Roll up a spring on a screw thread in a disk on rotation

[swansont]

1 hour ago, xyrth said:

Why ? for a small angle of study δ=1e-10rd, the distance increased by the spring is sqrt(3)*δ*R, it is 3.46e-11 m, the force is constant for that small difference of length.

I said "and for that" meaning for the situation I described (and you had originally described)

Quote

And it is possible in theory to have a "device" like a spring but with a constant force.

No, for a small angle of rotation δ. And it is possible to have a device composed of several springs to have a constant force:

But you did not say a constant force device. You said a spring.

 

I am now well past the point where I think this is a useful conversation. I'm trying to point out reasons why your analysis failed. It is incredibly frustrating for you to keep changing the example.

[xyrth]

Ok, I done calculations without any motor.

 

No external gravity. No friction. No external motor.
Rotational velocity of the disk at start: w0
The distance of the center of mass of the screw-thread from the center of the disk at start: d
The mass of the screw-thread: m
The mass of the disk: md = 0
Radius of the screw-thread: R
The inertia of the thread in its main own axis (when the thread rotates around itself) : Ir
The inertia of the thread in axis y: Iy
At start, time t = 0 s
The force of the device (I drew a spring but it is a composition of springs) composed of several springs is constant: F

The center of mass of the screw-thread moves away the center of the disk at the speed √3/2*2πR for each turn of the screw-thread on the disk. The rotational velocity of the screw-thread relatively to the disk is w(t)*cos(30°). So, the center of mass moves at v(t)=√3/2*2πR*w(t)*cos(30°)/2/π = √3/2*R*w(t)*cos(30°) = 3/4*R*w(t)
Like there is no external motor the rotational velocity of the disk is a function of time: w(t)=w0-d/(d+v(t)*t)
So the screw-thread is turning at ws(t)=√3/2*w(t) on the disk

When I attach at time ‘t0’ the spring the angular velocities are changed and I need to take in accound the inertia of the screw-thread. At time = t1, I stop to count the energies.

With the spring the angular velocity of the disk is changed:
wr(t)=w(t)-(√3/2+1/2)RF/Ir*(t-t0)

And the thread is also changed in:
wsr(t)=√3/2*wr(t)-1/2*RF/Iy*(t-t0)

For a small angle from the time t0 I attached the spring, (t1-t0) near equal to 0 so -(√3/2+1/2)RF/Ir*(t1-t0) near to 0 and -1/2*RF/Iy*(t1-t0) near to 0, the energies are counted from t0 to t1, the energies are:

The energy won by the spring is :

+(wsr(t1)+wrs(t0))/2*2*RF=√3/2*(wr(t1)+wr(t0))/2*2*RF

The energy lost from the torque of the force F on the disk, at distance R from the center of the disk is :

-(wr(t1)+wr(t0))/2*√3/2*RF

The energy lost from the force F on the axis ‘y’ on the disk is:

-(wr(t1)+wr(t0))/2*1/2*RF

The energy lost by the center of mass is:

 -3/4*RF*(wsr(t1)+wsr(t0))/2  = - 3/4*√3/2*RF*(wr(t1)+wr(t0))/2

The sum of energy is well at:

- 0.28*RF*(wr(t1)+wr(t0))/2

Edited September 30, 2018 by xyrth

[xyrth]

No external gravity. No friction. No external motor.

 

 

Rotational velocity of the disk at start: w0

The distance of the center of mass of the screw-thread from the center of the disk at start: d

The mass of the screw-thread: m

The mass of the disk: md = 0

Radius of the screw-thread: R

The inertia of the thread in its main own axis (when the thread rotates around itself) : Ir

The inertia of the thread in axis y: Iy

At start, time t = 0 s

The force of the device composed of several springs is constant: F

 

The center of mass of the screw-thread moves away the center of the disk at the speed √3/2*2πR for each turn of the screw-thread on the disk. The rotational velocity of the screw-thread relatively to the disk is w(t)*cos(30°). So, the center of mass moves at v(t)=√3/2*2πR*w(t)*cos(30°)/2/π = √3/2*R*w(t)*cos(30°) = 3/4*R*w(t)

Like there is no external motor the rotational velocity of the disk is a function of time: w(t)=w0-d/(d+v(t)*t)

So the screw-thread is turning at ws(t)=√3/2*w(t) on the disk

 

When I attach at time ‘t0’ the spring the angular velocities are changed and I need to take in accound the inertia of the screw-thread. At time = t1, I stop to count the energies.

