How does changing radius (constant mass) affect rotational kinetic energy? I'm trying to understand how rotational energy would change if radius shrinks but mass stays the same. The initial question was posed as follows: 

"The radius of a disk of matter forming around a new star shrinks by a
  factor of 2 without any additional mass being added into the system.
  The kinetic energy increases by what factor?"

As far as I understand, rotational kinetic energy should be inversely proportional to r2 (so I'm guessing the answer would be that rotational KE would decrease by a factor of 4). However, I'm having trouble parsing that out mathematically. Here's how far I've been able to get: 
RATIONALE A:
KE = ½Iω2


*

*I = ½ mr2

*ω = (v/r)
Combining this, you get KE = ½ mr2(v/r)2 or 
KE = ½ mv2  At this point, radius is not directly factored into the equation. If mass isn't changing, any increase in kinetic energy would have to be related to an increase in v, but I don't understand how to determine the change in v. 
RATIONALE B:
KE = ½Iω2


*

*L = Iω

*ω = (v/r)
Combining this, you get KE = ½(L)ω = ½(L)(v/r). If angular momentum (L) is conserved, any change in kinetic energy would be proportional to the change in v/r...and, once again, I'm at a road block. 
I'd greatly appreciate any advice that can help me figure out where my reasoning is going astray. Thanks!
 A: I have some additional information.
The way the original problem is phrased is awkward. "A disk of matter that shrinks".
Compare the rings of Saturn. (Those rings aren't solid of course, those rings consist of lots of small objects (I guess the size of pebbles or so)). The thing is: rings like that don't shrink.
Anyway, exactly the same physics can be presented in the following setup:
A satellite is in orbit around the Earth. The orbit is highly excentric, so much so that the point of farthest distance is twice as far away as the point of closest approach.
That means that as the object moves from farthest distance to closest approach the gravity from the Earth is accelerating the satellite. From the point of farthest distance to the point of closest approach, how much does the kinetic energy increase?
This way of presenting the problem illustrates why the kinetic energy increases. The kinetic energy increases because gravity is tugging at the satellite all the way down.
Now, as pointed out in another answer, there is a fast way to calculate the increase of velocity, and once you have that you can calculate how much the kinetic energy has increased. Still, it's good to keep the actual cause in the back of your head:  all the way down gravity is increasing the velocity of the satellite.
(And of course, on the part from closest approach to farthest distance the tug of gravity is decelerating the satellite.)
A: It is great that you are thinking carefully about which quantities would remain constant and are trying to find an expression for the rotational kinetic energy which depends only on the radius $r$ except for the conserved quantities--that is a nice way to think about such problems as far as I know. 
Since it is not considered that great to provide full answers to homework-like questions, I would lay down detailed hints which should suffice for you to solve the question yourself.


*

*Now, since the problem is a central force problem, the angular momentum of the disk is the obvious conserved quantity. Also, the problem states that the mass of the disk is another constant of motion.  Thus, it would be great if we can express the rotational kinetic energy in terms of the mass of the disk, its angular momentum, and finally, its radius. 

*In order to do so, let's recall, as you have already done in your post, that $$I=\frac{1}{2}mr^2$$$$L=I\omega$$$$K=\frac{1}{2}I\omega^2$$

*Now, notice that the expression for $I$ is already in terms of $m$ and $r$ so we don't need to tweak it for the rationale that I expressed in the first point. But, the expression for $K$ depends on $\omega$ which we would like to express in terms of constants of motion and the radius $r$. This can be easily done via expressing $\omega$ in terms of $I$ and $L$ using the expression for $L$ that I listed in the previous point. 

*Now, you would have an expression for $K$ in terms of constants of motion and $r$--enabling you to predict how $I$ varies with a variation in $r$ without worrying about how other quantities vary with a variation in $r$. 


Cheers! 

Edit
One should refer to the answer by @Cleonis to understand what is going on physically. It might be mentioned that the "replacement" of the disk with a satellite should be understood much more rigorously than just as an analogy via thinking in terms of the shrinking radius of gyration of the disk. One can imagine an object to be just a point mass situated at a distance from the axis of rotation given by its radius of gyration.
