Sum of torque

dlouth15

neufneufneuf wrote: »

You're right and it's for that I'm looking for a case in physics for understand where Algodoo find energy (maybe Algodoo is wrong but I hope no). I think about a case that I explain here:

http://physics.stackexchange.com/questions/143377/one-disk-ring-in-double-rotation-and-sum-of-energy

if a torque can exist on ring, it's won, the system create energy and my calculations shows trajectories can be like I think, but maybe it's an error...I'm waiting for an answer.

Build it. Physically construct the thing. You will have your answer.

neufneufneuf

but the energy won is very low...before construct something I prefer understand the theory and verify with maths if it's ok or not.

It's seems it's not possible to have trajectories like I want. But it's possible to add N system in parallel. Each system give two energy E1 in heating and energy from disk E2. Sum for each disk is 2E1+E2. With N system this will give N(2E1+E2). I need to add energy at 2 last systems, need 2E1, so the sum of energy is N(2E1+E2)-2E1.

Morbert

I'll let this thread run for a little longer but I'm beginning to question its appropriateness for this board.

dlouth15

neufneufneuf wrote: »

but the energy won is very low...before construct something I prefer understand the theory and verify with maths if it's ok or not.

I see someone's answered the question of the total energy of your system in another thread on stack exchange.

http://physics.stackexchange.com/questions/143715/is-this-expression-for-the-kinetic-energy-of-a-spinning-disk-revolving-about-a-s/144043#144043

[latex]\displaystyle{E_{total}&=\frac12 m (\omega_1 d)^2 + \frac12 m (\omega_1-\omega_2)^2 r^2 }[/latex]

To me this looks correct.

I think neufneufneuf's point on stackexchange is that if you attach some sort of braking system to the apparatus to lower the rotation of the wheel thereby reducing [latex]\omega_2[/latex], the rotation of the wheel relative to the arm, then the overall system appears to have gained energy according to that equation.

I think what is being left out here is conservation of angular momentum. If you apply a brake to lower [latex]\omega_2[/latex] then the angular momentum will be transferred to the arm assembly and [latex]\omega_1[/latex], the rotation of the arm, will also be decreased.

When you calculate the effect of this and plug the result back into the equation you find that the energy has actually gone down.

neufneufneuf

In this case, with:



Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Fdw3t with w3 the mean of w2

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforce=−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Fdw3t−2Fdw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Fdw3t

dlouth15

neufneufneuf wrote: »

In this case,

I don't see that as being any different from the simpler example. You are still using a braking mechanism (friction among the rings) to slow down the rotation of the rings. This will have an effect of also reducing the rate of rotation of the entire assembly due to conservation of angular momentum and overall the energy of the system will reduce.

You can counteract this by inputting mechanical energy to rings at the ends of the system but all you have done is turned mechanical energy into heat.

I think in general it is very unlikely that overunity will be found in mechanical systems such as this. If it were the case, it would have been found by now purely by accident.

A more promising avenue to explore might be renewable sources of energy such as wind, solar etc. They don't create energy out of nothing but they do make use of freely available natural sources. Another area would be energy conservation, making do with less energy.

neufneufneuf

With one disk, if a torque is applied energy is constant. Here with N systems there is energy from friction, with (N-1) coeficient, energy of kinetics disks increase with N coeficient, F1 and F2 don't need energy, there are only 2 magenta forces that need energy and it is not a coeficient with N. Even magenta forces need more energy (I'm not sure about my calculations) I can choose N at 1000.

I made a mistake in my calculations:

Works of forces

F is the value of green or magenta force

wdisks=+N1/2mr²((w1−w2i)²−(w1−w2f)²)

with w2f<w2i, w2f=w2 at final, w2i=w2 initial

Wfriction=2(N−1)Frw3t with w3 the mean of w2 **************************** I noted 'd' but it is 'r'

WF1=2dF−2dF=0
WF2=2dF−2dF=0

Wmagentaforces=−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−1)Frw3t−2Frw3t

Sum=+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t

Sum of energy:

Before t=0, the system (N disks) has the energy N(1/2md²w1²+1/4mr²(w1−w2)²)

At final, the system has the energy:

N(1/2md²w1²+1/4mr²(w1−w2)²)+N1/2mr²((w1−w2i)²−(w1−w2f)²)+2(N−2)Frw3t


This change the result but the sum is not constant. Maybe if you can tell me where in my equations there is a mistake ?

