Rotationnal velocity and acceleration

[Del8]

Hello,

I would like to be sure about my rotationnal velocities w1(t) and w2(t):

http://imagizer.imageshack.us/v2/800x600q90/912/c6X2du.png The black arm is turning clockwise at w1. The blue disk is turning at w2 with w2 < w1. w1 and w2 are in labo frame reference. The disk is turning around itself counterclockwise at w2'=w2-w1 in the arm frame reference.

I1: inertia of black arm+disk
I2: inertia of the disk
F: force from friction
r: radius of the disk
t: time.

I guess no friction elsewhere than between disk and black arm. I guess that the force F from friction is constant even rotationnal velocities are changing, just for simplify calculations. Friction gives forces F1 and F2.

For me, while w2<w1 and while the force F can be constant :

w1(t)=w1-Frt/I1

w2(t)=w2+Frt/I2-Frt/I1For you, is it ok ?

 

[Bystander]

You are missing a radius in the problem statement, are you not?

 

[Del8]

I defined the radius of the disk: 'r' and I take in account in w1(t) and w2(t) with Fr=F*r, I need another radius ?

 

[Bystander]

An arm radius? You're trying to conserve angular momentum.

 

[Del8]

I need it for the rotationnal velocities ? could you explain please ? There is friction, so energy of heating.

 

[zoki85]

What is this device?

 

[Bystander]

I need it for the rotationnal velocities ? could you explain please ?

You need it to calculate angular momentum, the quantity that must be conserved as the two angular velocities change.

Edit: Hang on a minute, that's built into "I." Let me double-check myself --- I keep thinking you need it to get the angular accelerations.

 

[Del8]

but there is energy of heating too, no ? angular momentum is conserved even there is energy from friction ?

@zoki: just a disk with an arm, and friction between when w2 < w1.

 

[Bystander]

The angular momentum is conserved. Total energy is conserved as that dissipated by friction plus the remaining rotational energy.

ω₁I₁ + ω₂I₂ = constant;

E = I₁ω₁² + I₂ω₂² - Frtω₂.

... and, one radius is all you need. My apologies.

What is this device?

You could call it a "tidal orrery" to demonstrate tidal friction --- doesn't change orbital radius of the moon, though.

 

[Del8]

my expressions of w1(t) and w2(t) are correct, or not ? If I use your formula this would say that w2(t)=w2+Frt/I2, true ? w2(t) is increasing because it receives a clockwise torque from F2/F3, but in the same time w1(t) is decreasing of -Frt/I1, for me w2(t) depends of the change of w1(t) because the disk is on the arm, no ?

Maybe I need to define moment of inertia:

I1=Md²/3+md², for the arm+disk, 'd' is the length of arm, 'M' the mass of arm
I2=mr²/2, for the disk, 'm' the mass of the disk

 

[Bystander]

From conservation of angular momentum, ω₁(t=0)I₁ + ω₂(t=0)I₂ = constant = ω₁(t)I₁ + ω₂(t)I₂.

Rearranging, ω₁(t) = ω₁(0) - (I₂/I₁)(ω2(t) - ω₂(0)).

The expression for conservation of energy is not strictly correct; the friction loss is a time dependent rate and must be integrated from t=0 to t of interest.

 

[Del8]

Rearranging, ω₁(t) = ω₁(0) - (I₂/I₁)(ω2(t) - ω₂(0)).

I need to have 2 equations: w1(t) and w2(t), so w1(t)=w1-Frt/I1 is correct ? Your equation is for find w2(t) ? Edit: So like this: w2(t)=w2+Frt/I2

 

[Bystander]

I need to have w1(t) and w2(t) so w1(t)=w1-Frt/I1 is correct ? Your formula is for find w2(t) ?

No, and no. That is ω₁(t) in terms of ω₁(0), ω₂(t), ω₂(0), I₁, and I₂.

 

[Del8]

Ok, but I have only one equation, and I would like to find w1(t) and w2(t), how can I do ?

 

[Bystander]

I've shown you a rearrangement for ω₁(t); you can rearrange in similar fashion for ω₂(t).

 

[Del8]

Ok, could you explain why I can't use the angular acceleration for have angular velocities:

http://en.wikipedia.org/wiki/Angular_acceleration

w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1 ?

 

[Bystander]

I₁ and I₂ are not necessarily equal from the information given.

 

[Del8]

Sure, but why I can't write:

w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
w2(t)=w2(0)-Torque/I2*t = w2(0)-Frt/I2

Why the formula of angular acceleration is not correct in this case ?

With:

I1=Md²/3+md², for the arm+disk, 'd' is the length of arm, 'M' the mass of arm
I2=mr²/2, for the disk, 'm' the mass of the disk, 'r' the radius of the disk

 

[Bystander]

Nor, are the two torques equal.