 

With the spring the angular velocity of the disk is changed:

wr(t)=w(t)-(√3/2+1/2)RF/Ir*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wr(t)*(t1-t0)

= (w(t)-(√3/2+1/2)RF/Ir*(t1-t0) ) * (t1-t0)

 

The major part of the angle is changed by w(t) not by (√3/2+1/2)RF/Ir*(t-t0)

 

And the angular velocity of the thread is also changed by the spring in:

wsr(t)=√3/2*wr(t)-1/2*RF/Iy*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wsr(t)*(t1-t0)

= (√3/2*wr(t)-1/2*RF/Iy*(t1-t0)) * (t1-t0)

 

The major part of the angle is changed by wr(t) not by 1/2*RF/Iy*(t-t0)

 

The energy won by the spring is :

 

+(wsr(t1)+wrs(t0))/2*(t1-t0)*2*RF

= +( √3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

= +( √3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

 

The energy lost from the torque of the force F on the disk, at distance R from the center of the disk is :

 

-(wr(t1)+wr(t0))/2*(t1-t0)*√3/2*RF

= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0)+w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))/2*(t1-t0)*√3/2*RF

 

The energy lost from the force F on the axis ‘y’ on the disk is:

 

-(wr(t1)+wr(t0))/2*(t1-t0)*1/2*RF
= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0) + w(t0)-(√3/2+1/2)RF/Ir*(t0-t0) )/2*(t1-t0)*1/2*RF

 

 

The energy lost by the center of mass is:

 

 -3/4*RF*(wsr(t1)+wsr(t0))/2*(t1-t0)

=  -3/4*RF*(√3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

= -3/4*RF*(√3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

 

 

For a small angle from the time t0 I attached the spring, (t1-t0) near equal to 0 but -(√3/2+1/2)RF/Ir*(t1-t0)² is 10 times smaller and -1/2*RF/Iy*(t1-t0)² is 10 times smaller too, the energies are counted from t0 to t1, the energies are:

 

 

The sum of energy is well at:

 

- 0.28*RF*(t1-t0)*(wr(t1)+wr(t0))/2

 

Edited September 30, 2018 by xyrth

[Strange]

Perhaps you need to think of some way of checking each step of your calculation individually to find the error. (You do know there is an error, yes?)

[xyrth]

28 minutes ago, Strange said:

Perhaps you need to think of some way of checking each step of your calculation individually to find the error. (You do know there is an error, yes?)

It is what I'm doing, I found a little mistake but it is more for the expression of the angular velocity of the disk due to the movement of the center of mass:

 

No external gravity. No friction. No external motor.

 

 

Rotational velocity of the disk at start: w0

The distance of the center of mass of the screw-thread from the center of the disk at start: d

The mass of the screw-thread: m

The mass of the disk: md = 0

Radius of the screw-thread: R

The inertia of the thread in its main own axis (when the thread rotates around itself) : Ir

The inertia of the thread in axis y: Iy

At start, time t = 0 s

The force of the device composed of several springs is constant: F

 

The center of mass of the screw-thread moves away the center of the disk at the speed √3/2*2πR for each turn of the screw-thread on the disk. The rotational velocity of the screw-thread relatively to the disk is w(t)*cos(30°). So, the center of mass moves at v(t)=√3/2*2πR*w(t)*cos(30°)/2/π = √3/2*R*w(t)*cos(30°) = 3/4*R*w(t)

Like there is no external motor the rotational velocity of the disk is a function of time: w(t)=d/(d+v(t)*t)*w0

The screw-thread is turning at ws(t)=√3/2*w(t) on the disk

 

When I attach at time ‘t0’ the spring the angular velocities are changed and I need to take in accound the inertia of the screw-thread. At time = t1, I stop to count the energies.

 

With the spring the angular velocity of the disk is changed:

wr(t)=w(t)-(√3/2+1/2)RF/Ir*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wr(t)*(t1-t0)

= (w(t)-(√3/2+1/2)RF/Ir*(t1-t0) ) * (t1-t0)

 

The major part of the angle is changed by w(t) * (t1-t0) not by (√3/2+1/2)RF/Ir*(t1-t0)²

 

And the angular velocity of the thread is also changed by the spring in:

wsr(t)=√3/2*wr(t)-1/2*RF/Iy*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wsr(t)*(t1-t0)

= (√3/2*wr(t)-1/2*RF/Iy*(t1-t0)) * (t1-t0)

 

The major part of the angle is changed by wr(t)* (t1-t0) not by 1/2*RF/Iy*(t1-t0)²

 

The energy won by the spring is :

 

+(wsr(t1)+wrs(t0))/2*(t1-t0)*2*RF

= +( √3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

= +( √3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

 

The energy lost from the torque of the force F on the disk, at distance R from the center of the disk is :

 

-(wr(t1)+wr(t0))/2*(t1-t0)*√3/2*RF

= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0)+w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))/2*(t1-t0)*√3/2*RF

 

The energy lost from the force F on the axis ‘y’ on the disk is:

 

-(wr(t1)+wr(t0))/2*(t1-t0)*1/2*RF
= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0) + w(t0)-(√3/2+1/2)RF/Ir*(t0-t0) )/2*(t1-t0)*1/2*RF

 

 

The energy lost by the center of mass is:

 

 -3/4*RF*(wsr(t1)+wsr(t0))/2*(t1-t0)

=  -3/4*RF*(√3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

= -3/4*RF*(√3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

 

 

For a small angle from the time t0 I attached the spring, (t1-t0) near equal to 0 but -(√3/2+1/2)RF/Ir*(t1-t0)² is 10 times smaller and -1/2*RF/Iy*(t1-t0)² is 10 times smaller, I can approximate the sum of energy.