You speak about conservation of angular momentum but it's possible to think only with forces and works.

dlouth15

neufneufneuf wrote: »

You speak about conservation of angular momentum but it's possible to think only with forces and works.

That is true but sometimes things can get too complicated to calculate in those terms. For example, in collisions the forces repelling objects might be very complex to model, but away from the collision itself, we know that the total momentum of the system (not angular in this case) must be constant throughout the process. If we know the trajectories of the particles prior to collision, then conservation of momentum (along with conservation of energy) can be used to calculate the trajectories after collision.

In this case we're using friction. We don't know the coefficient of friction involved so we can't work out the forces of friction. But we do know that friction will eventually slow the rings down. We also know that conservation of angular momentum means that this will have an effect on the rotation of the arms in the assembly. From this we can calculate the new energy and show that it is less than it would be if the rings were not under the influence of friction.

neufneufneuf

You're right in my calculations I suppose friction is constant, this give force F, in reality F will change with w2. For simplify the problem, and with a theoretical study it's possible to imagine the roughness is changing with time for have the same force F. It's not for change datas but only for simplify calculations. I know in physics you have habits to consider energy constant and angular momentum constant but I try to find a leak in physics so I calculated the sum of work. I don't use Algodoo, I explain an exercice and I tried to resolve it. Maybe you can help to find my error ?

dlouth15

neufneufneuf wrote: »

You're right in my calculations I suppose friction is constant, this give force F, in reality F will change with w2. For simplify the problem, and with a theoretical study it's possible to imagine the roughness is changing with time for have the same force F. It's not for change datas but only for simplify calculations. I know in physics you have habits to consider energy constant and angular momentum constant but I try to find a leak in physics so I calculated the sum of work. I don't use Algodoo, I explain an exercice and I tried to resolve it. Maybe you can help to find my error ?

Well, I think one mistake is that you have [latex]\omega_1[/latex], the rotation of the arm, constant. We know from ordinary experience that this will change. This is due to Newton's third law: For every action there is an equal and opposite reaction. You haven't shown how the forces cause this change and calculated the change.

neufneufneuf

If w1 decrease, there is a force somewhere I can draw that add a negative torque on arm. But I don't find this force, each disk have only a torque, no ? F1/F2 cancel themselves, for me it's all forces I see, no ?

dlouth15

neufneufneuf wrote: »

If w1 decrease, there is a force somewhere I can draw that add a negative torque on arm. But I don't find this force, each disk have only a torque, no ? F1/F2 cancel themselves, for me it's all forces I see, no ?

Think torque instead of linear forces. The friction is inducing a torque in the disks. This will, in turn, induce a torque of opposite direction in arms.

If you had just one disk and a fixed brake slowing it down rather than another wheel, you would still have two friction forces f1 and f2 acting always in oppositite directions. The friction causes a torque which changes the angular velocity of the disk. But newton's third law says that there will also be a torque produced of opposite direction in the arm.

Just because you have multiple disks doesn't change this.

neufneufneuf

This will, in turn, induce a torque of opposite direction in arms.

you're sure of that ? the center of gravity of each disk (so the green axis) receive the sum of external forces, no ?

dlouth15

neufneufneuf wrote: »

you're sure of that ? the center of gravity of each disk (so the green axis) receive the sum of external forces, no ?

Sit in a swivel chair and hold a heavy flywheel horizontally. Have someone spin up the flywheel to high speed. Then lift your feet from the floor and try to stop the flywheel. You will find yourself rotating in the chair.