 

[Del8]

Sorry, it is:

w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
w2(t)=w2(0)+Torque/I2*t = w2(0)+Frt/I2

Why the formula of angular acceleration is not correct in this case ?

 

[Bystander]

The forces are equal; the arm lengths, and, therefore, the torques are not.

 

[Del8]

Look at the image please:

http://imagizer.imageshack.us/v2/800x600q90/912/c6X2du.png

friction gives F1 and F2 forces. The disk receives F2, it applies F4 on arm, the reaction is F3 on disk, the torque on disk is F*r. For the arm, it has F1 and F4, so the torque is -F*r, no ? I'm wrong in my forces ?

 

[Bystander]

ω

Look at the image please:

http://imagizer.imageshack.us/v2/800x600q90/912/c6X2du.png

friction gives F1 and F2 forces. The disk receives F2, it applies F4 on arm, the reaction is F3 on disk, the torque on disk is F*r. For the arm, it has F1 and F4, so the torque is -F*r, (comment on highlighted item: torque changing ω₁ is (-)F x arm length, not disc radius) no ? I'm wrong in my forces ?

 

[Del8]

So, for you F4 doesn't exist ? The sum of forces must at 0.

 

[Bystander]

The torque about the fixed point is F₄ x arm length, not r.

 

[Del8]

Friction applies F1 on arm too I think. w2 < w1: the disk is turning around itself counterclockwise.

Edit: Torque on arm = -d*F4+(d-r)F1 = -rF

 

[Bystander]

Let's clarify your definitions of forces: you have F₁ and F₂ acting on the arm, and at the rim of the disc to retard the rotation of the disc. You have F₃ and F₄ acting at the end of the arm and on the center of the disc as reaction forces to F₁ and F₂.

I've been looking at your diagram and assuming one thing, and I think you've been meaning something else. For starters, specify the reaction pairs and their directions, so that I can understand what you mean.

 

[Del8]

I think the image is not clear enough, sorry, F1 and F4 acting on the arm (source of forces is on black part), F2 and F3 acting on the disk (source of forces is on blue part). Friction gives forces F1 and F2 forces. Is it ok like that ?

I define:

w1(0)=w1
w2(0)=w2

With:

w1(t)=w1-Torque/I1*t = w1-Frt/I1
w2(t)=w2+Torque/I2*t = w2+Frt/I2

I have the angular momentum constant I think:

w1(t)I1+w2(t)I2 = w1I1-Frt+w2I2+Frt = w1I1+w2I2 = K

Remember w2(t) is a labo frame reference angular velocityWithout friction between disk and arm, just for understand my problem:

1/ Fixed the arm, you apply an external torque Fr on the disk, the angular velocity will be: w2(t)= w2+Frt/I2
2/ Arm is not fixed, w2(t) is constant, you apply an external torque -Fr on the arm, why w2(t) don't take in account the change of the rotationnal velocity of the arm ? When w1(t) is changing, the rotationnal velocity of disk around itself don't change (no friction), so for me w2(t) must decrease.
3/ If an external torque Fr on disk and an external torque -Fr on arm are apply, why w2(t) don't take in account the change of the rotationnal velocity of the arm ?

With friction it is the case 3/.

For me, I don't understand why the velocity of the disk is not : w2(t) = w2+Frt/I2-Frt/I1 (and it can't) because in this case the angular momentum is not constant. Do you understand my problem ?

 

[Bystander]

That helps me. Give me a couple hours to work my way through it, and then I should be able to clarify it for you.

 

[Bystander]

I think the image is not clear enough, sorry, F1 and F4 acting on the arm (source of forces is on black part), F2 and F3 acting on the disk (source of forces is on blue part). Friction gives forces F1 and F2 forces. Is it ok like that ? (We'll make it work. I'm going to compact things just to save space.)
I define: w1(0)=w1; w2(0)=w2
With: (I'll insert a couple notational corrections in green for you, but otherwise you're good to go.)
w1(t)=w1-(Torque/I1)*t = w1-Frt/I1; w2(t)=w2+(Torque/I2)*t = w2+Frt/I2
I have the angular momentum constant I think: (This is where I've been getting confused and doing my best to confuse you.)
w1(t)I1+w2(t)I2 (angular momentum, good) = w1I1 (angular momentum, good) -Frt +w2I2 (angular momentum, good)+Frt (snip)

Without friction between disk and arm, just for understand my problem:
1/ Fixed the arm, you apply an external torque Fr on the disk, the angular velocity will be: w2(t)= w2+Frt/I2 (looks good)
2/ Arm is not fixed, w2(t) is constant, you apply an external torque -Fr on the arm, why w2(t) don't take in account the change of the rotationnal velocity of the arm ? (Frictional force isn't operating) When w1(t) is changing, the rotationnal velocity of disk around itself don't change (no friction), so for me w2(t) must decrease. (Relative to the arm, not relative to the laboratory so long as friction force is zero)(snip)
For me, I don't understand why the velocity of the disk is not : w2(t) = w2+Frt/I2-Frt/I1 (and it can't) because in this case the angular momentum is not constant. Do you understand my problem ?

