E_{heat}=-W_{friction}[/latex] so

[latex]E_{heat}=2Fr\omega_2\Delta t[/latex]

And [latex]E_{heat}+W_{friction}=0[/latex]

---

For your final example:


I think you want to know what is the work done by the force [latex]F[/latex] on the system.

Again the general principle is Work = Force x distance. We're given the force so we only need to work out the distance.

In a small interval of time [latex]\Delta t[/latex] the point where the force is applied will have moved

d = [distance due to rotation of the arm] + [distance due to rotation of the disk]

[latex]d=\left[\left(r_{1}+r_{2}\right)\omega_{1}\Delta t\right]+\left[-r_{2}\omega_{2}\Delta t\right][/latex]

where [latex]\omega_1[/latex] is the rotation of the arm and [latex]\omega_2[/latex] is the rotation of the disk. [latex]r_1[/latex] and [latex]r_2[/latex] are the radii of the arm and disk respectively.

Therefore the work is
[latex]\displayvalue{W=F\Delta t\left(\left(r_{1}+r_{2}\right)\omega_{1}-r_{2}\omega_{2}\right)}[/latex]

Again this assumes that there's some mechanism maintaining constant angular velocity of the disk and the arm.

The sum of energy is zero because this mechanism is doing work to maintain the constant angular velocity of the system.

[latex]W_{force}=-W_{mechanism}[/latex] and so

[latex]W_{force}+W_{mechanism}=0[/latex].

If there's no mechanism maintaining the angular velocities then there will be a change in the total kinetic energy [latex]\Delta E_{kin}[/latex] of the system and

[latex]W_{force}+\Delta E_{kin}=0[/latex].

---

I think that is correct. But if I am wrong it should be fairly easy to spot the error because I have explained my reasoning in the post and the principles involved.

Although you give answers which are sometimes wrong, in my opinion, it is very hard for me to say why they are wrong since I usually don't know the reasoning behind them. I don't know what principles you are employing to solve the problem. It may be that you don't have any and are simply making things up as you go along. That is why I was asking you to state the principles and explain your reasoning.

It is easier for me to ignore your answers and simply solve the problem directly. That is why I have ignored whatever answer you have given for the calculation work in this last example.

neufneufneuf

The black arm is turning clockwise at w1. The blue axis is fixed to the ground. The grey cruz (grey arms) is turning at w2, with w2 < w1, so the grey cruz is turning around itself counterclockwise on the black arm reference. Each red disk is turning at w3, with w3 < w2. For example, w1=10, w2=9 and w3=1. Friction from disk/disk gives black forces. The rotationnal velocity of each red disk on grey arm reference is w3' with w3'= w2-w3 = 8, so each disk is turning counterclockwise around itself on the grey arm reference. So w3 can give forces like I drawn. These black forces give a torque on each disk and give a torque on the grey cruz. Each red disk increases its kinetic energy, because the torque is clockwise and w3 < w1. The black arm don't have a torque, so w1 is constant. Green forces reduces w2. BUT, and this is The Idea: the inertia of grey cruz can be like I want and don't depend only of the disks, I can add a big mass in the center of the cruz for example. Remember, torque=inertia*acceleration and kinetic energy is 1/2*inertia*w², so kinetic energy is 1/2w²/inertia. If inertia of cruz is very high the rotationnal velocity of the cruz don't decrease (imagine a mass very high), or very few in practise. If I increase the inertia of the cruz, I change w1 ? no. I change w3 ? no. I change the work from friction ? no. It is an independant parameter.

a/ w1 is constant
b/ I can choose the inertia of the grey cruz as high I want (in the center of the cruz)
c/ The heating is a positive energy
d/ The kinetic energy of each red disk increases
e/ I can reduce the lost of kinetic energy from grey cruz by adding a mass in the center of the cruz


Cycle:

1/ Friction is OFF. Launch the device with w1, w2, w3 with external motor
2/ No external motor or force
3/ Set friction ON
4/ Measure the sum of energy, heating too, for me the energy increases, the heating and the kinetic energy.

dlouth15

neufneufneuf wrote: »

Thanks for your help Dlouth15. Take the last example, a torque is apply to the disk it works at Frtw2=Fdtw1 because I set w2=d/r*w1. F2 works on the arm at -Fdtw1. The sum is 0. But, the inertia of the disk is not the same than the inertia of the arm+disk, so with the formula of Newton: T=Iα, if the inertia is not the same the acceleration is not the same too. At start, ok, the sum of energy is at 0, but with time, rotationnal velocities are not the same and the work from torque are more and more different. Are you agree with that ?

