Rotationnal velocity and acceleration

[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.