# Flywheel problem with limited data? Is it possible?

1. Jun 9, 2012

### RichMortimer

A solid circular flywheel around which a string is wrapped. At the end of the string is a 2 kg mass. If the mass is dropped, the drum will rotate.

After the mass has fallen h metres, it has a velocity of v m/s and the wheel has a rotational velocity of ω rad/s

The flywheel radius is 120 mm. Use principle of conservation of energy to show that the distance fallen h, and the velocity v, are related by the equation h = v^2(0.051+1.77xI)

-----------------------------------------------------------------------------------------

Right then,

Firstly, please don't give the answer as I would like to work it out. However, I'm a little lost on the direction to go.

Conservation = mgh = 1/2mv^2+1/2Iω^2

We have no height, Inertia, radius of gyration or time... So how can we start?

I've thought about substitution but do I just pluck a number? See's to random to do that?!

I'm assuming I can substitute equations into the above equation but yet still don't seem to have the relevant information.

As I said, not after the answer, just a kick in the right direction if possible.

P.S. That's all the information.... no more!!

2. Jun 9, 2012

### cepheid

Staff Emeritus
I think the key insight here is that for every cm that the mass drops, this much rope unwinds. And for every cm of rope unwound, a point on the circumference of the wheel moves by this same distance. Can you see that?

If so, you have a relationship between the speed of the falling mass, and the speed moved by a point on the outside edge of the wheel. You also have a relationship between this tangential speed, and the rotational speed of the wheel.

3. Jun 9, 2012

### RichMortimer

Hello cepheid,

I initially thought something similar, using the radius. However, unless I "create" a height and time value I can't see how I relate to the function h = v^2(0.051+1.77xI)

I'm assuming I can somehow find a connection between the above values and a PE or KE combination but still looking!!!

:(

Thanks again.

4. Jun 9, 2012

### cepheid

Staff Emeritus

You have this equation:

All you have to do is solve it for h and plug in numbers. The only additional info that you need is how v (the speed at which the object falls) relates to ω (rotation speed of the wheel). I explained how in my first post.

5. Jun 9, 2012

### RichMortimer

So would you just pick a random height? Or try working backwards from the given equation?

I'll try again in the morning and re-post.

See where we get to!!

Thanks,

Night.

6. Jun 9, 2012

### cepheid

Staff Emeritus
I don't understand what you're asking. You don't need to pick any height. As stated in the problem, "h" is the distance by which the thing has fallen. So mgh (potential energy lost) = (1/2)mv2 + (1/2)Iω2 (kinetic energy gained by mass and flywheel). Solve for h.

7. Jun 10, 2012

### RichMortimer

A clear rested mind is a wonderful thing!!!

I don't need a value for h or v as they are unknown in the given equation!!

I'll have a go at the question again and feed back.

Once again, thank you!

8. Jun 10, 2012

### RichMortimer

Ok, sorry about quality, just easier than trying to type it all out!!!

See attached...

I feel like I'm nearly there but I'm not sure how to bring the 2*g*m*r^2 out from below. I know I'm on the right lines as I can equate the 0.051 required but as yet figured where the 1.77 has come from.

Please can you confirm that I'm right up until now.

I'll keep trying.

#### Attached Files:

• ###### working.jpg
File size:
17 KB
Views:
44
9. Jun 10, 2012

### RichMortimer

Haaaaaaaaaaaaa haaa!!!

Just got it!!!!

I must be a singular. Therefore there is a 1 in front. 1/0.565 = 1.77 :-D

Therefore:

h=(v^2(m*r^2+I)) / (2*g*m*r^2) is equal to h=v^2(0.051+1.77*I)

Wow.

Thanks again cepheid... :)

Last edited: Jun 10, 2012
10. Jun 10, 2012

### cepheid

Staff Emeritus
Glad to see that you figured it out.