Why don't you try and work it out yourself?

But I would recommend that you first state the problem you are trying to solve in mathematical form. What hypothesis are you trying to test? What quantities do you need to determine in order to test the hypothesis.

Secondly I would suggest that you clearly state the principles you are using to determine those quantities. An example of a principle might be torque = rate of change of angular momentum. Another might be work = force x distance. And so on.

Finally, you should explain how you are applying those principles to the solution. For example, if you are using work=force x distance, explain how you are defining the force or the distance in each situation where you are applying that principle.

I have tried to do this in my example earlier. It means that if there is an error, it is easier to pinpoint and correct. It also means that someone else reading it doesn't have to try to solve the problem independently as I've done most of the work.

Many forums don't allow people to post only problems and expect others to solve them. They want some evidence that the person has attempted to solve the problem themselves.

However just putting up a final answer isn't sufficient evidence. It could simply be a random guess. In order for others to figure out whether it is right or wrong, they have to solve the problem from scratch.

Better therefore that you show your working in detail. You will find you get much better responses from people.

neufneufneuf

Thanks for your help. Before to try I prefer to understand how the physics works, but I understood the equation and I know how to calculate the sum of energy with multiple rotations. I need to explain more sure and give calculations in the same time. Even I try to understand physics without formulas, I think it's important if I want other people want to verify and it allow to understand the "device" like I thought it is.

Sorry, I edited the last question.

dlouth15

neufneufneuf wrote: »

Thanks for your help. Before to try I prefer to understand how the physics works, but I understood the equation and I know how to calculate the sum of energy with multiple rotations. I need to explain more sure and give calculations in the same time. Even I try to understand physics without formulas, I think it's important if I want other people want to verify and it allow to understand the "device" like I thought it is.

The problem is that showing that there's no change in overall energy of the system requires a lot of work. It can only be done mathematically. I might believe that there's unaccounted for energy created or lost in a system, but it means nothing until I have analyzed everything mathematically. There is no real understanding of the physics that doesn't involve mathematics.

I am prepared to help with the simple example of the disk on the end of the arm. I have redrawn it below:




The force [latex]F[/latex] of constant magnitude is applied over an extended period of time. The point of application is always at the far point of the disk from the axis of the arm and follows the arm-disk assembly as it moves. The disk is of mass [latex]m[/latex].

Positive [latex]\omega[/latex] represents rotation in the clockwise direction. It doesn't mean all rotations have to be clockwise but rather that counterclockwise rotations will have a negative value for [latex]\omega[/latex].

Now the problem you are trying to solve, as I understand it, is whether or not the sum of energy is constant over time. To solve it, therefore, we need to calculate the work done [latex]W\left(t\right)[/latex], the rotational kinetic energy of the disk [latex]E_{disk}\left(t\right)[/latex] and the rotational kinetic energy of the arm with disk attached, [latex]E_{arm}\left(t\right)[/latex], all as a function of time.

The hypothesis is then that

[latex]\displayvalue{W\left(t\right)+E_{disk}\left(t\right)+E_{arm}\left(t\right)=\mathrm{constant}}[/latex]

Of course conservation of energy says that the sum of these must always be constant, but we can't assume this. We have to arrive at our conclusion using forces, torques, angular momentum and so on.

So why don't you have a go at calculating these values. I will help spot errors if you explain your working as you go along. I will also try to do bits of the problem when I have time. When it is complete we can apply the same principles to your more complex diagram with the four disks on the arm and friction. But we need to be able to do this simpler one first.

neufneufneuf

Ok, I calculated all energies for the device with 4 red disks.

Edit: I just see you're editing you're question, sure a lot of time I'm wrong with the trajectory, if trajectory is false, force can't be in this direction.

1/ I define the device:



Blue axis is fixed to the ground.

