# Double rotation, friction and sum of energy

1. Nov 7, 2014

### V711

Hi,

I posted my question on another forum:

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

but it is "on hold" and nobody knows where is the error, so I try to post here if you are agree ? I can understand if you close the question because I asked on another forum. I ask here because I asked to a colleague teacher of physics this morning and he can't find the error. Maybe you can see where is my error.

http://imagizer.imageshack.us/v2/800x600q90/674/ZCCxsA.png [Broken]

At time t=0

At time $t=0$, I rotate a grey disk or ring with a rough outer surface at angular velocity $w_1$ around the blue axis (clockwise) and at $w_2$ around the green axis (counterclockwise) as shown in the figure. $w_2$ is relative to black arm.

Now, the kinetic energy of the system is:

$K_E = \frac{1}{2}md²w_1²+\frac{1}{2}mr²(w_1-w_2)²$

for a ring or

$K_E = \frac{1}{2}md²w_1²+\frac{1}{4}mr²(w_1-w_2)²$

for a disk, with $d$ the lenght of the arm, $m$ the mass of disk and $r$ radius of disk. Note again that $w_2$ is relative to black arm, the expression of kinetics is here.

I built N systems like that:

http://imagizer.imageshack.us/v2/800x600q90/910/RvX9fg.png [Broken]

Noted blue axis are fixed to the ground. There is friction between disks. Like black arms are turning at the same rotationnal velocity, disks are always in contact. I count all energies including the energy of friction. For simplify the study I consider friction from disk/disk give the same force F even $w_2$ decreasing, it's possible to imagine a theoretical problem or in reality it's possible to imagine oil between disks and something remove oil more and more when $w_2$ decrease.

Friction can be obtain, for example, with two forces $F1/F2$: disks must be side to side. These forces don't work. An example with 2 positions at 2 diffrent times:

http://imagizer.imageshack.us/v2/800x600q90/905/3niK6h.png [Broken]

Forces that need energy are magenta forces (purple). But these forces need the same energy for 3 systems than for 10 or more. In the contrary, with 10 systems, heating is higher and energy of each disk increase too.

I guess no friction at green axis and at blue axis. At start, for prevent shock, I guess friction is added more and more like magenta forces. I guess friction is low, like that $w_2$ decrease slowly.

Works of forces

$F$ is the value of green or magenta force

$w_{disks} = +N \frac{1}{2}mr²((w_1-w_{2i})²-(w_1-w_{2f})²)$

with $w_{2f}<w_{2i}$
with $w_{2f}$: $w_2$ at final and $w_{2i}$: $w_2$ initial.

$W_{friction}=2(N-1)Frw_{2m}t$ with $w_{2m}$ the mean of $w_{2}$
$W_{F1}=2dF-2dF=0$
$W_{F2}=2dF-2dF=0$

$W{magentaforce}= -2Fdw_{2m}t$

$Sum=+N \frac{1}{2}mr²((w_1-w_{2i})²-(w_1-w_{2f})²)+2(N-2)Frw_{2m}t$

Sum of energy

At $t=0$, the system (N disks) has the energy $N(\frac{1}{2}md²w_1²+\frac{1}{4}mr²(w_1-w_2)²)$

At final, the system has the energy:

$N(\frac{1}{2}md²w_1²+\frac{1}{4}mr²(w_1-w_2)²)+N \frac{1}{2}mr²((w_1-w_{2i})²-(w_1-w_{2f})²)+2(N-2)Frw_{2m}t$

For resume

If I define $2E_1$ the energy that each disk give to heat. And I define $E_2$ the energy won by each disk because $w_2$ decrease. The energy needed by two last systems is $2E_1$. With N systems, the sum of energy won by the system is $(N-2)2E_1+NE_2$.

So :

a) My reasoning is false ?
b) I'm wrong somewhere in my calculations ?
c) I forgot a force or a torque, in particular on black arm ?
d) A movement can't be like I describe ?
e) The sum of energy is not constant in this case ?