(Finally got the picture. The wrong person has been helping you )

We'll look further at your case 1): the arm is fixed, and the disc is spinning at a constant angular velocity of ω₂, and we'll apply the frictional force between the edge of the disc and the arm at time t = 0; the torque applied to the disc is Fr, the rate of change of angular velocity of the disc is dω₂/dt, or "ω₂ dot," and "ω₂ dot" = Fr/I₂, and ω₂(t) = ω₂(0) + "ω₂ dot" x t. At the same time, the frictional force is applied in the opposite direction to the arm at a distance d-r from the center pivot (and here's where I went off track trying to explain things to you) and the torque, still Fr, is trying to rotate the arm in the same direction as the disc rotation about the center of the disc. The arm is prevented from rotating by it's fixed pivot point, and by the disc pivot. If we free the arm from the fixed pivot it will accelerate about the disc pivot from a zero angular velocity in the same direction of rotation as the disc at F₁r/I_arm, I_arm < I₂. The force necessary to prevent the arm from moving away from the fixed pivot is found from the lever law, F_al_a = F_bl_b where the short lever is r, distance from disc center to rim, and the long lever is d, distance from fixed pivot to disc center, acting through a fulcrum at the disc center. Or, torque acting about the axis of the disc equals torque acting about the axis of the fixed pivot. Relating these to your diagram, F_pivotd = F₂r, where F_pivot = F₃ in the diagram, and F₃ = F₂r/d. Stated another way, you have a lever with one arm length the disc radius r, and the other arm the arm length between the pivot and the disc center acting across a fulcrum at the disc center; the lever can be straight, bent, or reflex, as long as the fulcrum is at the disc center, you get the same mechanical advantage.

Moment of inertia of the arm plus the disc about the fixed pivot is m_armd²/3 + m_discd² since the center of mass of the disc is at distance d from the pivot. Moment of inertia of the disc is what you've stated, I₂ = m_discr² if the mass is concentrated at the radius.

(snip)
I1=Md²/3+md², for the arm+disk, 'd' is the length of arm, 'M' the mass of arm
I2=mr²/2, for the disk, 'm' the mass of the disk

We agree.

Then, torque on the disc = F_frictionr, and torque about the fixed pivot = -F_frictionrd/d.

ω₁(t) = ω₁(0) - F_frictionrdt/(dI₁), and ω₂(t) = ω₂(0) + F_frictionrdt/(dI₂); the d cancels in both, ω₁(t) = ω₁(0) - F_frictionrt/(I₁), and ω₂(t) = ω₂(0) + F_frictionrt/(I₂)

I₁ω₁(t) = I₁ω₁(0) - F_frictionrt, and I₂ω₂(t) = I₂ω₂(0) + F_frictionrt; adding the two expressions

I₁ω₁(t) + I₂ω₂(t) = I₁ω₁(0) - F_frictionrt + I₂ω₂(0) + F_frictionrt. Applying the conservation of angular momentum between t=0 and later time t, 0=0, which is good. Does this answer this question?

For me, I don't understand why the velocity of the disk is not : w2(t) = w2+Frt/I2-Frt/I1 (and it can't) because in this case the angular momentum is not constant.

Your third answer,

(snip)
w1(t)=w1-Frt/I1
w2(t)=w2+Frt/I2-Frt/I1 (snip)

Sure, but why I can't write:
w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
w2(t)=w2(0)-Torque/I2*t = w2(0)-Frt/I2
(snip)

(snip)
w1(t)=w1(0)-Torque/I1*t = w1(0)-Frt/I1
w2(t)=w2(0)+Torque/I2*t = w2(0)+Frt/I2(snip)

and my 97th, but we finally agree.

I was getting obsessed with the fact that the two angular accelerations and velocities are different, and generalized that to forces and distances from the rotational centers. My bad. Pick a sign for the friction force, and then assign opposite sign to the reaction force, and be careful with the labelling.

Hopefully this helps rather than confuses things for you.

