Sum of torque

dlouth15 · 23-11-2014 11:45am

neufneufneuf wrote: »

I'm agree with you, and in my calculations I don't take the black arm in the equations. I done all calculations with only 4 red disks and the grey cruz. I find 0 for one disk, and for 4 disks but I have a little error in my last sum if I consider I2 not infinite. I think the integrale is correct because I found 0 for one disk, no ?

Integrals are for adding up an infinite amount of infinitesimal quantities. You are using them to add up non-infinitesimal quantities.

I don't think integration is necessary in these examples. The general method is that you take a very small interval of time [latex]\Delta t[/latex]. Calculate the corresponding changes in the system over [latex]\Delta t[/latex]. This gives quanties like [latex]\Delta W[/latex], [latex]\Delta \omega_1[/latex], [latex]\Delta E_1[/latex], [latex]\Delta E_2[/latex] and so on.

If the total of the small changes in total energy add up to 0, over the small period of time [latex]\Delta t[/latex], then you have shown that the sum of energy over a long period of time does not change.

neufneufneuf · 23-11-2014 11:53am

You are using them to add up non-infinitesimal quantities.

But if it is ok for one disk, the method must work for 4 disks, no? Could you explain please ?

The calculation of kinetic energy of the grey cruz is:

[latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex]

Is it good ? because the 8 is at square, it gives 64/2 and I can't find the term in friction where there is 32.

neufneufneuf · 24-11-2014 8:42am

I changed my message #146, I give a summary with all calculations:

For one disk:

I give my calculations with force, torque, kinetic energy. Like that I can understand where come from my error. I used the same definitions of velocities than in the 4 disks: w3 and I3=>the disk, w2 and I2=>the cruz.

The additionnal kinetic energy for the disk:

[latex]\frac{1}{2}I_3(\omega_3[/latex]+[latex]\frac{Fr}{I_3}t)^2-\frac{1}{2}I_3\omega_3^2=\frac{1}{2}I_3(\omega_3^2[/latex]+[latex]\frac{F^2r^2t^2}{I_3^2}[/latex]+[latex]2\omega_3\frac{Frt}{I_3})-\frac{1}{2}I_3\omega_3^2=Frt\omega_3[/latex]+[latex]\frac{F^2r^2t^2}{2I_3}[/latex]

The additionnal kinetic energy for the arm+disk:

[latex]\frac{1}{2}I_2(\omega_2-\frac{Fr}{I_2}t)^2-\frac{1}{2}I_2\omega_2^2=\frac{1}{2}I_2(\omega_2^2[/latex]+[latex]\frac{F^2r^2t^2}{I_2^2}-2\omega_2\frac{Frt}{I_2})-\frac{1}{2}I_2\omega_2^2=-Frt\omega_2[/latex]+[latex]\frac{F^2r^2t^2}{2I_2}[/latex]

The energie from friction:

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]\int_0^t{Fr((\omega_2-\frac{Frt}{I_2})-(\omega_3+\frac{Frt}{I_3})dt}[/latex] = [latex]\int_0^t{Fr\omega_2-\frac{F^2r^2t}{I_2}-Fr\omega_3-\frac{F^2r^2t}{I_3}dt}[/latex] = [latex]Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3}[/latex]

The difference is well at 0.

For 4 disks, the cruz but not the black arm:

[latex]w_2[/latex]: the rotationnal velocity of the cruz
[latex]w_3[/latex]: the rotationnal velocity of each red disk
[latex]r[/latex]: the radius of a red disk
[latex]I_3[/latex]: Inertia of each red disk
[latex]I_2[/latex]: Inertia of the cruz

There is friction between red disks. [latex]\omega_3 < \omega_2[/latex]. Each red disk is turning around itself counterclockwise. For simplify calculations, I guess the force from friction is always the same F even rotationnal velocity of red disks decreases.