In the labo frame reference:

[latex]w_1[/latex]: the rotationnal velocity of the black arm
[latex]w_2[/latex]: the rotationnal velocity of the grey cruz
[latex]w_3[/latex]: the rotationnal velocity of each red disk

There is friction between red disks. Like [latex]\omega_2 < w_1[/latex] and [latex]\omega_3 < \omega_2[/latex]. The grey cruz is turning around itself counterclockwise in the black arm reference and each red disk is turning around itself counterclockwise in the grey cruz reference.

For example, I can have [latex]\omega_1=10[/latex], [latex]\omega_2=9[/latex] and [latex]\omega_3=1[/latex].

There are black forces on each disk like that:



Each disk receives a torque and the rest is for the grey cruz: I drawn green forces the sum of forces from friction that the grey cruz receives.

Look if there isn't a torque on black arm, for me no, but maybe I'm wrong. I think I can have no torque on black arm, only if I adjust friction at each time. Because outer green force work more than at inner radius. Even I can't have no torque on black arm, I calculated the kinetic energy of the black arm. At start I can take w2=w1, like that the grey cruz don't turn in the black arm reference.

The cycle:

1/ Friction is OFF between disks. I launch the device with [latex]\omega_1[/latex], [latex]\omega_2[/latex], [latex]\omega_3[/latex] with an external motor
2/ I cut off the external motor, so there is no external force.
3/ I Set friction ON between disks but no friction elsewhere
4/ I Measure the sum of energy, the heating too, all the energy must be constant but for me the energy increases, the heating and the kinetic energy.

Note, for simplify calculations, I guess the force from friction is always the same F even rotationnal velocity of red disks decreases in the grey cruz reference.

[latex]\omega_{2'}[/latex] is the rotationnal velocity of the grey cruz in the black arm frame reference
[latex]\omega_{3'}[/latex] is the rotationnal velocity of a disk in the grey cruz frame reference

I define :

Inertia of each red disk : [latex]I_3[/latex]
Inertia of the grey cruz is [latex]I_2=1000000I_3[/latex], look later why I choose [latex]I_2[/latex] very high
inertia of the black arm is [latex]I_1[/latex]
[latex]N[/latex], the number of disk = [latex]4[/latex]
[latex]r[/latex] : the radius of a red disk
[latex]d[/latex] : the length of the black arm

---

Friction

---

The energy from friction between each disk is : [latex]F*d[/latex], the force by length (here d is a generic letter not the length of the arm)

[latex]2Fr\omega_{3'm}[/latex], with [latex]\omega_{3'm}[/latex] the mean of [latex]\omega_{3'}[/latex]

For 4 disks it is :

[latex]4*2Fr\omega_{3'm} = 8Fr\omega_{3'm}[/latex]

I don't calculated the mean of [latex]\omega_{3'm}[/latex], look at the final difference of energy why.

---

Kinetic energy of red disks

---

Each disk receive a torque : [latex]Fcos(\frac{\pi}{4})2rcos(\frac{\pi}{4}) = Fr[/latex]

The rotationnal velocity of each disk change like :

[latex]\omega_3(t) = \omega_3[/latex] + [latex]\frac{Fr}{I_3}t[/latex]

The kinetic energy of a red disk change like :

[latex]Ek=+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_3}t)^2[/latex]

For 4 disks it is :

[latex]Ek=4*(+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_3}t)^2)[/latex]

---

Kinetic energy of the grey cruz

---

The grey cruz receive from each disk a torque of : [latex]2Fcos(\frac{\pi}{4})d [/latex]

The grey cruz receive from 4 red disks a torque of : [latex]8Fcos(\frac{\pi}{4})d [/latex]

The rotationnall velocity of the grey cruz change like :

[latex]\omega_2(t)=\omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t[/latex], note with [latex]I_2[/latex] very high, [latex]\omega_2[/latex] is decreasing very few.

The kinetic energy of the grey cruz change like :

[latex]Ek=\frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2[/latex], note the kinetic energy can be near the same if [latex]I_2[/latex] is very high.

---

Black arm

---

The rotationnall velocity of the black arm change like :

[latex]\omega_1(t)=\omega_1-\frac{F_1}{I_1}t[/latex], note with [latex]I_1[/latex] very high, [latex]\omega_1[/latex] is decreasing very few.

