Solving 2 Rotating Discs Homework: Angular Velocities

[person123]

Homework Statement

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.

Homework Equations

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

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?

 

[haruspex]

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

 

[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.

 

[TSny]

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.

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

 

[person123]

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

 

[scottdave]

The Attempt at a Solution

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

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?

So how are you relating mass m1 and m2 ?

 

[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

 

[scottdave]

I knew the mass of disc 2 is proportional to the square of the mass of disc 1. All other quantities (eg. density, thickness, pi) cancel out leaving me with r^2r^2=r^4

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.

 

[person123]

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

 

[scottdave]

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

 

[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.

 

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

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

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

 

[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.

 

[scottdave]

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.

I substituted m₂ with (4*m₁)

 

[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.

 

[scottdave]

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

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

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

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

 

[person123]

Where did I write that?

 

[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.

 

[TSny]

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

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.

 

[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?

 

[haruspex]

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.

Quite so, you cannot use conservation of angular momentum either.
So let's go back to something more fundamental: action and reaction being equal and opposite.
What can you say about the force each exerts on the other?

 

[haruspex]

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

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

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

As I pointed out in post #2, you cannot use conservation of energy here.
E.g. consider two identical discs rotating clockwise. On their edges being brought together they will eventually come to a stop.

 

[scottdave]

I have not found your answer on YouTube. I have heard, that if he hides it from public view, then you will still be able to see it, because it is your comment.

 

[person123]

Quite so, you cannot use conservation of angular momentum either.
So let's go back to something more fundamental: action and reaction being equal and opposite.
What can you say about the force each exerts on the other?

Yes—coincidentally, I was trying to work on that before I saw this.
I know that the force which disc 1 applies on disc 2 is the same as the force disc 2 applies on disc 1. However, I would not say the torques are the same, as the same force can apply a larger torque if it's applied from a greater distance. So $τ_{1/2}=F_{f}r_{2}$ and $τ_{2/1}=F_{f}r _{1}$, correct?
What I would like to do next is multiply that by the time to get the change in angular momentum, and then use that to find the change in angular velocity. However, I was having a bit of trouble with that.

 

[haruspex]

What I would like to do next is multiply that by the time to get the change in angular momentum, and then use that to find the change in angular velocity. However, I was having a bit of trouble with that.

A good plan. Where are you stuck?

 

[Hiero]

I know that the force which disc 1 applies on disc 2 is the same as the force disc 2 applies on disc 1.

In magnitude, yes. Keeping directions in mind, though, is important.

However, I would not say the torques are the same, as the same force can apply a larger torque if it's applied from a greater distance. So $τ_{1/2}=F_{f}r_{2}$ and $τ_{2/1}=F_{f}r _{1}$, correct?

This is true, provided you are aware that you are taking the torque about two different points (the center of each disc) for two different systems (each respective disc).

The other approach which stands out is to only take the torque about a single point with your system being both discs.

 

[haruspex]

The other approach which stands out is to only take the torque about a single point with your system being both discs.

That would involve a torque from an unknown force at an axle.

 

[Hiero]

That would involve a torque from an unknown force at an axle.

Ah, you are right that I ignored this, my apologies. Taking that into account makes that approach no simpler.

 

[scottdave]

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

I just rewatched the video, but I don't see him specify a ratio (like twice) between r2 and r1.

 

[person123]

Alright. I finally think I have the answer. I wrote two equations, one for the change in momentum of disc 2, the other for change in momentum of disc 1. They both follow the format of $IΔω=F_{f}rt$ . I divided one by the other, leaving me with:

$\frac {r_{2}^4(ω_{2}-ω_{1})} {r_{1}^4ω_{1}}=-\frac {r_{2}} {r_{1}}$

. When simplifying and substituting the values for r1 and r2, it gave me:

$ω_{2}=\frac {ωr_{2}^2} {r_{1}^2+r_{2}^2}$
$ω_{1}=\frac {r_{2}^3ω} {r_{1}^3+r_{1}r_{2}^2}$

I just rewatched the video, but I don't see him specify a ratio (like twice) between r2 and r1.

No, he didn't. I have no idea why I thought he did. I decided to keep using that ratio on the forum to avoid confusion, but now I left it out.

 

[scottdave]

Alright. I finally think I have the answer... They both follow the format of $IΔω=F_{f}rt$ . I divided one by the other, leaving me with:

$\frac {r_{2}^4(ω_{2}-ω_{1})} {r_{1}^4ω_{1}}=-\frac {r_{2}} {r_{1}}$

. When simplifying and substituting the values for r1 and r2, it gave me:

$ω_{2}=\frac {ωr_{2}^2} {r_{1}^2+r_{2}^2}$
$ω_{1}=\frac {r_{2}^3ω} {r_{1}^3+r_{1}r_{2}^2}$

I arrived at the same answer, which I posted on his YouTube video, this morning. As of 3:30 PM central time today, he has not released it to the public view. So I'm going to assume it is probably correct.

 

[person123]

I arrived at the same answer, which I posted on his YouTube video, this morning. As of 3:30 PM central time today, he has not released it to the public view. So I'm going to assume it is probably correct.

At last! I'm just going to wait a while to make sure I got it, and then I'll close the thread.

 

[person123]

Well, I generalized it so that disc 1 could have any angular velocity as well:

https://www.desmos.com/calculator/discbz9gf1