--------------------------------------------------------------------------------------------------------------

I added more cases for watch different positions of the system:

http://imagizer.imageshack.us/v2/800x600q90/661/IjNCnC.png [Broken]

Following image shows blue axis fixed to the ground:

http://imagizer.imageshack.us/v2/800x600q90/540/S555kL.png [Broken]

Last edited by a moderator: May 7, 2017
2. Nov 7, 2014

### Staff: Mentor

1. There is no reason to have the systems revolve around the blue point and indeed, nothing to make that happen. You started the problem by assuming they were moving (magically adding kinetic energy to the system for no reason), but with no force to make it move. Since it is constant speed motion (you should avoid trying scenarios where it changes) with no inputs or outputs, it serves no purpose but to add complexity and confuse you.

2. Since the speeds are constant, there is no reason to bother with kinetic energy here. It doesn't mater.

3. The forces applied to make the wheels spin and the forces of friction must be equal. You've added them together wrong. In the scenario with 3 wheels, the middle wheel doesn't move and all four forces are equal to each other.

Last edited: Nov 7, 2014
3. Nov 8, 2014

### V711

Hi Russ_watters,

At start I give 2 rotations, $w_2$ and $w_1$, all disks are turning around blue axis and around green axis. I counted this energy (that I need to give at start) in the sum. Like disks are turning they have inertia and it's possible to have friction. I added forces F1 and F2 for have friction. Why I can't have 2 rotations at start just before set friction ON ?

I think $w_2$ is not constant (relative to black arm), it's decreasing more and more because each disk receive a torque (green and magenta forces). I set magenta forces at the same value than green forces. I guess force F from friction is at the constant value F for simplify the study, it's possible to imagine a rough surface that change for increase roughness when $w_2$ decrease, it's only for simplify equations.

I don't understand. Disks turn before friction is ON. The middle disk is turning at $w_2$ and at $w_1$ at start. It receives 2 forces not 4, a green force from bottom disk and a green force from top disk. Could you explain these 4 forces ?

Bye

4. Nov 8, 2014

### Staff: Mentor

Let's back up a sec:
What, exactly, is your goal here? It looks to me like you are trying to compare input and output power. Is that correct? See, as this is your scenario you can set it up however you want, but the question is whether the setup and conduct of the problem helps you find what you are looking for or not.

 Actually I did make an error there with my #3: since the forces on the middle wheel are in opposite directions, it won't just stop rotating, it'll rotate in the opposite direction from what is shown! (which then also makes the forces unbalanced)

Last edited: Nov 8, 2014
5. Nov 8, 2014

### V711

Not input/output power. I give energy at start for give $w1$ and $w2$and I let all the system "live" and I watch how energy is converted (kinetic to heat, etc.) and the sum.

Each disk has at every moment the same torque on it, it is drawn with green/green or green/magenta forces so I don't understand why you speak about the middle disk. Could you explain ?

Think with small friction, disks have inertia so there is friction. I'm interesting about the transcient analysis only, from start at $w_2=3$ to $w_2=1$ for example.

Last edited: Nov 8, 2014
6. Nov 8, 2014

### Staff: Mentor

I guess we need to go even slower, because this still isn't making sense to me:

I thought the magenta forces are externally applied and the green ones are reactions due to friction? Or all they all externally applied?

And when you say they are all under the same torque do you mean that there is an additional torque you are applying at the rotation axis of each or are you CONCLUDING that the torques from the green/magenta forces are equal?

7. Nov 8, 2014

### V711

Yes, it's that.

I choose magenta forces for have only a torque. Like magenta forces are applied from external system, I can set like I want.

For the middle disk: green forces have the same value (absolute value) so the middle disk has only a torque too.

8. Nov 8, 2014

### Staff: Mentor

Ok, so for the top wheel, at the start of the scenario, when you turn off your start-up power systems and apply the magenta forces:

1. The top wheel is spinning counterclockwise and the magenta force applies a clockwise torqe.
2. The bottom wheel is spinning counterclockwise and the magenta force applies a clockwise torque.
3. Friction with the middle wheel provides a clockwise torque to the top and bottom wheels.
4. Friction with the top and bottom wheels provides a clockwise torque to the middle wheel.

Is this how you see it?

