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 ?