Note that as you try to stop the flywheel with friction from your hand there will be two linear forces involved and these will cancel out. Yet you still rotate.

This is because you are applying torque to the flywheel. And there is also a reactive torque which causes you and the chair to rotate.

The linear forces cancelling out means that the chair doesn't move across the room. But it is not enough to describe the system. After dealing with the linear forces we must also deal with the torques.

neufneufneuf

I understand your example, but here it's not the arm that give the torque (green forces), green forces come from outside each disk, no ?

dlouth15

neufneufneuf wrote: »

I understand your example, but here it's not the arm that give the torque (green forces), green forces come from outside each disk, no ?

Actually I think I was wrong earlier in my assertion that reducing the rotation of the disks will cause a reduction in the rotation of the arms. There will be a transfer of torque but this will have the effect of trying to change the orientation of the assembly of disks from the vertical to something that isn't vertical. However the disks are fixed in the vertical so they can't change.

neufneufneuf

There will be a transfer of torque

How a torque is tranfert ? Could you explain, please ?

dlouth15

neufneufneuf wrote: »

How a torque is tranfert ? Could you explain, please ?

I mean that a reactive torque will be induced in the thing that is applying the torque.

neufneufneuf

If green axis has no friction, how a torque can be induced ? If there is a torque I can draw forces, no ?

dlouth15

neufneufneuf wrote: »

If green axis has no friction, how a torque can be induced ? If there is a torque I can draw forces, no ?

I think that if there's friction between the disks that causes them to slow down relative to one another, then there will be a reactive torque in the opposite direction causing the centre line of the disk to be non-vertical. However the disks although they can move, are always vertical so they can't move in such a way that they aren't vertical. So this reactive torque does no work. It is like pushing against a wall. I'm not sure it is easy to draw simple forces to illustrate this.

I thought earlier that reducing the rotation of the disks through friction would cause a change in the rotation of the arms however I now think that this is incorrect.

I have do a few things over the next few days so I won't be able to respond on this thread. I will try to respond again in a week or so if I can.

neufneufneuf

ok, I will try to find help from others forums, thanks for your help dlouth15

neufneufneuf

I understood my error, it's green forces, they can't be like that because trajectories are not in this direction, so torque is in the other direction, and sum of energy is constant.

neufneufneuf

I'm not sure about my forces and my sum of energy but in the following image, grey disk don't turn around itself. It turns only at w1 around red axis. Purple disk is fixed to the ground. There is friction between grey disk and purple disk, so there are forces F1 and F2 and F3. The energy from heating H=+FRtw1, and torques give the energy +Frtw1-F(R+r)tw1, the sum is 0.



The same case, but grey disk turns around itself counterclockwise. The energy from heating is H=+FRtw2. The energy from torque is +Frtw2-F(R+r)tw1, here the sum is Ft(R+r)(w2-w1), with w2<w1 it's not 0. I'm wrong in the enegy from heating or it's in the work of torque ?



Maybe H=Ft((R+r)w1-w2) ? And in this case I have the sum at 0. It's possible because the energy from heating is F*d and d=v*t and v=Rw1+rw1-rw2

But in this case, if I add another disk (but no heating dissipation, no sliding, just like a gear):



I can't find the sum at 0 because the work from torque is : Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1 and if I add H the sum is not at 0.

dlouth15

OK, this is a reformulation of the second case, posted above. Instead of the smaller disk rotating around the bigger, it is fixed about its axis. But the same physics applies. We can vary the rotational speed of either as we require.



The disks are in contact with one another and experience a frictional force of magnitude [latex]F[/latex] acting against their rotation.

What is the work done by the rotating disks against friction?