9. Nov 8, 2014

### V711

No, maybe it's my error, I count differently forces:

For me, each wheel receive 2 forces: green/green or green/magenta

1. The top wheel is spinning counterclockwise and the magenta force and the green force apply a clockwise torque
2. The bottom wheel is spinning counterclockwise and the magenta force and the green force apply a clockwise torque
3. The middle wheel is spinning counterclockwise and 2 green forces apply a clockwise torque

Each wheel receive the same clockwise torque.

Is it ok for you ?

Last edited: Nov 8, 2014
10. Nov 8, 2014

### Staff: Mentor

That doesn't appear to me to be different from what I said, you just arranged it differently...

In any case, the result is that all three wheels initially decelerate at the same rate, right?

11. Nov 8, 2014

### V711

right :)

But they decelerate on arm frame reference but accelerate in lab frame reference, no ?

Note: |w1| > |w2|

Last edited: Nov 8, 2014
12. Nov 8, 2014

### Staff: Mentor

I was taking the disks by themselves, but OK:

So you are saying that because you have the magenta forces not pointing at the center of the blue axes, they apply a torque about the blue axes?

13. Nov 8, 2014

### V711

No, for me there is no torque on blue axes, because each disk receive a torque and only a torque (there is no net force). But I'm saying each disk accelerate in lab frame (not arm frame refererence) it's because $w_2$ decrease ($w_2$ is relative to black arm). But the rotational velocity of a disk in lab frame reference is $w_1-w_2$, if $w_1=10$ and if $w_2=3$ then the $w_2lab=10-3=7$, if $w_2$ decrease $w_2lab=10-2=8$ for example, no ?

Note: |w1| > |w2|

Last edited: Nov 8, 2014
14. Nov 8, 2014

### Staff: Mentor

Oh, you mean the "arm frame" is fixed to the arm, so it rotates with respect to the lab? Seems like an unnecessary complication to me since the forces are always in the same direction in the lab frame. So rather than reference both the disk's own rotation and the applied forces to the lab frame, you add add an additional rotation to each. Is there a reason for that? Seems like an unnecessary complication.

15. Nov 8, 2014

### V711

I'm not sure I understood your last message. Let's back up a sec: are you agree that $w_2$ is a "black arm reference" ? I wrote on first message. For me it's easier to think like that and the expression of the kinetic energy take in account $w_2$ is a "black arm reference".

In the contrary, kinetic energy is:

$KE=1/2md^2w_1^2+1/2mr^2w_2^2$

edit: If you tought $w_2$ is lab reference, I'm not agree when you said $w_2$ decrease, for me it increases. Like |w1| > |w2|, when a torque is applied to a disk its rotationnal velocity goes to w1, no ?

Last edited: Nov 8, 2014
16. Nov 8, 2014

### Staff: Mentor

It doesn't seem easier to me. You have the disks rotating with respect to the arm, but the forces always pointed in the same directions with respect to the lab. Since the orientation of the wheels with respect to each other is also always the same with respect to the lab, it is that rotation that causes the friction forces.

So it appears to me that you have the friction forces pointed in the wrong directions.

17. Nov 8, 2014

### V711

green forces ? could you explain please ?

18. Nov 8, 2014

### Staff: Mentor

The directions of the friction forces depend on the rotation of the wheels with respect to the lab frame, which is clockwise. So the friction forces act counterclockwise.

19. Nov 8, 2014

### V711

If I take 2 points near wheel/wheel, p1 and p2:

http://imagizer.imageshack.us/v2/800x600q90/540/RdJkK0.png [Broken]

$p1$ goes to the right and $p2$ goes to the left, no ? for me, like $p1$ and $p2$ rotate at $w_1$ no matter $w_1$. If $w_1=0$ forces are like I drawn, but not if $w_1<>0$ ? It's very difficult to understand that I drawn forces in the wrong direction.

Last edited by a moderator: May 7, 2017
20. Nov 8, 2014

### Staff: Mentor

No. You said the rotation about the blue axis is faster than the rotation about the green axis in the rotating frame. That means in the lab frame, the disks are rotating CLOCKWISE.

That makes P1 go to the left and P2 go to the right with respect to each other.

Last edited by a moderator: May 7, 2017