Friction

The energy from friction between each disk is : [latex]F*d[/latex], the force by length (here d is a generic letter not the length of the arm)

[latex]2Fr\omega_{3'(t)}[/latex]

For 4 disks it is :

[latex]4*2Fr\omega_{3'(t)} = 8Fr\omega_{3'm}[/latex]

I can use the same last calculation :

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]8\int_0^t{Fr((\omega_2-\frac{Frt}{I_2})-(\omega_3+\frac{Frt}{I_3})dt}[/latex] = [latex]8\int_0^t{Fr\omega_2-\frac{F^2r^2t}{I_2}-Fr\omega_3-\frac{F^2r^2t}{I_3}dt}[/latex] = [latex]8(Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3})[/latex]

Kinetic energy of red disks

Each disk receive a torque : [latex]Fcos(\frac{\pi}{4})2rcos(\frac{\pi}{4}) = 2Fr[/latex]

The rotationnal velocity of each disk change like :

[latex]\omega_3(t) = \omega_3[/latex] + [latex]\frac{2Fr}{I_3}t[/latex]

The kinetic energy of a red disk change like :

[latex]Ek=+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2[/latex]

For 4 disks it is :

[latex]Ek=2I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2[/latex]

Kinetic energy of the cruz

The cruz receive from each disk a torque of : [latex]2Fcos(\frac{\pi}{4})d = 2Fr[/latex] because [latex]cos(\frac{\pi}{4})d = r[/latex]

The cruz receive from 4 red disks a torque of : [latex]8Fr[/latex]

The rotationnall velocity of the cruz change like :

[latex]\omega_2(t)=\omega_2-\frac{8Fr}{I_2}t[/latex]

The kinetic energy of the cruz change like :

[latex]Ek=\frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2[/latex]

The difference of energies is:

[latex]Ekf = 8(Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3} [/latex]) + [latex](2I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2- 2I_3\omega_3^2)[/latex] + [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex]

[latex]Ekf = 28\frac{F^2r^2t^2}{I_2}[/latex]+[latex]4\frac{F^2r^2t^2}{I_3} [/latex]

The difference is not at 0

dlouth15 · 24-11-2014 3:59pm

neufneufneuf wrote: »

I changed my message #146, I give a

Friction

The energy from friction between each disk is : [latex]F*d[/latex], the force by length (here d is a generic letter not the length of the arm)

[latex]2Fr\omega_{3'(t)}[/latex]

For 4 disks it is :

[latex]4*2Fr\omega_{3'(t)} = 8Fr\omega_{3'm}[/latex]

I can use the same last calculation :

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]8\int_0^t{Fr((\omega_2-\frac{Frt}{I_2})-(\omega_3+\frac{Frt}{I_3})dt}[/latex] = [latex]8\int_0^t{Fr\omega_2-\frac{F^2r^2t}{I_2}-Fr\omega_3-\frac{F^2r^2t}{I_3}dt}[/latex] = [latex]8(Frt\omega_2-\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3})[/latex]

Mistake here with this integral. The earlier integral in the case of the 1 disk that you posted was:

[latex]\int_{0}^{t}Fr((\omega_{2}-\frac{Frt}{I_{2}})-(\omega_{3}+\frac{Frt}{I_{3}})dt[/latex]

Now we need to modify it for the case where the force is 2F as this one is for F. To do this we replace all the F's with 2F in the integral.

So the integral becomes:

[latex]\int_{0}^{t}2Fr((\omega_{2}-\frac{2Frt}{I_{2}})-(\omega_{3}+\frac{2Frt}{I_{3}})dt[/latex]

Now for 4 disks it is

[latex]4\int_{0}^{t}2Fr((\omega_{2}-\frac{2Frt}{I_{2}})-(\omega_{3}+\frac{2Frt}{I_{3}})dt[/latex]

This is not the same mathematically as the integral you wrote.

Expanding we have

[latex]\displayvalue{8\int_{0}^{t}\left(Fr\omega_{2}-2\frac{F^{2}r^{2}t}{I_{2}}-Fr\omega_{3}-2\frac{F^{2}r^{2}t}{I_{3}}\right)\, dt}[/latex]

which I think evaluates to

[latex]\displayvalue{8\left(Frt\omega_{2}-\frac{F^{2}r^{2}t^{2}}{I_{2}}-Frt\omega_{3}-\frac{F^{2}r^{2}t^{2}}{I_{3}}\right)}[/latex]

This is a different result. The second and fourth terms are different to what you got.

neufneufneuf · 24-11-2014 4:10pm

The integral must have the term of w2, so there is another error:

For 4 disks, the cruz but not the black arm:

[latex]w_2[/latex]: the rotationnal velocity of the cruz
[latex]w_3[/latex]: the rotationnal velocity of each red disk
[latex]r[/latex]: the radius of a red disk
[latex]I_3[/latex]: Inertia of each red disk
[latex]I_2[/latex]: Inertia of the cruz

There is friction between red disks. [latex]\omega_3 < \omega_2[/latex]. Each red disk is turning around itself counterclockwise. For simplify calculations, I guess the force from friction is always the same F even rotationnal velocity of red disks decreases.