 

 

The sum of energy is well at:

 

- 0.28*RF*(t1-t0)*(wr(t1)+wr(t0))/2

[xyrth]

I corrected the comparison about the factor (t1-t0) and (t1-t0)², the text in red color. I'm looking for another mistake.

 

No external gravity. No friction. No external motor.

 

 

Rotational velocity of the disk at start: w0

The distance of the center of mass of the screw-thread from the center of the disk at start: d

The mass of the screw-thread: m

The mass of the disk: md = 0

Radius of the screw-thread: R

The inertia of the thread in its main own axis (when the thread rotates around itself) : Ir

The inertia of the thread in axis y: Iy

At start, time t = 0 s

The force of the device composed of several springs is constant: F

 

The center of mass of the screw-thread moves away the center of the disk at the speed √3/2*2πR for each turn of the screw-thread on the disk. The rotational velocity of the screw-thread relatively to the disk is w(t)*cos(30°). So, the center of mass moves at v(t)=√3/2*2πR*w(t)*cos(30°)/2/π = √3/2*R*w(t)*cos(30°) = 3/4*R*w(t)

Like there is no external motor the rotational velocity of the disk is a function of time: w(t)=d/(d+v(t)*t)*w0

The screw-thread is turning at ws(t)=√3/2*w(t) on the disk

 

When I attach at time ‘t0’ the spring the angular velocities are changed and I need to take in accound the inertia of the screw-thread. At time = t1, I stop to count the energies.

 

With the spring the angular velocity of the disk is changed:

wr(t)=w(t)-(√3/2+1/2)RF/Ir*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wr(t)*(t1-t0)

= (w(t)-(√3/2+1/2)RF/Ir*(t1-t0) ) * (t1-t0)

 

The major part of the angle is changed by w(t) * (t1-t0) not by (√3/2+1/2)RF/Ir*(t1-t0)²

 

And the angular velocity of the thread is also changed by the spring in:

wsr(t)=√3/2*wr(t)-1/2*RF/Iy*(t-t0)

 

For a delta time t1-t0, the angle is :

 

wsr(t)*(t1-t0)

= (√3/2*wr(t)-1/2*RF/Iy*(t1-t0)) * (t1-t0)

 

The major part of the angle is changed by wr(t)* (t1-t0) not by 1/2*RF/Iy*(t1-t0)²

 

The energy won by the spring is :

 

+(wsr(t1)+wrs(t0))/2*(t1-t0)*2*RF

= +( √3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

= +( √3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0) ) /2 *(t1-t0) *2*RF

 

The energy lost from the torque of the force F on the disk, at distance R from the center of the disk is :

 

-(wr(t1)+wr(t0))/2*(t1-t0)*√3/2*RF

= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0)+w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))/2*(t1-t0)*√3/2*RF

 

The energy lost from the force F on the axis ‘y’ on the disk is:

 

-(wr(t1)+wr(t0))/2*(t1-t0)*1/2*RF
= -(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0) + w(t0)-(√3/2+1/2)RF/Ir*(t0-t0) )/2*(t1-t0)*1/2*RF

 

 

The energy lost by the center of mass is:

 

 -3/4*RF*(wsr(t1)+wsr(t0))/2*(t1-t0)

=  -3/4*RF*(√3/2*wr(t1)-1/2*RF/Iy*(t1-t0) + √3/2*wr(t0)-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

= -3/4*RF*(√3/2*(w(t1)-(√3/2+1/2)RF/Ir*(t1-t0))-1/2*RF/Iy*(t1-t0) + √3/2*(w(t0)-(√3/2+1/2)RF/Ir*(t0-t0))-1/2*RF/Iy*(t0-t0))/2*(t1-t0)

 

 

For a small angle from the time t0 I attached the spring, (t1-t0) near equal to 0 but -(√3/2+1/2)RF/Ir*(t1-t0)² is near 0 at square and for -1/2*RF/Iy*(t1-t0)² it is the same, I can approximate the sum of energy. For example if t1-t0=0.01 then (t1-t0)²=0.0001.

 

 

The sum of energy is well at:

 

- 0.28*RF*(t1-t0)*(wr(t1)+wr(t0))/2

 

In fact, it is logical because the angular velocity for the disk is w(t)=w0-k*t with 'k' a constant and the angle of rotation is w(t)*t = (w0-k*t)*t and if t is small the factor kt² is negligible. So, one work I calculated is wrong.

Edited September 30, 2018 by xyrth