Rotationnal velocity and acceleration

[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.