Friction

The energy from friction between each disk is : [latex]F*d[/latex], the force by length (here d is a generic letter not the length of the arm)

[latex]2Fr\omega_{3'(t)}[/latex]

I can use the same last calculation :

[latex]\omega_3'=\omega_2-\omega_3[/latex]

[latex]8Fr\int_0^t{((\omega_2-\frac{8Frt}{I_2})-(\omega_3+\frac{2Frt}{I_3})dt}[/latex] = [latex]8Fr\int_0^t{\omega_2-\frac{8Frt}{I_2}-Fr\omega_3-\frac{2Frt}{I_3}dt}[/latex] = [latex]8(Frt\omega_2-4\frac{F^2r^2t^2}{I_2}-Frt\omega_3-\frac{F^2r^2t^2}{2I_3})[/latex]

Kinetic energy of red disks

Each disk receive a torque : [latex]Fcos(\frac{\pi}{4})2rcos(\frac{\pi}{4}) = 2Fr[/latex]

The rotationnal velocity of each disk change like :

[latex]\omega_3(t) = \omega_3[/latex] + [latex]\frac{2Fr}{I_3}t[/latex]

The kinetic energy of a red disk change like :

[latex]Ek=+\frac{1}{2}I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2[/latex]

For 4 disks it is :

[latex]Ek=2I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2[/latex]

Kinetic energy of the cruz

The cruz receive from each disk a torque of : [latex]2Fcos(\frac{\pi}{4})d = 2Fr[/latex] because [latex]cos(\frac{\pi}{4})d = r[/latex]

The cruz receive from 4 red disks a torque of : [latex]8Fr[/latex]

The rotationnall velocity of the cruz change like :

[latex]\omega_2(t)=\omega_2-\frac{8Fr}{I_2}t[/latex]

The kinetic energy of the cruz change like :

[latex]Ek=\frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2[/latex]

The difference of energies is:

[latex]Ekf = 8(Frt\omega_2-8\frac{F^2r^2t^2}{2I_2}-Frt\omega_3-\frac{2F^2r^2t^2}{2I_3} [/latex]) + [latex](2I_3(\omega_3[/latex] + [latex]\frac{2Fr}{I_3}t)^2- 2I_3\omega_3^2)[/latex] + [latex]( \frac{1}{2}I_2( \omega_2-\frac{8Fr}{I_2}t )^2-\frac{1}{2}I_2 \omega_2^2)[/latex]

[latex]Ekf = 0[/latex]

I found 0 !! but now, I ask me why I calculated w2-w3 in the integrale, why w2 change something in the work from friction ? The integrale must have only w3', no ?

dlouth15 · 24-11-2014 5:37pm

Well done.

neufneufneuf · 24-11-2014 6:26pm

I have a question, why the friction depends of w2 ?

dlouth15 · 24-11-2014 6:38pm

neufneufneuf wrote: »

I have a question, why the friction depends of w2 ?

In the situation with the rotating arm rotating at rate w2, what is the value of w3 for no friction?

neufneufneuf · 24-11-2014 6:40pm

It is w3'=0 so w2=w3, true ?

dlouth15 · 24-11-2014 6:41pm

neufneufneuf wrote: »

It is w3'=0 so w2=w3, true ?

Yes. So if w3 = w2 there is no friction. Friction depends on the difference between w3 and w2.

neufneufneuf · 24-11-2014 6:47pm

At start, but after when w2 changes it for all disks, no ? Imagine the device with 100 small disks, contact disk/disk will be at the perpendicularity of w2, no ?

dlouth15 · 24-11-2014 6:54pm

" but now, I ask me why I calculated w2-w3 in the integrale, why w2 change something in the work from friction ? The integrale must have only w3', no ?"

The answer to this is simply that w3' = w3 - w2. You have stated this yourself.

If a formula depends on w3' then it depends on w2 since w3' = w2 - w3. You want to use w2 - w3 rather than w3' because otherwise you won't know that the sum of energy = 0 as the other terms use absolute rotational velocities.

If w3 was zero, i.e. no absolute rotation of the disks, there would still be friction caused by the rotations of the arms forcing the disks around one another.

Don't worry, you have solved the problem correctly.

neufneufneuf · 24-11-2014 7:21pm

Look at the image:

I gave trajectories for 2 points (contact disk/disk), there is a little difference because I separated 2 points but in reality it is the same point. How w2 can affect the velocity of the point from w3' ?