The kinetic energy of the black arm change like :

[latex]Ek=\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex], note the kinetic energy can be near the same if [latex]I_1[/latex] is very high.

I don't calculated [latex]F_1[/latex], but even I can't have [latex]F_1=0[/latex], the energy lost by black arm is very low

---

The difference of energy

---

At final the difference of energy is :

[latex]Ekf = + 8Fr\omega_{3'm}[/latex] + [latex](4\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_1}t)^2- 4\frac{1}{2}I_3\omega_3^2)[/latex] + [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex] + [latex]\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex] - [latex]\frac{1}{2}I_1 \omega_1^2[/latex]

That I see it's the last term : [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2) [/latex] + [latex]\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex] - [latex]\frac{1}{2}I_1 \omega_1^2[/latex] can be very small due to [latex]I_2[/latex]. I calculated with numerical values, I think something is wrong here. Even, I need to calculate [latex]\omega_{3'm}[/latex], the last term can be ignored and heating gives energy and red disks increase their kinetic energy.

dlouth15

neufneufneuf wrote: »

Ok, I calculated all energies for the device with 4 red disks.

Edit: I just see you're editing you're question, sure a lot of time I'm wrong with the trajectory, if trajectory is false, force can't be in this direction.

Yes, I noticed that the diagram was not appearing, so I put that in. Without the diagram it would not be clear which system I was talking about. I would prefer if possible to deal with the simpler example, however I will look at your work for the four disks some time tomorrow.

neufneufneuf

Ok, I will calculate for the simpler case tomorrow morning and post my result. I changed my last message because I'm not sure if there is a torque on black arm or not, even there is a torque, for me the kinetic energy don't change if I1 is very high.

dlouth15

I'll just make some comments in red. In another post I will put my working of the problem.

neufneufneuf wrote: »

---

Friction

---

The energy from friction between each disk is : [latex]F*d[/latex], the force by length (here d is a generic letter not the length of the arm)

[latex]2Fr\omega_{3'm}[/latex], with [latex]\omega_{3'm}[/latex] the mean of [latex]\omega_{3'}[/latex]

For 4 disks it is :

[latex]4*2Fr\omega_{3'm} = 8Fr\omega_{3'm}[/latex]

I don't calculated the mean of [latex]\omega_{3'm}[/latex], look at the final difference of energy why.

I think this is OK.

---

Kinetic energy of red disks

---

Each disk receive a torque : [latex]Fcos(\frac{\pi}{4})2rcos(\frac{\pi}{4}) = Fr[/latex]

The rotationnal velocity of each disk change like :

[latex]\omega_3(t) = \omega_3[/latex] + [latex]\frac{Fr}{I_3}t[/latex]
Ok I think

The kinetic energy of a red disk change like :

[latex]Ek=+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_3}t)^2[/latex]

For 4 disks it is :

[latex]Ek=4*(+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_3}t)^2)[/latex]
OK i think.

---

Kinetic energy of the grey cruz

---

The grey cruz receive from each disk a torque of : [latex]2Fcos(\frac{\pi}{4})d [/latex]

The grey cruz receive from 4 red disks a torque of : [latex]8Fcos(\frac{\pi}{4})d [/latex]
This could well be correct but I'm not sure.


The rotationnall velocity of the grey cruz change like :

[latex]\omega_2(t)=\omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t[/latex], note with [latex]I_2[/latex] very high, [latex]\omega_2[/latex] is decreasing very few.
I think this is probably OK if the previous formula is OK.

The kinetic energy of the grey cruz change like :

[latex]Ek=\frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2[/latex], note the kinetic energy can be near the same if [latex]I_2[/latex] is very high.

Again this is probably ok if the formula involving the cos is OK.

---

Black arm

---

The rotationnall velocity of the black arm change like :

[latex]\omega_1(t)=\omega_1-\frac{F_1}{I_1}t[/latex], note with [latex]I_1[/latex] very high, [latex]\omega_1[/latex] is decreasing very few.

The kinetic energy of the black arm change like :

[latex]Ek=\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex], note the kinetic energy can be near the same if [latex]I_1[/latex] is very high.