[latex]\mathrm{Work}=\mathrm{Force\times\mathrm{Distance}}[/latex]

For disk 1, the force is [latex]F[/latex] and the distance is [latex]\omega_{1}r_{1}t[/latex]. So

[latex]\displaystyle{W_{1}=F\times\omega_{1}r_{1}t}[/latex]

For disk 2, the force is also [latex]F[/latex] but the distance is [latex]\omega_{2}r_{2}t[/latex] so

[latex]\displaystyle{W_{2}=F\times\omega_{2}r_{2}t}[/latex]

Therefore the total work is

[latex]\displaystyle{W=W_{1}+W_{2}=F\omega_{1}r_{1}t+F\omega_{2}r_{2}t}[/latex]

and so

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

What then is the energy produced due to the heat of friction?

It is

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

Note that this is the same as the work formula. That is because the energy expended in doing the work is given off as heat. This can be experimentally verified using a calorimeter.

neufneufneuf

Thanks for your example. So, in my second case H=Ft((R+r)w1-w2) ? It's logical because the velocity is v=Rw1+rw1-rw2, no ?

dlouth15

neufneufneuf wrote: »

Thanks for your example. So, in my second case H=Ft((R+r)w1-w2) ? It's logical because the velocity is v=Rw1+rw1-rw2, no ?

H=Ft((R+r)w1-w2)

That has to be wrong because you are adding (R+r)w1 to w2. (R+r)w1 is a velocity and w2 is a rate of rotation. Also is w2 an absolute rate of rotation or is it relative to the rotating arm?

neufneufneuf

Oh, yes, w1 and w2 are labo frame reference (all velocities in 3 cases)

dlouth15

neufneufneuf wrote: »

Oh, yes, w1 and w2 are labo frame reference (all velocities in 3 cases)

OK, in that case my [latex]\omega_2[/latex] = your [latex]\omega_2 + \omega_1[/latex]

Therefore for the second example:

[latex]W=\displaystyle{Ft\left(\omega_{1}\left(R+r\right)+\omega_{2}r\right)}[/latex]

And the heat produced

[latex]E=\displaystyle{Ft\left(\omega_{1}\left(R+r\right)+\omega_{2}r\right)}[/latex]

For the first example, your [latex]\omega_2=0[/latex], so

[latex]\displaystyle{W=E_{heat}=Ft\omega_{1}\left(R+r\right)}[/latex]

But it is only complicating matters to have a rotating arm. It doesn't introduce any new physics.

neufneufneuf

Ok, if w2 < w1, then H=Ft((R+r)w1-w2) ? or the sign of w2 is negative with your equation ?

dlouth15

neufneufneuf wrote: »

Ok, if w2 < w1, then H=Ft((R+r)w1-w2) ? or the sign of w2 is negative with your equation ?

In your example, which I will not be dealing with further, positive W1 is clockwise as indicated on your diagram. Positive W2 is anti-clockwise as indicated on your diagram. The formula I gave in my last post, [latex]Ft\left(\omega_{1}\left(R+r\right)+\omega_{2}r\right)
[/latex], assumes that. But that is the last I will be dealing with rotating arms.

OK, I'm only dealing with my example from now on (see my diagram). Both omegas represent clockwise rotation. If omega1 is a positive number, then disk one is rotating clockwise. Same with omega2.

If either omegas are negative then that means an anti-clockwise rotation.

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

You can see then that if [latex]\omega_1=-\omega_2[/latex] and [latex]r_1=r_2[/latex], i.e., they are two disks of identical size but rotating in opposite directions, then there would be work done, no friction, and so no heat.

dlouth15

I don't think the size of the disks makes any difference to the physics so we may as well make [latex]r_1=r_2=r[/latex]. Also we can have a single [latex]\omega[/latex] representing the rotation of the Disk 1. In this example, positive [latex]omega[/latex] indicates rotation in the counterclockwise direction.



We can adjust the rotation of the other disk by varying the value of [latex]k[/latex]. So [latex]k=-1[/latex] means the disk is rotating at the same rate but in the opposite direction. [latex]k=1/2[/latex] means that the first disk is rotating twice as fast as the second but in the same direction. [latex]k=-1/3[/latex] means the first disk is rotating 3 times as fast as the second but in the opposite direction. And so on. By varying [latex]k[/latex] between -1 and +1, we can get any combination of rotations between the disks.