It's possible to think like that : there is w2 and w3 at start. No friction between disks. If you change w2 (with your hand), you change w3' ? no, you change w3. The equation w3'=w2-w3 don't take in account the cause and the consequence. The true formula is w3=w2-w3'.

With friction, w2 changes ? true. w3' changes ? true. But the friction don't come with the w3 but with w3'. for me the integrale must be:

[latex]8\int_0^t{Fr((\omega_2-\omega_3-\frac{2Frt}{I_3})dt} = 8Frt\omega_2-8Frt\omega_3-8\frac{F^2r^2t^2}{I_3}[/latex]

dlouth15 · 24-11-2014 8:55pm

neufneufneuf wrote: »

The true formula is w3=w2-w3'.

With friction, w2 changes ? true. w3' changes ? true. But the friction don't come with the w3 but with w3'. for me the integrale must be:

Ok so you are saying now that w3=w2-w3'

That means that w3' = w2 - w3.

That means if w3 increases, w3' decreases, and and if w3' increases, w3 decreases. That does not make sense.

w3' is simply w3 when viewed from the rotating frame of reference. So if w3' increases then so should w3.

What you had earlier is correct. This w3=w2-w3' is wrong.

neufneufneuf · 24-11-2014 9:13pm

An example, at start: the cruz is at w2=10rd/s, a disk is at w3=2 rd/s, so w3'=8 rd/s, true ?

If w3' increases with friction, w3'=9, if I consider w2=10=constant, w3=1, yes, if w3' increases => w3 decreases, why that does not make sense ?

But wait a minute, w2 is clockwise, w3 is clockwise and w3' is counterclockwise, true ?

dlouth15 · 24-11-2014 9:22pm

neufneufneuf wrote: »

yes, if w3' increases => w3 decreases, why that does not make sense ?

Because, and I am going to have to repeat myself here, w3' is just w3 when viewed from a rotating frame of reference. If you have the disk rotating at a rate of w3 and you increase the rate of rotation, then it will increase from the point of view of the arm which is also rotating. If you have w3, you just subtract the speed of rotation of the arm w2 to get w3' which is how the rotation of the disk looks from the point of view of the rotating arm.

So w3' = w3 - w2.

neufneufneuf · 24-11-2014 10:01pm

Because you take in account the sign ? w3'<0 true ? I don't thought like that because I defined w3' counterclockwise, but you're right, it's ok with the sign, this don't change nothing in the sum.

A question about the cause and the consequence:

Imagine, the device is turning with w2 and w3, w3<w2, no friction. With your hand: you change w2, does w3' is changing ? no, you change w3. The equation w3'=w2-w3 don't take in account the cause and the consequence. The true formula is w3=w2-w3'.

With friction, w2 changes ? true. w3' changes ? true. But w2 can't change w3', it changes w3. The friction don't come with the w3 but with w3'. For me the integrale must be:

[latex]8\int_0^t{Fr((\omega_2-\omega_3-\frac{2Frt}{I_3})dt} = 8Frt\omega_2-8Frt\omega_3-8\frac{F^2r^2t^2}{I_3}[/latex]

Even with one disk I was wong:

Arm:
[latex]w_1(t)=w_1-\frac{Frt}{I_1}[/latex]

[latex]\frac{1}{2}I_1(w_1-\frac{Frt}{I_1})^2-\frac{1}{2}I_1w_1^2=-Frtw_1[/latex] +[latex] \frac{F^2r^2t^2}{2I_1}[/latex]

Disk:
[latex]w_2(t)=w_2[/latex] +[latex] \frac{Frt}{I_2}-\frac{Frt}{I_1}[/latex]

[latex]\frac{1}{2}I_2(w_2[/latex]+[latex]\frac{Frt}{I_2}-\frac{Frt}{I_1})^2-\frac{1}{2}I_2w_2^2=\frac{F^2r^2t^2}{2I_2}[/latex]+[latex]\frac{I_2F^2r^2t^2}{2I_1^2}[/latex]+[latex]Frtw_2-\frac{I_2Frtw_2}{I_1}-\frac{F^2r^2t^2}{I_1}[/latex]

Friction:
[latex]Fr\int_0^t{w_1-w_2-\frac{Frt}{I_2}dt}=Frw_1t-Frw_2t-\frac{F^2r^2t^2}{2I_2}[/latex]

Difference of energy:

[latex]-\frac{I_2Frtw_2}{I_1}[/latex]+[latex]\frac{I_2F^2r^2t^2}{2I_1^2}[/latex][latex]-\frac{F^2r^2t^2}{2I_1}[/latex]

neufneufneuf · 29-11-2014 10:45am

Just for understand half part of my problem: without friction between the disk and the arm. The disk is turning at [latex]w_2[/latex], with [latex]w_2 < w_1[/latex]. The arm is turning at [latex]w_1[/latex]. [latex]w_1[/latex] and [latex]w_2[/latex] are in labo frame reference. I let the disk turn at [latex]w_2'[/latex] counterclockwise around itself without friction. I apply an external clockwise torque on the arm with the force [latex]F_a[/latex].

