Sum of torque

Maybe like that:



All red disks will slow down, so like energy is (w1-w2)², energy increase. Except for 4 corners. But N can be great and there are always 4 corners. Before start, give energy for turn red disks around themselves and around green axis. After, friction is ON, this increase energy of the system.



With several layers for cancel problems at corners:






here all torques are reported for superior layer but each layer I win energy, 4 disks (4 corners) for first layer, 4 disks second layers, etc. Each layer 4 disks have won. There is a difference of energy.


With one layer only: 4 corners give 4 torques but 4 disks at corners don't decrease their rotationnal velocity around their center of gravity.
With 2 layers only: the same for 4 next corners, previous torque are reported at this layer but 4 disks at layers one decrease rotationnal velocity =>energy is bigger
Wit 3 layers: 4 disks at corners of the second layer win energy

R=Radius of first square layer (a side of inside square = 2R)
r= radius of red disk
F=basic force from friction
t=time

1/ With first layer only, there are:
- 20 torques work 20*T*(w1-w2)t = 2*20*Fr*w2t = 40Fr(w1-w2)t
- Friction 1 = 40Frw2t
- 8 torques back (R-r)Fw1t, with R=6r for example, this give -48Frw1t+8Frw1t
Sum = 40Fr(w1-w2)t + 40Frw2t -48Frw1t+8Frw1t = 0 so no energy from first layer if alone


2/ With second layer only, there are:
- 28 torques work 28*T*w2t
- Friction 2
- 8 torques back (R+2r-r)Fw1t
Sum is 0 too like first layer

3/ With first AND second layers only, there are:
- 20+28+8 torques work 52*T*(w1-w2)t = 2*56Fr(w1-w2)t so that add 16Frw1t-16Frw2t
- Friction 1+Friction2 + 8Frw2t
- 8 torques back (R-r)Fw1t + 8 torques back (R+2r-r)Fw1t

The difference is 16Frw1t-8Frw2t, the sum is not 0

Edit : other case:
I think it's possible to cover a sphere wih a lot of small disk (with different radius), Launch the sphere at w1 with all disk at w2. At t=0, friction is ON. Half disks slow down, other accelerate, but the enery is (w1-w2-dw2)² and (w1+w2+dw2)². With dw2 delta rotationnal velocity won of lost. So, the sum of energy is not constant due to the square function.

I compute work from torques:

1/ +24+32+8 torques on red disks = Frt(48(w1−w2)+64(w1−w′2)+16(w1−w2))
2/ Friction : +24Fr(w2+w2)t+32Fr(w′2+w′2)t+8Fr(w2+w′2)t
3/ -8 torques layer1 - 8 torque layer2 =−8(R+2r)Fw1t−8(R+2r)Fw1t=−8(6r+2r)Fw1t−64Frw1t=−128Frw1t

The sum of works from torque = 128Frtw1−64Frtw2−64Frtw′2 +56Frtw2+72Frtw′2 −128Frw1t

The sum = −8Frtw2+8Frtw′2=8Frt(w′2−w2) it's different of 0

Maybe this idea ?




Before t=0, turn at w1 and w2 disk. We need to give energy for that. W1>W2. Ar t=0, walls of disk has friction with air only at red point, this give forces F1 and F2 and reduce w2, like w1>w2, the energy increase, no ?

Maybe the formula: Energy of disk = 1/2md²w1²+1/2mr²(w1−w2)² is only true for a ring not a disk but the same idea can be use with a ring.

I drawn trajectories in doted lines. It's possible to have trajectories like that:



or trajectories like that if w1 and w2 are well choosen (depends of the radius):



The trajectory must be adapt for have a torque, better than red point, choose P1/P2 points :




and adjust the mass of object to shock for have the same force in value.

I think it's possible to have a disk with a part of mass and another part without mass. The goal is to find 2 points where trajectories are parallel and in other direction like

Like that ?

I made a mistake R1 must be < R2, so value can be:

Vleft=-R1w2+(R1−d)w1
Vright=-R2w2+(R2−d)w1 With:
R1=0.5
R2=15
w1=10
w2=8
d=4
I find:
Vleft=-49
Vright=-20

Add a translation in the good direction for all the system : +30

Vleft=-19
Vright=10

There is a torque only on disk. Energy increase.

I forgot a 2 forces, so there is a torque on black arm, this decrease energy. But the force (F12+F22) on blue axis move all the system, this must increase a little its energy, no ? What's the difference between fixed blue axis and free ? Fixed blue axis F12+F22 don't works and free F12+F22 works.

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

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.

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 ?