Then
[latex]\displaystyle{W_{1}=Fr\omega t}[/latex] and [latex]\displaystyle{W_{2}=Frk\omega t}[/latex].

So the total work is therefore

[latex]\displaystyle{W=W_{1}+W_{2}=Fr\omega t+Frk\omega t=Fr\omega t\left(1+k\right)}[/latex]

A positive [latex]omega[/latex] and [latex]-1\leq k\leq1[/latex] covers all possible combinations of rotations between two disks. And provided we meet those conditions the work done, [latex]W[/latex], will never be negative and therefore there will never be a gain in energy from the system.

Special cases of [latex]k[/latex]

[latex]k=-1[/latex]: The two disks are rotating at the same rate but in opposite directions. [latex]W=0[/latex]. This is as expected as there's no friction.

[latex]k=1[/latex]: The disks are rotating in the same direction and at the same rate. [latex]W=2Fr\omega t[/latex]

[latex]k=0[/latex]: One of the disk is not rotating. [latex]W=Fr\omega t[/latex]. This is as expected. Only one of the disks are rotating so we're getting half of that expected with two disks rotating.

locum-motion

Jeez, dlouth, don't encourage him!

neufneufneuf

In your example, which I will not be dealing with further, positive W1 is clockwise as indicated on your diagram. Positive W2 is anti-clockwise as indicated on your diagram.

Maybe there is something I don't understand when I must draw directions of rotation.

Example: I want that arm turns at +10 rd/s clockwise and grey disk turns at 8 rd/s clockwise, all in labo frame reference (arm turns clockwise at +10 rd/s and grey disk turns around itself (arm reference) at 2 rd/s counterclockwise), I must draw direction of arm clockwise but the direction of the arrow of the disk must be clockwise or counterclockwise ?

Edit: I'm agree with your last message.

dlouth15

neufneufneuf wrote: »

Maybe there is something I don't understand when I must draw directions of rotation.

Example: I want that arm turns at +10 rd/s clockwise and grey disk turns at 8 rd/s clockwise, all in labo frame reference (arm turns clockwise at +10 rd/s and grey disk turns around itself (arm reference) at 2 rd/s counterclockwise), I must draw direction of arm clockwise but the direction of the arrow of the disk must be clockwise or counterclockwise ?

Edit: I'm agree with your last message.

In general, if you have an symbol representing a quantity and an associated arrow, then the arrow represents the direction assuming the quantity is positive. If the quantity is negative then the arrow represents the opposite direction.

In your example, the arm should have a clockwise arrow, and if the grey disk is turning counterclockwise it should have a counterclockwise arrow.

In my example, positive omega and positive k*omega represents counterclockwise rotation of the disks, but I allow one of the disks to rotate clockwise by making k negative. I don't have to draw a new arrow for this as the convention is that negative values reverse the direction of the arrow. This way I cover all possible combinations of rotation with one diagram.

neufneufneuf

Grey disk is turning counterclockwise at 2rd/s with the arm frame reference, but in the labo frame reference it is turning clockwise at 8rd/s because the arm is turning clockwise at 10 rd/s. So the arrow for the grey disk don't take in account the rotationnal velocity of the arm ?

dlouth15

neufneufneuf wrote: »

Grey disk is turning counterclockwise at 2rd/s with the arm frame reference, but in the labo frame reference it is turning clockwise at 8rd/s because the arm is turning clockwise at 10 rd/s. So the arrow for the grey disk don't take in account the rotationnal velocity of the arm ?

Consider using only the laboratory frame for rotations. Use arrows accordingly. State this in your description.

But why do you have rotating disks on rotating arms at all? It is only a complicated way of producing friction between rotating disks and makes analysis difficult. Instead have the arm fixed and adjust the rotation of the disks.

neufneufneuf

dlouth15 wrote: »

Consider using only the laboratory frame for rotations. Use arrows accordingly. State this in your description.