If I guess [latex]w_2[/latex] constant, I don't see a torque on the disque for change [latex]w_2'[/latex]. If [latex]w_2'[/latex] is constant and like [latex]w_2(t)=w_1(t)[/latex]+[latex]w_2'(t)[/latex], if [latex]w_1(t)[/latex] increase then [latex]w_2(t)[/latex] must increase, no ? The disk receive on its axis [latex]F_b[/latex] and it's not a torque for turn the disk around itself. I forgot a force ?

Edit: I'm not sure about the energy needed for the force Fa, so I deleted this part. I will develop this later. [latex]E_k=\frac{1}{4}mr^2w_2^2[/latex]+[latex]\frac{1}{2}md^2w_1^2[/latex], if [latex]w_2[/latex] is increasing the kinetic energy increases too.

dlouth15 · 29-11-2014 2:03pm

neufneufneuf wrote: »

if [latex]w_2[/latex] is increasing the kinetic energy increases too.

Where is the torque which makes [latex]\omega_2[/latex] increase?

neufneufneuf · 29-11-2014 6:39pm

Forget the last case, it was a bad idea, I'm trying to understand but it's not easy to explain. Just for verify and to convince me, I imagine this case:

I rotate the disk alone at w2'(t) counterclockwise, with w2'<0, after I move in translation the disk at velocity V. No friction. I need energy [latex]\frac{1}{2}mV^2[/latex]+[latex]\frac{1}{2}mr^2\omega_2'^2[/latex]. After, I take a rope without mass and rotate the disk like before, with w1>0 clockwise. The rotationnal velocity of the rope is [latex]\omega_1=\frac{V}{d}[/latex], [latex]d[/latex] is the length of the rope.Now the energy is [latex]\frac{1}{4}mr^2\omega_2^2[/latex]+[latex]\frac{1}{2}md^2\omega_1^2[/latex]=[latex]\frac{1}{4}mr^2(\omega_1 [/latex]+[latex] \omega_2')^2[/latex]+[latex]\frac{1}{2}mV^2[/latex] the difference is [latex]\frac{1}{2}mr^2\omega_2'^2[/latex]-[latex]\frac{1}{4}mr^2( \omega_1[/latex]+[latex]\omega_2')^2[/latex], how I can have 0 ? Or w2'(t) change very quickly because I attached a rope to the disk ?

dlouth15 · 29-11-2014 8:10pm

neufneufneuf wrote: »

Forget the last case, it was a bad idea, I'm trying to understand but it's not easy to explain. Just for verify and to convince me, I imagine this case:

1. Imagine a simpler case. A point particle of mass m is travelling with velocity V when it is caught by the massless rope as in the diagram. What is the kinetic energy before and after the capture?

2. Now consider the disk of mass m moving at velocity V. Is the disk's rotation about itself (in the laboratory frame) affected by the capture of the rope?

neufneufneuf · 29-11-2014 8:13pm

1/ the same

2/ that's my problem I can't understand, for me w2'(t) is always the same, even the rope take the disk in rotation. For example, before attached the disk, w2'=-8, with w1=10, why w2 will be at 8 ? It's very strange for me.

dlouth15 · 29-11-2014 8:18pm

neufneufneuf wrote: »

1/ the same

OK.

2/ that's my problem I can't understand, for me w2'(t) is always the same, even the rope take the disk in rotation. For example, before attached the disk, w2'=-8, with w1=10, why w2 will be at 8 ? It's very strange for me.

Forget about the maths. Also forget about rotational frames of reference. Does the disk's rotational velocity change?

neufneufneuf · 29-11-2014 8:21pm

Does the disk's rotational velocity change?

for me the disk is turning around itself at -8 and when the rope take the disk, the rope turns at w1 (and the disk in on the rope), the rope turn at 10, so the disk turn at w2=2 for me

NB do you note the difference 1/2 in first equation and 1/4 in second ?

dlouth15 · 29-11-2014 8:27pm

neufneufneuf wrote: »

for me the disk is turning around itself at -8 and when the rope take the disk, the rope turns at w1 (and the disk in on the rope), the rope turn at 10, so the disk turn at w2=2 for me

NB do you note the difference 1/2 in first equation and 1/4 in second ?

