Double rotation, friction and sum of energy

In summary, the conversation discusses a physics problem involving a system of rotating disks and the calculation of kinetic energy and friction. The person posting the question is seeking help in understanding the problem and the calculations involved. The responder provides explanations and clarifications on the problem, pointing out potential errors in the calculations and suggesting simplifications for better understanding. The conversation also addresses the issue of forces and torques involved in the problem.
  • #1
V711
61
0
Hi,

I posted my question on another forum:

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

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

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 length 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

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

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 Following image shows blue axis fixed to the ground:

http://imagizer.imageshack.us/v2/800x600q90/540/S555kL.png
 
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  • #2
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.
 
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  • #3
Hi Russ_watters,

Thank for your help.

russ_watters said:
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.

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 ?

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

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.

russ_watters said:
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.

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
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.

[edit] 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)
 
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  • #5
russ_watters said:
What, exactly, is your goal here? It looks to me like you are trying to compare input and output power. Is that correct?
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.
russ_watters said:
since the forces on the middle wheel are in opposite directions
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.
 
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  • #6
V711 said:
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 ?
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
russ_watters said:
I thought the magenta forces are externally applied and the green ones are reactions due to friction?
Yes, it's that.

russ_watters said:
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?
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
V711 said:
Yes, it's that. [magenta forces are externally applied, green forces are not]
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
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 ?
 
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  • #10
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
right :)

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

Note: |w1| > |w2|
 
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  • #12
V711 said:
right :)

But they decelerate on arm frame reference but accelerate in lab frame reference, no ?
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
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|
 
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  • #14
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
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 ?
 
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  • #16
V711 said:
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".
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
russ_watters said:
So it appears to me that you have the friction forces pointed in the wrong directions.
green forces ? could you explain please ?
 
  • #18
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
If I take 2 points near wheel/wheel, p1 and p2:

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

##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.
 
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  • #20
V711 said:
If I take 2 points near wheel/wheel, p1 and p2:

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

##p1## goes to the right and ##p2## goes to the left, no ?
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.
 
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  • #21
Great, you're right, I understood my error. ;)

If I take a system with N Wheels, N/2 wheels are turning at ##w_2##, with ##|w_1|>|w_2|## anf N wheels are turning at ##w_3## with ##|w_3|>|w_1|##. I set one disk at ##w_2## next at ##w_3##, next at ##w_2## etc. N/2 wheels lost kinetic energy, N/2 disk win kinetic energy but there is N friction and work of magenta forces don't depend of N, here how the sum of energy can be constant ? It is not possible to have friction like I drawn in this case too ?

http://imagizer.imageshack.us/v2/800x600q90/674/7FSHBY.png

Thanks for your help
 
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  • #22
V711 said:
Great, you're right, I understood my error. ;)

If I take a system with N Wheels, N/2 wheels are turning at ##w_2##, with ##|w_1|>|w_2|## anf N wheels are turning at ##w_3## with ##|w_3|>|w_1|##. I set one disk at ##w_2## next at ##w_3##, next at ##w_2## etc. N/2 wheels lost kinetic energy, N/2 disk win kinetic energy but there is N friction and work of magenta forces don't depend of N, here how the sum of energy can be constant ? It is not possible to have friction like I drawn in this case too ?
You haven't given enough information. The direction of the friction forces depends on which direction the wheel edges are sliding past each other. Now that the wheels are rotating at different speeds wrt the rotating frame and opposite directions (with respect to the lab frame), you can't know the direction of the friction force unless you know in which direction the wheel edges are sliding past each other. Or, even, if they aren't sliding past each other at all but are just rotating at the same speed in the lab frame, in opposite directions, friction force may be zero.

Why make the problem even more complicated when you haven't completed your analysis of the first system yet? I'm not interested in going through potentially endless iterations of "find the error in this scenario" unless I have a good reason to and I don't see any value in what we're doing here. So please tell me: what is the point of all of this?
 
  • #23
russ_watters said:
when you haven't completed your analysis of the first system yet?
It's ok, I understood my error, it was only the direction of points (trajectories), if forces are wrong, all calculations are wrong, my reasoning is good but I started with bad green forces ! Maybe it's because I'm thinking with relative rotationnal velocity.

russ_watters said:
So please tell me: what is the point of all of this?
Just study double rotation and where goes energy (friction, kinetic, etc). Don't worry it's not a perpetual mobile or something like that.

russ_watters said:
You haven't given enough information. The direction of the friction forces depends on which direction the wheel edges are sliding past each other. Now that the wheels are rotating at different speeds wrt the rotating frame and opposite directions (with respect to the lab frame), you can't know the direction of the friction force unless you know in which direction the wheel edges are sliding past each other. Or, even, if they aren't sliding past each other at all but are just rotating at the same speed in the lab frame, in opposite directions, friction force may be zero.
It's for that I asked the question, trajectories are not so easy than that. I thought if I take :

##w_1=-10##
##w_2=8##
##w_3=12##

it was enough for have green forces like I drawn, but I need to compute trajectories for that.

russ_watters said:
you can't know the direction of the friction force unless you know in which direction the wheel edges are sliding past each other
This point is very important.
 