I don't calculated [latex]F_1[/latex], but even I can't have [latex]F_1=0[/latex], the energy lost by black arm is very low

I don't think the black arm's rotational velocity should change at all, since all the forces balance out. As such it won't lose or gain any energy.

---

The difference of energy

---

At final the difference of energy is :

[latex]Ekf = + 8Fr\omega_{3'm}[/latex] + [latex](4\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{Fr}{I_1}t)^2- 4\frac{1}{2}I_3\omega_3^2)[/latex] + [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex] + [latex]\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex] - [latex]\frac{1}{2}I_1 \omega_1^2[/latex]

I think the problem here is that the terms of the equation are made up of different variables. For example the first term is using relative angular velocities whereas the other terms are using absolute angular velocities. Even if the equation balanced out you would not see this from the equation as written above.

That I see it's the last term : [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fcos(\frac{\pi}{4})d}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2) [/latex] + [latex]\frac{1}{2}I_1( \omega_1-\frac{F_1}{I_1}t )^2[/latex] - [latex]\frac{1}{2}I_1 \omega_1^2[/latex] can be very small due to [latex]I_2[/latex]. I calculated with numerical values, I think something is wrong here. Even, I need to calculate [latex]\omega_{3'm}[/latex], the last term can be ignored and heating gives energy and red disks increase their kinetic energy.

The next post will give my thinking on the problem.

dlouth15

Ok, this is my working of the problem of the four disks. Instead of dealing with four disks and then at every stage multiplying by 4, it is easier to deal with one disk. If there's no energy lost with one disk then there won't be with 4.

Also since the long arm doesn't change its velocity we can forget about that and not include it in the diagram.

The problem then reduces to the diagram below. The disk is subjected to a constant torque [latex]\tau[/latex] by some mechanism in the arm. It is similar but not identical to a diagram I put up earlier.



Work done by torque in the interval [latex]\Delta t[/latex]

For a single rotating disk subjected to torque [latex]\tau[/latex] in a short interval of time [latex]\Delta t[/latex] is

[latex]\displaystyle{\Delta W=-\tau_{disk}\omega_{disk}\Delta t }[/latex]

However since the mechanism which is doing the work is itself rotating we need to adjust for this this:

[latex]\displayvalue{\Delta W=-\tau_{disk}\left(\omega_{disk}-\omega_{arm}\right)\Delta t}[/latex]

It is a negative figure because it is work input into the system.

Change in angular momentum of the components of the system.

Torque = rate of change of angular momentum, so for the disk

[latex]\displaystyle{\tau=\frac{\Delta L_{disk}}{\Delta t} }[/latex]

and so

[latex]\displaystyle{\Delta L_{disk}=\tau\Delta t }[/latex]

From the conservation of angular momentum we can also say that

[latex]\displaystyle{\Delta L_{arm}+\Delta L_{disk}=0 }[/latex],

And therefore
[latex]\displaystyle{ \Delta L_{arm}=-\Delta L_{disk} }[/latex]

We will use these results later.

Change in kinetic energy of the system

The change can be expressed as

[latex]\displaystyle{ \Delta E=E\left(t+\Delta t\right)-E\left(t\right) }[/latex]

i.e. the change is the the difference between the energy before after the small interval of time [latex]\Delta t[/latex] and the energy before. This kinetic energy of the system is composed of the kinetic energy of the disk rotating about itself [latex]E_{disk}[/latex] and also of the assembly of the rotating arm and non-rotating disk [latex]E_{arm}[/latex].

Therefore the change in kinetic energy is

[latex]\displaystyle{ \Delta E=\left[E_{disk}\left(t+\Delta t\right)-E_{disk}\left(t\right)\right]+\left[E_{arm}\left(t+\Delta t\right)-E_{arm}\left(t\right)\right]=\Delta E_{disk}+\Delta E_{arm}. }[/latex]

We will work out the terms of this equation below.