Let's consider the physical situation before looking at the maths.

You are saying that when the disk is taken by the rope, the disk's rotation slows down in the laboratory frame. Is that correct?

neufneufneuf · 29-11-2014 8:28pm

yes, it is

dlouth15 · 29-11-2014 8:31pm

neufneufneuf wrote: »

yes, it is

So the rope prevents rotation of the disk except that after capture the disk must rotate the same way as the rope. i.e. after capture the rotation of the disk = rotation of the rope. Is that correct?

neufneufneuf · 29-11-2014 8:34pm

not really, an example:

before capture:

w2'=-8

After:

w2'=-8
w1=10 (for example)
w2=10-8=2

For me the rotation of the rope change the labo frame reference w2, but I don't understand why w2' would change. For me, the rope can't change the rotation w2'.

dlouth15 · 29-11-2014 8:38pm

neufneufneuf wrote: »

not really, an example:

before capture:

w2'=-8

After:

w2'=-8
w1=10 (for example)
w2=10-8=2

For me the rotation of the rope change the labo frame reference w2, but I don't understand why w2' would change. For me, the rope can't change the rotation w2'.

Does the rope apply a torque to the disk when it captures it?

neufneufneuf · 29-11-2014 8:44pm

No, and for that's the reason w2' is constant for me. If I take a disk that rotates in my hand (I have a gyroscope and I use it like a disk not like a gyroscope), if I move in translation w2' don't change and if I rotate it, w2' don't change or I don't see the change of rotationnal velocity (the disk turns very slowly, not like a gyroscope). Do you done the test ?

dlouth15 · 29-11-2014 8:53pm

neufneufneuf wrote: »

No, and for that's the reason w2' is constant for me. If I take a disk that rotates in my hand (I have a gyroscope and I use it like a disk not like a gyroscope), if I move in translation w2' don't change and if I rotate it, w2' don't change or I don't see the change of rotationnal velocity (the disk turns very slowly, not like a gyroscope). Do you done the test ?

So we now have answers to the earlier questions.

1. Imagine a simpler case. A point particle of mass m is travelling with velocity V when it is caught by the massless rope as in the diagram. What is the kinetic energy before and after the capture?

Answer: the same.

2. Now consider the disk of mass m moving at velocity V. Is the disk's rotation about itself (in the laboratory frame) affected by the capture of the rope?

Answer: There is no torque so the disk's rotation velocity about itself is not affected by the capture of the rope.

Do you agree with the second answer?

neufneufneuf · 29-11-2014 9:03pm

1/ Yes, because it is a point particule and r=0 so kinetic energy is 0

2/ No, if you said "Is the disk's rotation about itself (in the rope frame) affected by the capture of the rope?" I'm agree. Maybe I need to see a video for look the test.

A question:

The disk don't turn around itself, after I turn the rope (or an arm), for you w2' change ? because for me the disk don't rotate around itself.

dlouth15 · 29-11-2014 9:09pm

neufneufneuf wrote: »

1/ Yes, because it is a point particule and r=0 so kinetic energy is 0

2/ No, if you said "Is the disk's rotation about itself (in the rope frame) affected by the capture of the rope?" I'm agree. Maybe I need to see a video for look the test.

A question:

The disk don't turn around itself, after I turn the rope (or an arm), for you w2' change ? because for me the disk don't rotate around itself.

Please leave out the arm frame of reference. Please tell me what happens to the disk. How is it attached to the rope? Is it still free to rotate independently of the rope after it has been attached?

neufneufneuf · 29-11-2014 9:15pm

Is it still free to rotate independently of the rope after it has been attached?

yes, it is free to rotate around itself, there is an axis of rotation, I spoke about a rope but with an arm I drawn an axis where the disk can rotate around itself. I drawn an image:

It is a side view

dlouth15 · 29-11-2014 9:18pm

Ok then the arm or rope exerts no torque on the disk about the disk's centre of rotation when the disk becomes attached.

If there is no torque then there is no change in the absolute rotational velocity of the disk. The only way a rotational velocity can change is through torque.

