# Homework Help: Two Rotating Discs

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1. Jun 19, 2017

### person123

1. The problem statement, all variables and given/known data
There are two discs with equivalent density and thickness. One has radius r1 while the other has radius r2. r2 is twice as great as r1 The larger disc has an initial angular velocity ω. The two discs then come in contact with one another and friction causes them to rotate with the same linear velocity. What are the angular velocities of the two discs?

For everything related to the larger disk, besides the initial angular velocity which is just ω, I'll subset it with 2. For everything related to the smaller disk, I'll subset it with 1.

2. Relevant equations
$v=ωr$
$KE=1/2Iω^2$
$I_{disc}=1/2mr^2$
$E_{i}=E_{f}$

3. The attempt at a solution

$½I_{1}ω^2=½I_{1}ω_{1}^2+½I_{2}ω_{2}^2$
$½(½m_{1}r_{1}^2)ω^2=½(½m_{1}r_{1}^2)ω_{1}^2+½(½m_{2}r_{2}^2)ω_{2}^2$
$r_{1}^4ω^2=r_{1}^4ω_{1}^2+r_{2}^4ω_{2}^2$

Since $r_{2}=2r_{1}$:
$r_{1}^4ω^2=r_{1}^4ω_{1}^2+16r_{1}^4ω_{2}^2$

By knowing that $v=ω_{1}r_{1}=ω_{2}r_{2}⇒ω_{1}=2ω_{2}$ I could solve for the angular velocity of the two discs.

For $ω_{1}$:

$r_{1}^4ω^2=r_{1}^4ω_{1}^2+16r_{1}^4\frac{ω_{1}^2} 4$
$r_{1}^4ω^2=r_{1}^4ω_{1}^2+4r_{1}^4ω_{1}^2=5r_{1}^4ω_{1}^2$
$ω^2=5ω_{1}^2⇒ω_{1}=\frac {ω} {\sqrt5}$

For $ω_{2}$ :

$r_{1}^4ω^2=r_{1}^4(2ω_{2})^2+16r_{1}^4ω_{2}^2$
$r_{1}^4ω^2=4r_{1}^4ω_{2}^2+16r_{1}^4ω_{2}^2=20r_{1}^4ω_{2}^2$
$ω^2=20ω_{2}^2⇒ω_{2}=\frac {ω} {2 \sqrt{5}}$

When I posted these answers on Walter Lewin's youtube channel, he posted it which means it's incorrect. I've checked it several times but I can't find any errors. Does anyone know where I went wrong?

Last edited: Jun 20, 2017
2. Jun 20, 2017

### haruspex

You cannot assume work is conserved. What else can you do?

3. Jun 20, 2017

### person123

The only other thing I could think of is conservation of angular momentum. But because the gears are not attached to the same center axle, I don't know how to go about doing that.

4. Jun 20, 2017

### TSny

What condition(s) must be satisfied in order for the angular momentum of a system to be conserved about some point (origin)?

5. Jun 20, 2017

### person123

The sum of the angular momentum of all the parts (mr^2ω or mrv) around the origin must be constant, right?

6. Jun 20, 2017

### scottdave

So how are you relating mass m1 and m2 ?

7. Jun 20, 2017

### person123

I knew the mass of disc 2 is proportional to the square of the radius. All other quantities (eg. density, thickness, pi) cancel out leaving me with r^2r^2=r^4

Last edited: Jun 20, 2017
8. Jun 20, 2017

### scottdave

If you have a circle of radius 3, and the other circle is radius 6 (double the first), how do the areas relate? Are you saying that the bigger disc has 9 times the area of the first?

BTW, I've started watching his videos, recently. Pretty interesting.

9. Jun 20, 2017

### person123

No, it would be four times the area—it was my fault as I meant the square of the radius.

10. Jun 20, 2017

### scottdave

Regardless of what value r1 is, Area2 will be 4 times Area1

11. Jun 20, 2017

### person123

Yes. And the moment of inertia of disc one will be 16 times times the moment of inertia of disc two, as I wrote.

12. Jun 20, 2017

### scottdave

But you skipped a step.

$½(½m_{1}r_{1}^2)ω^2=½(½m_{1}r_{1}^2)ω_{1}^2+½(½m_{2}r_{2}^2)ω_{2}^2$

Should go to this:

$½(½m_{1}r_{1}^2)ω^2=½(½m_{1}r_{1}^2)ω_{1}^2+½(½(4)m_{1}r_{2}^2)ω_{2}^2$

Which gives us:

$r_{1}^4ω^2=r_{1}^4ω_{1}^2+4r_{2}^4ω_{2}^2$

Then:

Since $r_{2}=2r_{1}$:

$r_{1}^4ω^2=r_{1}^4ω_{1}^2+4(2r_{1})^4ω_{2}^2$

$r_{1}^4ω^2=r_{1}^4ω_{1}^2+64r_{1}^4ω_{2}^2$

13. Jun 20, 2017

### person123

I already multiplied by 4 once because of the change in mass and 4 again because of the change in radius, giving me 16. Why did you multiply by 4 for a third time?

Your first and second line are identical except you multiplied by 4 in the second.

14. Jun 20, 2017

### scottdave

I substituted m2 with (4*m1)

15. Jun 20, 2017

### person123

Oh, my bad...but I'm pretty sure you still made an error. The radius was only to the power of 4 because the mass was still in terms of r2. You did the same step twice.

16. Jun 20, 2017

### scottdave

OK, I copied your formula. So it should be:

$r_{1}^2ω^2=r_{1}^2ω_{1}^2+4r_{2}^2ω_{2}^2$

$r_{1}^2ω^2=r_{1}^2ω_{1}^2+4(2r_{1})^2ω_{2}^2$

$r_{1}^2ω^2=r_{1}^2ω_{1}^2+16r_{1}^2ω_{2}^2$

17. Jun 20, 2017

### person123

Where did I write that?

18. Jun 20, 2017

### scottdave

So you get the same thing for omega as your original solution. I'm going to go look at the YouTube video to see it.

19. Jun 20, 2017

### TSny

Once you pick an origin, the angular momentum of the system about that origin will be conserved only if the net external torque (about that origin) is zero. So, you need to determine if there are any external forces acting on the system when the disks are brought together.

20. Jun 20, 2017

### person123

Only if you include the force of friction which the two discs impart on one another, but isn't that still part of the system, and therefore there are no external net forces?