  • #24
Like I understood calculations too. I will compute all trajectories to be sure trajectories can't give green forces at post #1 with my first case and others parameters. You're right I need to finish to study my first case :s, I want to learn too many things in the same time !

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

I just tested on a simple simulator for watch position of point p1 and p2 defined before and I can see trajectories like I thought, and green forces like I thought too, you're sure that green forces can't be like I drawn even ##|w_1| > |w_2|## ?It's late maybe I'm tired :confused:.
 
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  • #25
V711 said:
I just tested on a simple simulator for watch position of point p1 and p2 defined before and I can see trajectories like I thought, and green forces like I thought too, you're sure that green forces can't be like I drawn even ##|w_1| > |w_2|## ?It's late maybe I'm tired :confused:.
And from your PM:
...it's like w1 don't change nothing.
If w1 doesn't change anything, then maybe you have the frame of reference for w2 wrong...? If you are running it in a simulator, it might not allow you to have a rotating frame of reference for the rotation of the disks.
 
  • #26
Hi Russ_watters,

You're right. Sorry, I don't want to change my problem but it's difficult for me to say it's only a problem of trajectory. Because it's possible to add disks for change rotationnal velocity:

It's possible to use intermediate magenta disks (with no mass) for increase rotationnal velocity, like that forces can be like I drawn. Magenta/magenta disk gives friction. Green disk forced magenta disk like gear can do (but it's not gear because I need friction to magenta/magenta disk interface). Magenta disks are turning at ##w3##, and it can be very high if radius of magenta disk is very small:

http://imagizer.imageshack.us/v2/800x600q90/673/J3e3fR.png

Here velocities of 2 points ##P1## and ##P2## are:

http://imagizer.imageshack.us/v2/800x600q90/901/idmO5a.png

With ##r'## the radius of magenta disk: ##P1## move at right at ##(d-r')w1-r'w3## and ##P2## move at right at ##(d+r')w1+r'w3##, ##P2## move more at right than ##P1##, so forces ##F1## and ##F2## can be like that (forces ##F3## to ##F10## are consequences of ##F1## and ##F2##):

http://imagizer.imageshack.us/v2/800x600q90/908/d61Ud4.png

Magenta disks are turning at clockwise and receive a counterclockwise torque, they slow down, so green disks accelerate (in lab frame reference), no ? If magenta disks have no mass, no kinetic energy is lost by magenta disks, in the contrary each green disk won kinetic energy. What's wrong here ?
 
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  • #27
Sorry, I'm really not interested in analyzing ever-more complex devices for no reason. It just isn't worth my time/effort. It seems like you are trying to get the device to do something it can't do.
 

1. What is double rotation and how does it affect friction?

Double rotation refers to the movement of an object in two different directions, often at the same time. This can affect friction in two ways. First, the frictional force acting on the object will depend on the direction of its movement. Second, if the object is rotating, the frictional force will also depend on the speed of rotation.

2. How does friction affect the sum of energy in a system?

Friction is a non-conservative force, meaning that it dissipates energy. This means that as an object moves due to forces such as friction, its kinetic energy decreases over time. As a result, the total energy in a system that includes friction will decrease over time.

3. Can friction be ignored in the calculation of energy in a system?

No, friction should not be ignored in energy calculations. While it may sometimes be negligible, friction always acts in the opposite direction to the motion of an object and therefore affects the energy of a system. Neglecting friction can lead to inaccurate predictions and calculations.

4. How does the coefficient of friction affect the energy of a system?

The coefficient of friction is a measure of the amount of friction between two surfaces. It affects the energy of a system by determining the magnitude of the frictional force acting on an object. A higher coefficient of friction will result in a greater frictional force and therefore a greater decrease in energy.

5. Is there a way to reduce the effect of friction on energy in a system?

While it is not possible to completely eliminate friction, it is possible to reduce its effects on energy in a system. One way to do this is by using lubricants to reduce the friction between two surfaces. Additionally, designing objects with smooth surfaces and reducing the amount of contact between surfaces can also minimize the impact of friction on energy.

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