Do you agree with this?

neufneufneuf · 29-11-2014 9:28pm

Yes I'm agree, it is just me that thought it was noted w2' but it's true it is w2. It is w2' that depend of w1, it's that ?

dlouth15 · 29-11-2014 9:38pm

Right so we have:

So we now have answers to the earlier questions.

1. Imagine a simpler case. A point particle of mass m is travelling with velocity V when it is caught by the massless rope as in the diagram. What is the kinetic energy before and after the capture?

Answer: the same.

2. Now consider the disk of mass m moving at velocity V. Is the disk's rotation about itself (in the laboratory frame) affected by the capture of the rope?

Answer: There is no torque so the disk's rotation velocity about itself is not affected by the capture of the rope.

You now agree finally with these answers.

Now please do the following.

Tell me the energy of the system prior to capture by the rope. It will be in two forms parts. 1) energy due to translational velocity of the disk. and 2) energy due to the rotational velocity of the disk. Remember. Just these two quantities. Prior to capture.

neufneufneuf · 29-11-2014 9:46pm

Yes, we're agree. And yes, there are only 2 quantities before capture.

I need :

Rotation:
[latex]\frac{1}{2}mr^2\omega_2^2[/latex]

Translation:
[latex]\frac{1}{2}mV^2[/latex]

After capture I was wrong:
[latex]\frac{1}{4}mr^2\omega_2^2[/latex]+[latex]\frac{1}{2}md^2\omega_1^2[/latex]=[latex]\frac{1}{4}mr^2( \omega_2)^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

There is a problem with 1/4

Edit: maybe I don't understand the question, you wanted only before, not after capture ?

dlouth15 · 29-11-2014 9:53pm

OK. and since [latex]V[/latex] before capture = [latex]d\omega_1[/latex] after capture, there is no change in the overall energy of the system before and after capture of the rope. Energy is conserved.

neufneufneuf · 29-11-2014 9:54pm

There is 1/4, no ?

dlouth15 · 29-11-2014 10:04pm

On the left hand side you have

[latex]\frac{1}{4}mr^2\omega_2^2[/latex]+[latex]\frac{1}{2}md^2\omega_1^2[/latex]

On the right hand side:

[latex]\frac{1}{4}mr^2( \omega_2)^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

The first terms:

[latex]\frac{1}{4}mr^2\omega_2^2[/latex] = [latex]\frac{1}{4}mr^2( \omega_2)^2[/latex]

These are OK and are both the same. The 1/4 is OK.

The second terms on each side:

[latex]\frac{1}{2}md^2\omega_1^2[/latex] = [latex]\frac{1}{2}mV^2[/latex] because [latex]V=d\omega_1[/latex].

So left hand side = right hand side. No change in energy. Energy is conserved.

Well done.

neufneufneuf · 29-11-2014 10:11pm

I gave the energy after capture, it is not before/after.

The kinetic energy of a disk is not [latex]\frac{1}{2}mr^2\omega^2[/latex] ?

Before capture it is:
[latex]\frac{1}{2}mr^2 \omega_2^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

After capture it is:
[latex]\frac{1}{4}mr^2\omega_2^2[/latex]+[latex]\frac{1}{2}md^2\omega_1^2[/latex]=[latex]\frac{1}{4}mr^2 \omega_2^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

Edit: I'm wrong, kinetic energy is 1/2 of Iw², all is fine, it's ok.

dlouth15 · 29-11-2014 10:23pm

neufneufneuf wrote: »

I gave the energy after capture, it is not before/after.

The kinetic energy of a disk is not [latex]\frac{1}{2}mr^2\omega^2[/latex] ?

Rotational kinetic energy of a disk:
[latex]\frac{1}{2}I\omega^{2}=\frac{1}{2}\left(\frac{1}{2}mr^{2}\right)\omega^{2}=\frac{1}{4}mr^{2}\omega^{2}[/latex]

Yes.

Before capture it is:
[latex]\frac{1}{2}mr^2 \omega_2^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

Should be

[latex]\frac{1}{4}mr^2 \omega_2^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

before capture.

After capture it is:
...
[latex]\frac{1}{4}mr^2 \omega_2^2[/latex]+[latex]\frac{1}{2}mV^2[/latex]

I agree with that.