Thanks

dlouth15 wrote: »

But why do you have rotating disks on rotating arms at all? It is only a complicated way of producing friction between rotating disks and makes analysis difficult. Instead have the arm fixed and adjust the rotation of the disks.

You're right it's more complicated. I tested an example like that because at start, I thought it was easier ! For the first case and the second I understood but the third case I don't find the energy from heating equal to the energy from torques. I guess |w2| < |w1| < |w3|. There is a relation between w2 and w3 at start because there is no slinding between grey disk and brown disk. In the arm frame reference w2 = -w3 at start. For example, w1=10, w2=8 and w3=12. With good arrow it is:



The heating is the same than in the second case: H=Ft((R+r)w1-rw2) but the energy from the torque is : Ft ( rw2-1/2rw2-3/2(R+r)w1+1/2(R+3r)w1-1/2rw2 ) = -FRtw1. So the sum is Frt(w1-w2).

dlouth15

neufneufneuf wrote: »

You're right it's more complicated. I tested an example like that because at start, I thought it was easier ! For the first case and the second I understood but the third case I don't find the energy from heating equal to the energy from torques. I guess |w2| < |w1| < |w3|. There is a relation between w2 and w3 at start because there is no slinding between grey disk and brown disk. In the arm frame reference w2 = -w3 at start. For example, w1=10, w2=8 and w3=12. With good arrow it is:

I still don't see why you have the arm rotating and a fixed purple disk. Why don't you have the arm fixed and the purple disk rotating? Much easier to analyze and less chance of error.

neufneufneuf

Why don't you have the arm fixed and the purple disk rotating?

Because, I would like to study with forces F1 and F2 like I drawn. I define w4 the rotationnal velocity of purple disk. If I want forces F1 and F2 like I drawn, I need to have w1R+(w1-w2)r-w4R > 0. In this case, H=Ft((R+r)w1-rw2-Rw4). The energy form torque is -FRw1t+FRw4t, the sum is the same Frt(w1-w2).

dlouth15

neufneufneuf wrote: »

Because, I would like to study with forces F1 and F2 like I drawn.

Are they not simply the friction forces? You can still draw them in if you have the arm not rotating but the purple disk rotating since you still have two surfaces slipping over one another. The point of this is that the problem reduces to the example I gave earlier:



The work done against friction is:

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

Although I haven't drawn in the forces, there's a force acting upwards on Disk 1 and downward on Disk 2 at the point of contact of the two disks of magnitude F.

neufneufneuf

In this case I'm agree with you, the sum of energy from heating and torque =0, I can compare your case with the first of mine (post #74), all is fine. But in my case (with an arm and 3 disks, one can be fixed), even it's a little more complicated, I find heating different of work of torque.I think my calculation about heating is good because in the second case I found the sum at 0. I think it's a force I drawn in the bad direction because my calculation of work from torque is good.

dlouth15

neufneufneuf wrote: »

In this case I'm agree with you, the sum of energy from heating and torque =0, I can compare your case with the first of mine (post #74), all is fine. But in my case (with an arm and 3 disks, one can be fixed), even it's a little more complicated, I find heating different of work of torque.I think my calculation about heating is good because in the second case I found the sum at 0. I think it's a force I drawn in the bad direction because my calculation of work from torque is good.

Why not have the arm holding the three disks fixed and adjust the rotation of the purple disk to compensate?

At the end of the day all the work done is due to friction between disk 1 and disk 2. Disk 3 doesn't produce any friction. We only need to find out how much sliding has occurred in time t between the two surfaces of disks 1 and 2. As such it doesn't matter whether the purple disk is rotating or the arm is rotating.

neufneufneuf

Why not have the arm holding the three disks fixed and adjust the rotation of the purple disk to compensate?

