# Rotationnal velocity and acceleration

1. Nov 28, 2014

### 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 [Broken]

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
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/I1

For you, is it ok ?

Last edited by a moderator: May 7, 2017
2. Nov 28, 2014

### Bystander

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

3. Nov 28, 2014

### 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 ?

4. Nov 28, 2014

### Bystander

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

5. Nov 28, 2014

### Del8

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

6. Nov 28, 2014

### zoki85

What is this device?

7. Nov 28, 2014

### Bystander

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.

Last edited: Nov 28, 2014
8. Nov 28, 2014

### 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.

9. Nov 28, 2014

### 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.

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

10. Nov 28, 2014

### 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

Last edited: Nov 28, 2014
11. Nov 28, 2014

### 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.

12. Nov 28, 2014

### Del8

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

13. Nov 28, 2014

### Bystander

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

14. Nov 28, 2014

### Del8

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

15. Nov 28, 2014

### Bystander

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

16. Nov 28, 2014

### Del8

17. Nov 28, 2014

### Bystander

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

18. Nov 28, 2014

### 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

19. Nov 28, 2014

### Bystander

Nor, are the two torques equal.

20. Nov 28, 2014

### 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 ?

21. Nov 28, 2014

### Bystander

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

22. Nov 28, 2014

### Del8

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

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 ?

Last edited by a moderator: May 7, 2017
23. Nov 28, 2014

### Bystander

ω

Last edited by a moderator: May 7, 2017
24. Nov 28, 2014

### Del8

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

25. Nov 28, 2014

### Bystander

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