So energy before capture = after capture because [latex]d\omega_2=V[/latex]

neufneufneuf · 29-11-2014 10:26pm

Great, thanks for your help I understood a lot of things

neufneufneuf · 29-11-2014 10:53pm

I come back about the signs of works. We noted the work from friction positive (absolute value), ok ? And with math, we know that the energy is conserved: kinetic + heating (thank again to you). But if I take only kinetic energy, it's not possible to have a positive value because there are positives terms and negative term. At start, I calculated energy from friction and I found something negative because I apply maths, but after I think friction is always positive and I give the absolute value. Don't consider friction, just kinetic energy:

Arm:
[latex]-Frt\omega_1[/latex]+[latex]\frac{F^2r^2t^2}{2I_1}[/latex]

Disk
[latex]Frt\omega_2[/latex]+[latex]\frac{F^2r^2t^2}{2I_2}[/latex]

The sum:
[latex]-Frt\omega_1[/latex]+[latex]\frac{F^2r^2t^2}{2I_1}[/latex]+[latex]Frt\omega_2[/latex]+[latex]\frac{F^2r^2t^2}{2I_2}[/latex]

For example, if I2 is very small ? The sign can't be positive ?

With :

F=1000
R=1
T=1
I1=10
I2=1
w1=10
w2=2

The sum is positive.

Edit: even I take t very small the sum is positive with these values and if necessary decrease I2

dlouth15 · 29-11-2014 11:27pm

Rather than work out this which could take a long time, i will just give general advice.

1. Think about the physical process before trying to model it with mathematics. In the last example you got confused with relative rotational velocities and the like when in fact the problem was very simple.

2. Try to simplify the model. Is the physics demonstrated with a simpler model? For example the four disks on the crossed arm which was also on a longer arm: the relevant physical properties are illustrated with a single disk on a single arm with the disk undergoing a torque.

3. Avoid relative rotational velocities, i.e., rotational velocities when viewed from a rotating frame. Particularly avoid accelerating rotating frames of reference. You should deal with absolute rotations where possible.

4. If several quantities are varying then use a small time interval [latex]\Delta t[/latex] to determine small quantities of energy [latex]\Delta E[/latex]. Use the Greek letter Delta [latex]\Delta[/latex] for this purpose. If the sum of energy is zero for this small interval it will be zero over time. No need for integration.

5. Use torque or some force instead of friction where possible. Friction reverses direction in order to oppose relative motion of surfaces and this can be confusing. It can be hard to determine which direction it is acting when you have lots of rotations.

6. Let positive values for rotations, torques etc represent clockwise rotations. Let negative values represent counterclockwise. Or the reverse, but be consistent.

7. Don't, if possible, depend on some quantities being very small or very large, or torques being clockwise or counterclockwise (i.e. positive or negative). Try to produce results as general as possible.

8. You will find given enough work, that energy is always conserved but it may not always seem so initially.

neufneufneuf · 30-11-2014 10:31am

If I want this:

[latex]-Frt\omega_1[/latex]+[latex]\frac{F^2r^2t^2}{2I_1}[/latex]+[latex]Frt\omega_2[/latex]+[latex]\frac{F^2r^2t^2}{2I_2}>0[/latex]

It's not possible with one disk nor even 4 disks because for have forces like I want I have an expression between w1(t) and w2(t): w1(t) > w2(t). Just a question about this case:

The only problem in the last equation is -Frtw1 because it is negative. If I place the cruz on the arm and I rotate the black arm it's not possible to reduce w1 for reduce -Frtw1 ? We know that the arm don't receive a torque. w2 change if there is an arm ?

dlouth15 · 30-11-2014 10:43am

neufneufneuf wrote: »

The only problem in the last equation is -Frtw1 because it is negative. If I place the cruz on the arm and I rotate the black arm it's not possible to reduce w1 for reduce -Frtw1 ? We know that the arm don't receive a torque. w2 change if there is an arm ?

Does the black arm exert a torque on the cross in any way?

neufneufneuf · 30-11-2014 11:02am

No. The arm don't change the rotationnal velocity of the cruz (with equation of rotationnal velocity) ?

Edit: if I define w0 the rotationnal velocity of the arm, can I take w1=0 and w0 like w1 ?

dlouth15 · 30-11-2014 12:01pm

neufneufneuf wrote: »

No. The arm don't change the rotationnal velocity of the cruz (with equation of rotationnal velocity) ?

So that means that we can remove it from the diagram. It doesn't change the rotation of the rest of the system.

Edit: if I define w0 the rotationnal velocity of the arm, can I take w1=0 and w0 like w1 ?

Not sure what you mean.

Why not just ignore [latex]\omega_1[/latex]? The black arm does not affect the rotation of the cross, and nor does the rotation of the cross affect the rotation of black arm. Why not just get rid of it?

Sum of torque

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