# Rotational Inertia of a disc

1. May 25, 2012

### DreamChaser77

Now I have a disc that has mass on each corner(N,W,S,E) it has a total mass of 1 Killo now with the small object spread on its edges. I want to increase the speed of that disc with the same amount of input energy I should put the object that weight on the disc on the center where the disc is being spinned? If so could I keep the objects at the same spot and just add equal mass on the center or double that?

lets we have a disc that weighs at 10 killo's and I added 14 metal bars on the edges of the disc that all weight at 0.50 killo's x 14 = 7Kilo's. Now the speed would be slower than if I added those bars in the center of the rotational axis.Now If I had those bars on the edges and got a 8Killo'd object and added it on the center rotational axis would it increase speed or not...?

My goal is to increase rotational speed from mass with the same amount of energy!

Thanks

Dream,

2. May 25, 2012

### DreamChaser77

I wanted to add is that: If the speed is increased will the torque also increase?

3. May 25, 2012

### Ken G

You can't increase speed by adding mass, you have to move mass inward, like a skater pulling in his/her arms during a spin. There is also no direct connection between the speed and the torque, torque increases the speed but the speed doesn't tell you the torque.

4. May 26, 2012

### DreamChaser77

So adding weight on the center won't do anything but cause drag? The only was to increase speed of the disc is to bring the bars inward?

5. May 26, 2012

### DreamChaser77

If we added a 3rd wheel putting mass outward and inward together what would be the result?

6. May 26, 2012

### Infinitum

If the total mass put is double the mass put on a single(original) wheel, it would be slower than both. The reason is, since you have mass both 'inward' and 'outward', the total moment of inertia of the wheel is higher that either of the two cases shown in that video. Meaning, more energy is going into rotation, and lesser into translational kinetic energy.

With similar reasoning, if the total mass put on the wheel is same as the first and second wheels, then its translational kinetic energy would be in between the two.

7. May 26, 2012

### tiny-tim

Hey DreamChaser77!
if the axle is fixed:

KE = 1/2 Iω2,

so ω = √2KE/I,

so to increase ω you decrease I …

so put the mass as close as possible to the centre

(if the disc is rolling, KE = 1/2 (I + mr22 … same result)

8. May 26, 2012

### DreamChaser77

So what... the disc will rotate more and more or will it be difficult to rotate it due to its mass?
I feel my disc if both mass is inward and outward it will be more of a flywheel because the kinetic energy...?

I would probably try to make the radius less and bring the weight's more closer to the cenrtial axes how about that? Make it a smaller disc and add more weight if I needed to because the speed would increase is that good?

9. May 26, 2012

### tiny-tim

Yes

You want the opposite of a flywheel …

a flywheel is intended to store as much energy as possible with minimum speed, so nearly all its mass is in the rim.

10. May 26, 2012

### DreamChaser77

Thanks Tim!

Well I want to achieve a fast spinning wheel/disc with the same input of energy so decreasing the radius is the solution, I will add more weight on the center so all in all I kinda made a perfect design to it now. If my disc current is going 100RPM's could you estimate how fast it would go with the weight inward very close to the central axis?

Last edited: May 26, 2012
11. May 27, 2012

### tiny-tim

i don't understand … what is "disc current"? … how is it different from speed?

12. May 27, 2012

### DreamChaser77

Ow sorry I ment if the disc was rotating @ 100rpms by bring the mass inward how fast do you it would increase?(just a simple guess)

13. May 27, 2012

### Ken G

It depends on how you set up the system. If you are a skater, then you inject a given angular momentum into your body, and you pull your arms in. Since angular momentum is conserved, we have angular momentum J = I*omega, so omega is inversely proportional to I. Whatever you do to the moment of inertia, the opposite will happen to the rotation rate.

However, in your description it sounds like you are talking about building something first, and then injecting angular momentum and energy into it. If both angular momentum and energy are things that you are injecting, then the key relation is
omega = 2*E/J, and that is independent of I. If you can only put in so much E and so much J, then in that situation, omega does not care what the I is. So it depends on what you are doing.

14. May 27, 2012

### DreamChaser77

Thanks for that Ken.G

Thing is I thought if I had a disc rotating at certain speeds and got the mass inward I would maybe double or triple the speed that would be great! Because right now while I'm typing this down I'm rotating my chair while my arms are spread outward with mass then pull it back inward I nearly fell a few minutes ago and the push was AMAZING felt like something very very strong kicked the hell of the chair hahaha

But its really interesting so far and I have nothing to do so I'll research it more and more :)

15. May 27, 2012

### Ken G

Yes, that's a favorite demonstration-- it works even better if you have barbells in your hands (but beware, you can hurt yourself if you fly off the chair!). In case you wonder where the rotational energy comes from, it takes work to pull in your arms or legs.

16. May 28, 2012

### DreamChaser77

I WILL defiantly not do that lol its dangerous!
I'm aware of the inputed energy pretty well thanks for that!

Dream,