1. Dec 14, 2011

### Phystud

Im trying to attempt this problem, its not a assigned hw problem but it has 2 stars beside it(which means its one of the harder ones) in the book so Im trying to solve it but i don't even know how to begin and have a crack at it. I really could use some help. Thank you everyone who can chime in!
A pizza chef tosses a spinning ball of dough into the air. The ball of dough may be considered a solid sphere(mass uniformly distributed throughout its volume) with a radius of 5.80 cm, a density of 1100 kg/m3, and it is initially spinning around an axis of rotation passing through its center at rate of 113 rad/s. When the chef catches it, the dough has stretched out into a disk 18 cm in radius with a uniform thickness. The disk continues to rotate around an axis passing through its center and perpendicular to the plane of the disk. How fast is the disk rotating just before the chef catches it?

2. Dec 14, 2011

### e.bar.goum

Do you know how moments of inertia work?

3. Dec 14, 2011

### Phystud

I believe so. An object can speed up or slow down in a rotational manner, it's linked with mass(inertia). I also know that it depends on the mass of the object, the point of rotation, and its shape.

4. Dec 14, 2011

### e.bar.goum

Yep, and knowing how the moment of inertia will change from a ball to a disk will be crucial here.

(Hint, you can just look them up online)

Then, do you remember the formula for rotational kinetic energy?

5. Dec 15, 2011

### Phystud

KErot=1/2 ω⋅LC=1/2 ICω2 i found this one on this site for kinetic energy.

Ball: I=2m(r(squared))/5

Disk: i=m(r(squared))/2

6. Dec 15, 2011

### e.bar.goum

That looks like the moment of inertia for an infinitely thin disk, not one with thickness - you'll need a (slightly) different equation. Otherwise, fine.

Now, kinetic energy will be conserved, so from those three equations, you should be able to work it out.

7. Dec 15, 2011

### Phystud

So it is not KE=Iw^2, im looking at it in my book and it says this one!

8. Dec 15, 2011

### e.bar.goum

As far as I'm aware, it's Ek = 1/2 I ω2

ETA: And it looks like you've got the right moment of inertia for the disk, sorry. The thickness terms would cancel out.

9. Dec 15, 2011

### Phystud

Aha, my book says KE, but ill check with my instructor. So far I have solved for the inertia for the ball and disk: ball:74008 and disk:178200. How do I approach the next step? Do I add both inertias and solve for KE(Ek)?

10. Dec 15, 2011

### Phystud

I believe i made a mistake, i need to first convert centimeters into meters!
Its 7.4008 for the ball and for the disk: 17.82

11. Dec 15, 2011

### Phystud

So do I just use the disk answer and plug in like this: KE=1/2 x 17.82 x (113rad/s)^2. But, this will get me the rot. kinetic energy and not how fast its rotating!?

Last edited: Dec 15, 2011
12. Dec 15, 2011

### Phystud

Or can I just use the spheres value and solve for rotational kinetic energy and then solve for angular speed since I have the kinetic energy now???

13. Dec 15, 2011

### e.bar.goum

Sounds reasonable to me!

14. Dec 15, 2011

### Phystud

If you could just verify this for me id greatly appreciate it, even tho im thankful already for all your help.
1st KE=1/2(7.4008)(113)^2=4.725x10^4
2nd KE=1/2 Iw^2 4.725x10^4=1/2(17.82)w^2 w=51.5 rad/s

I really hope I did this right! And thank you sooo much for your help.

15. Dec 15, 2011

### e.bar.goum

I'm not going to do it all, sorry, but there's something wrong with (at least) your last step
www.wolframalpha.com/input/?i=solve+4.725x10^4+%3D+1%2F2(17.82)w^2

16. Dec 15, 2011

### Phystud

I forgot to multiply 1/2 to 17.82 which equals to 8.91

I then: 4.725x10^4 divide by 8.91 and get the anwser and square root it and my answer would be w=72.82 rad/s. How about now? I believe I have found my mistake.

17. Dec 15, 2011

### Phystud

It that the correct answer? If anyone has any input on this! Greatly appreciated.