 

[Del8]

Ok, that helps me a lot, thank. I isolated the last problem: with w2(t). If I respect the conservation of the angular momentum I'm agree with w2(t)=w2+Frt/I2, but when I'm thinking only with rotationnal velocities: w2(t)=w2+Frt/I2-Frt/I1, because w2 is a labo frame reference velocity and like the disk is on the arm, when the arm is decelerating, the disk must decelerate too, no ? I can't understand that the rotationnal velocity of w2(t) is not a factor with the new change of w1(t). w2(t) increases its rotationnal velocity with Frt/I2 but in the same time the arm is decelerating, the arm decelerates all point on the disk.

Edit: Remember: w2(t) is a labo frame reference and w2'(t) is an arm frame reference. w2(t) < w1(t). And w2'(t)=w1(t)-w2(t). w2'(t) is counterclockwise.

Maybe it's easier to think with w'2(t) = constant, without friction. Let the disk turn all the time at w2' around itself. The arm is turning at w1. With your hand: apply a torque on the arm. Now, what is the new w2' ? For me it's w2', the rotationnal velocitity of the disk around itself don't change (I can't see how it can change, what forces). w2(t)=w1(t)+w2'(t), if w1(t) is changing and w2'(t) is constant, for me, w2(t) must "follow" w1(t), if w1(t) is decreasing => w2(t) is decreasing.

 

[Bystander]

w2'=w2-w1 in the arm frame reference.

What's ω₁'(t) in the "arm frame reference?"

 

[Del8]

What's ω₁'(t) in the "arm frame reference?"

0

 

[Bystander]

Right --- always and forever, and never changing.

Maybe it's easier to think with w'2(t) = constant, without friction. Let the disk turn all the time at w2' around itself. The arm is turning at w1. With your hand: apply a torque on the arm. Now, what is the new w2' ? For me it's w2', the rotationnal velocitity of the disk around itself don't change (I can't see how it can change, what forces). w2(t)=w1(t)+w2'(t), if w1(t) is changing and w2'(t) is constant, for me, w2(t) must "follow" w1(t), if w1(t) is decreasing => w2(t) is decreasing.

ω_2' = ω₂ - ω₁, and only ω₂ remains constant in the laboratory reference frame when no friction is acting. When you change ω₁, you change ω₂'.

 

[Del8]

I would like to understand with forces. Without friction, I apply a force Fa with my hand on the arm, I can give only a force Fb on blue axis (axis of the disk), true ? in this case, how w2' can change ? If w2' is changing, it should be a torque on disk, I don't see it.

http://imagizer.imageshack.us/v2/800x600q90/673/UPZdj7.png

 

[Bystander]

You're changing ω₁ and ω₂ is constant, so that ω₁ - ω₂ = ω₂' changes.

 

[Del8]

w2: is in the labo frame reference and w2' is in the arm frame reference, we're ok with that ?

w2' is constant: the disk is turning around itself without friction, why you say w2 is constant ?

 

[Bystander]

ω₂' is defined only in the arm reference frame; when the angular velocity of the arm changes and there is no frictional coupling to the disc, ω₂ can't change, and the difference in angular velocities between the disc relative to the lab, and the arm relative to the lab which is the angular velocity of the disc relative to the arm changes.

 

[Del8]

I asked another question for that point because it's not logical for me, maybe let others users reply for have several points of view. I would like to understand with torque, and I don't see a torque on the disk.

 

[Del8]

You're right because in the contrary the energy is not conserved. But I don't understand where is the torque. For change a rotation a torque is needed, no ?

 

[Bystander]

There is no change in the "fixed" reference frame of the lab; if we place it in the reference frame of the arm, that frame can move at a constant or accelerating angular velocity without changing the disc's velocity in the fixed frame, because there is no torque applied between the moving frame and the disc. There is a torque applied between the fixed frame and the arm to accelerate the arm, but with the friction turned off, it cannot affect the rotation of the disc relative to the fixed frame.

 

[Del8]

Ok, I understood, it was difficult for me the difference of the frame reference. Thank. The work from friction is : \int_0^t \omega_1(0)-\omega_2(0)-\frac{Frt}{I_2} dt ? For me, it don't depend of -Frt/I1 because when the arm is decelerating, all points on the disks decelerate and it is the same for the arm.

 

[Bystander]

∫ ( ω₁(t))dω₁ + ∫ ( ω₂(t))dω₂ = Fr ∫ ( ω₂'(t) )dt
from ω₁(0) to ω₁(t), and from ω₂(0) to ω₂(t), and from 0 to t.

The friction energy is just that dissipated by the frictional force over however much distance the disc covers from t = 0 to time t. This is where ω₂'(t) comes in, since it describes the velocity of the disc relative to the arm.

 

[Del8]

With the last integrale the energy is negative ?

 

[Bystander]

As written, yes.