What Is the Rotation Speed of a Disk After Being Tossed?

[Phystud]

How fast a disk is rotating? Please help me

Im trying to attempt this problem, its not a assigned homework problem but it has 2 stars beside it(which means its one of the harder ones) in the book so I am 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?

 

[e.bar.goum]

Do you know how moments of inertia work?

 

[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.

 

[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?

 

[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

 

[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.

 

[Phystud]

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

 

[e.bar.goum]

As far as I'm aware, it's E_k = 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.

 

[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)?

 

[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

 

[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!?

 

[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?

 

[e.bar.goum]

Sounds reasonable to me!

 

[Phystud]

If you could just verify this for me id greatly appreciate it, even tho I am 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.

 

[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

 

[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.

 

[Phystud]

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