Sum of torque

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.

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.