The rotational kinetic energy for the disk at time [latex]t[/latex] is given by

[latex]\displaystyle{ E_{disk}\left(t\right)=\frac{1}{2}I_{disk}\omega_{disk}^{2} }[/latex]

and at time [latex]t+\Delta t[/latex] it is

[latex]\displaystyle{ E_{disk}\left(t+\Delta t\right)=\frac{1}{2}I_{disk}\left(\omega_{disk}+\Delta\omega_{disk}\right)^{2} }[/latex]
So combining these two and expanding the square

[latex]\displaystyle{\Delta E_{disk}=E_{disk}\left(t\right)-E_{disk}\left(t+\Delta t\right)=I_{disk}\left(\omega_{disk}\Delta\omega_{disk}+\left(\Delta\omega_{disk}\right)^{2}\right) }[/latex]

as the time interval becomes very small, [latex]\Delta\omega_{disk}[/latex] becomes very small and so [latex]\left(\Delta\omega_{disk}\right)^{2}[/latex] vanishes. So

[latex]\displaystyle{\Delta E_{disk}=I_{disk}\omega_{disk}\Delta\omega_{disk} }[/latex]

Since [latex]I_{disk}\Delta\omega_{disk}=\Delta L_{disk}[/latex] ,

[latex]\displaystyle{\Delta E_{disk}=\Delta L_{disk}\omega_{disk} }[/latex]

Similarly for the arm + non rotating disk component of the kinetic energy

[latex]\displaystyle{\Delta E_{arm}=\Delta L_{arm}\omega_{arm} }[/latex]

but from conservation of angular momentum, [latex]\Delta L_{arm}=-\Delta L_{disk}[/latex] and so

[latex]\displaystyle{\Delta E_{arm}= -\Delta L_{disk}\omega_{arm} }[/latex]

We have now worked out all the terms of the earlier equation for the kinetic energy of the system and so

[latex]\displaystyle{\Delta E=\Delta E_{disk}+\Delta\omega_{arm}=\Delta L_{disk}\omega_{disk}-\Delta L_{disk}\omega_{arm}=\Delta L_{disk}\left(\omega_{disk}-\omega_{arm}\right) }[/latex]

Since [latex]\Delta L_{disk}=\tau_{disk}\Delta t[/latex]

[latex]\displaystyle{ \Delta E=\tau_{disk}\Delta t\left(\omega_{disk}-\omega_{arm}\right) }[/latex]

Now going right back to the start recall that [latex]\Delta W=-\tau_{disk}\left(\omega_{disk}-\omega_{arm}\right)\Delta t[/latex]

[latex]\displaystyle{ \Delta W+\Delta E=-\tau_{disk}\left(\omega_{disk}-\omega_{arm}\right)\Delta t+\tau_{disk}\Delta t\left(\omega_{disk}-\omega_{arm}\right)=0 }[/latex]

Conclusion

The work put into the system equals the increase in kinetic energy of the system as expected. The sum of work plus kinetic energy over time is constant.

This result holds for any value of rotational velocity or torque. Each can be either positive or negative. Rotations can be clockwise or counterclockwise as can torque.

Generalizing to the 4 disk diagram

Although we haven't calculated the torque based on friction we know that each disk will be subject to a torque. This torque will be the same for all disks. Since the result we found earlier is valid for all torques whatever the value, the result is also valid for the situation with four disks.

neufneufneuf

I don't know how to used the angular momentum. I never used it. I will learn the method of angular momentum for understand your calculations, but I need a little time.

---

---

For one disk:

I give my calculations with force, torque, kinetic energy. Like that I can understand where come from my error. I used the same definitions of velocities than in the 4 disks: w3 and I3=>the disk, w2 and I2=>the arm.

The additionnal kinetic energy for the disk:

[latex]\frac{1}{2}I_3(\omega_3[/latex]+[latex]\frac{Fr}{I_3}t)^2-\frac{1}{2}I_3\omega_3^2=\frac{1}{2}I_3(\omega_3^2[/latex]+[latex]\frac{F^2r^2t^2}{I_3^2}[/latex]+[latex]2\omega_3\frac{Frt}{I_3})-\frac{1}{2}I_3\omega_3^2=Frt\omega_3[/latex]+[latex]\frac{F^2r^2t^2}{2I_3}[/latex]

The additionnal kinetic energy for the arm+disk:

[latex]\frac{1}{2}I_2(\omega_2-\frac{Fr}{I_2}t)^2-\frac{1}{2}I_2\omega_2^2=\frac{1}{2}I_2(\omega_2^2[/latex]+[latex]\frac{F^2r^2t^2}{I_2^2}-2\omega_2\frac{Frt}{I_2})-\frac{1}{2}I_2\omega_2^2=-Frt\omega_2[/latex]+[latex]\frac{F^2r^2t^2}{2I_2}[/latex]

The energie from friction:

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]\int_0^t{Fr((\omega_2-\frac{Frt}{I_2})-(\omega_3+\frac{Frt}{I_3})dt}[/latex] = [latex]\int_0^t{Fr\omega_2-\frac{F^2r^2t}{I_2}-Fr\omega_3-\frac{F^2r^2t}{I_3}dt}[/latex] = [latex]Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3}[/latex]

The sum is well at 0.