Ok, in this case for have forces F1 and F2 like I want, I need to rotate w1 counterclockwise, true ? Like the arm is fixed, no work from it. Heating H is the same. Purple disk works at -FRtw1, grey disk works at Frtw2 - 1/2Frtw2 and brown disk works at -1/2Frtw2, so the sum of energy is :

Ft((R+r)w1-rw2) + ( -FRtw1 +Frtw2 - 1/2Frtw2 -1/2Frtw2 ) = Frt(w1-w2)

What's wrong ?

dlouth15

neufneufneuf wrote: »

Ok, in this case for have forces F1 and F2 like I want, I need to rotate w1 counterclockwise, true ? Like the arm is fixed, no work from it. Heating H is the same. Purple disk works at -FRtw1, grey disk works at Frtw2 - 1/2Frtw2 and brown disk works at -1/2Frtw2, so the sum of energy is :

Ft((R+r)w1-rw2) + ( -FRtw1 +Frtw2 - 1/2Frtw2 -1/2Frtw2 ) = Frt(w1-w2)

What's wrong ?

Yes, counterclockwise. If the arm was clockwise then purple disk is now counter clockwise.

If the purple disk is counterclockwise at [latex]\omega_1[/latex] and the grey disk is clockwise at [latex]\omega_2[/latex] then the work should be

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

I'm treating the rates of rotation as being measured in the laboratory frame.

dlouth15

Here's the setup I envisage. I've drawn in the forces due to friction on the two surfaces. The arrows are a bit funny but they should be pointing in the direction of the vector lines.



In this setup

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

neufneufneuf

You're right for the case with the arm fixed. I know what's wrong in the case with the arm fixed, rotationnal velocities can't be like I guess with no sliding between the grey disk and the brown disk. For example, at start w2=-2 and w3=2 (labo frame reference). With the arm fixed, w2 will increase and w3 will decrease, but it's not possible to have w2=-2.1 and w3=1.9. In the case with the arm is turning, it's possible to have at start w1=10, w2=8 and w3=12. w2 will increase at 9, and w3 will decrease at 11. In the arm frame reference, w2=-w3 and I can say no sliding between the grey disk and the brown disk. What do you think about that ?

dlouth15

neufneufneuf wrote: »

You're right for the case with the arm fixed. I know what's wrong in the case with the arm fixed, rotationnal velocities can't be like I guess with no sliding between the grey disk and the brown disk. For example, at start w2=-2 and w3=2 (labo frame reference). With the arm fixed, w2 will increase and w3 will decrease, but it's not possible to have w2=-2.1 and w3=1.9. In the case with the arm is turning, it's possible to have at start w1=10, w2=8 and w3=12. w2 will increase at 9, and w3 will decrease at 11. In the arm frame reference, w2=-w3 and I can say no sliding between the grey disk and the brown disk. What do you think about that ?

It doesn't matter because having a rotating arm is the same as rotating the entire system or viewing the system from a rotating frame of reference. Physically nothing of importance changes.

neufneufneuf

With fixed arm, you think it's possible to have no sliding between the grey disk and the brown disk, with w2 is increasing and w3 is decreasing ? For me if w2=-2 and w3=2 at start, w2 move up at -2.1 and w3 move down at 1.9, how no sliding is possible ?

dlouth15

neufneufneuf wrote: »

With fixed arm, you think it's possible to have no sliding between the grey disk and the brown disk, with w2 is increasing and w3 is decreasing ? For me if w2=-2 and w3=2 at start, w2 move up at -2.1 and w3 move down at 1.9, how no sliding is possible ?

Didn't you say that the grey disk acts like a gear? No sliding, no friction, no heat. Therefore both grey and brown disks rotate at the rate of [latex]\omega_2[/latex] but in opposite directions.

neufneufneuf

Yes, it's possible to have the grey disk like that :



And study during a little time.

No sliding, no friction, no heat. Therefore both grey and brown disks rotate at the rate w2 of but in opposite directions.

In the case with the arm fixed, I'm agree, but with the arm in rotation, it's possible to have :

w1=10
w2=8
w3=12

and if w2 increases w3 can decrease.