---

---

For 4 disks:

I can use the same last calculation :

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]8\int_0^t{Fr((\omega_2-\frac{Frt}{I_2})-(\omega_3+\frac{Frt}{I_3})dt}[/latex] = [latex]8\int_0^t{Fr\omega_2-\frac{F^2r^2t}{I_2}-Fr\omega_3-\frac{F^2r^2t}{I_3}dt}[/latex] = [latex]8(Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3})[/latex]

Without the black arm it is :

[latex]Ekf = 8(Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3} [/latex]) + [latex](2I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2- 2I_3\omega_3^2)[/latex] + [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex]



[latex]Ekf = 28\frac{F^2r^2t^2}{I_2}[/latex]+[latex]4\frac{F^2r^2t^2}{I_3} [/latex]

Note [latex]d*cos(\frac{\pi}{4})=r[/latex]

Thanks again for your help Dlouth15

dlouth15

neufneufneuf wrote: »

I don't know how to used the angular momentum. I never used it. I will learn the method of angular momentum for understand your calculations, but I need a little time.

For one disk:

I give my calculations with force, torque, kinetic energy. Like that I can understand where come from my error. I used the same definitions of velocities than in the 4 disks: w3 and I3=>the disk, w2 and I2=>the arm.

I don't know why you are bothering with the black arm. If you imagine the 4 disks on the crossed arms on their own without the black arm. If you apply friction, you would not expect this assembly to move in any particular direction. Therefore when it is attached to the black arm, you would not expect the assembly to affect the rotation of the black arm. Therefore its contribution to the kinetic energy of the system is constant over time. When we are looking at changes in kinetic energy due to friction between the disks it will not be a factor in any calculation and can be dropped.

Once you have shown that the work due to friction equals the change in kinetic energy for one disk, you have shown that it is also true for four disks.

neufneufneuf

w2 and I2=>the arm.

sorry it's w2 and I2=>the grey cruz. I don't consider the black arm, with or without it the result is the same. I done all calculations, and I changed my message a lot of times because there were a lot of errors, but now I find the sum at 0 with one disk. With 4 disks too, but only if I consider I2 very high, if not there is a little error in my calculations I think.

dlouth15

neufneufneuf wrote: »

sorry it's w2 and I2=>the grey cruz. I don't consider the black arm, with or without it the result is the same. I done all calculations, and I changed my message a lot of times because there were a lot of errors, but now I find the sum at 0 with one disk. With 4 disks too, but only if I consider I2 very high, if not there is a little error in my calculations I think.

See my remark about the arm. It shouldn't change its angular velocity during the friction process, so it is the same before and after friction is applied. So it doesn't need to be considered in calculations of the change in energy regardless of its moment of inertia.

dlouth15

neufneufneuf wrote: »

I don't know how to used the angular momentum. I never used it. I will learn the method of angular momentum for understand your calculations, but I need a little time.

The general idea is conservation of angular momentum. If you have a system which causes a torque - and therefore a change in angular momentum in one of its components, then there must be a corresponding change in the other direction in one or more of the other components in order that the total angular momentum of the system remains unchanged. In the single disk example, the angular momentum of the disk is changed by the torque so we must have an equal and opposite change in the angular momentum the arm.

Once we have done this, we can work out the new rotation rates of arm and the disk and therefore the new total kinetic energy of the system.

neufneufneuf

I'm agree with you, and in my calculations I don't take the black arm in the equations. I done all calculations with only 4 red disks and the grey cruz. I find 0 for one disk, and for 4 disks but I have a little error in my last sum if I consider I2 not infinite. I think the integrale is correct because I found 0 for one disk, no ?