Can some one explain to me why, when I include translational kinetic energy ($\frac{1}{2}mv^2$), I run into problems dealing with rotational motion? Overall, the problems I know when to use it are when there is something like a rotating pulley disk which has a string wrung around it and a block at the end falling down. Another place I know when to include it is when there is a ball rolling down a hill, and I need to take into account rotational and translational kinetic energy.
However, there are some instances where I don't understand why they include it. For example, if there is a rotating disk with radius $R$ at rest, and I glue a point mass to the side, and it asks for the angular velocity of the disk when the point mass reaches the bottom. In this case, should we include rotational and translational kinetic energy?
 A: In the ball rolling down the hill example, both are included, as the energy would be different to a stationary ball spinning at the same rate.
For the point mass at the side of the rotating disk example, it's spinning but not moving sideways or up or down, so just change the moment of inertia of the disk, by including the point mass glued to it and just deal with the rotational kinetic energy of the combined disk.
A: 
Overall, the problems I know when to use it are when there is something like a rotating pulley disk which has a string wrung around it and a block at the end falling down. Another place I know when to include it is when there is a ball rolling down a hill, and I need to take into account rotational and translational kinetic energy.


However, there are some instances where I don't understand why they include it. For example, if there is a rotating disk with radius R at rest, and I glue a point mass to the side, and it asks for the angular velocity of the disk when the point mass reaches the bottom. In this case, should we include rotational and translational kinetic energy?

Excellent question. This had 'troubled' me for quite a long time, and I found my peace after reading Resnick Halliday's take on it.
So, suppose we have a disc rolling down and we wanted to find the kinetic energy of it, it we would have to run the following integral:
$$ K = \frac{1}{2} \int v^2 dm$$
This becomes a bit complicated complicated to do, but we can make it simple by considering the point of contact of the disc with the ground. This due to the interesting property of this point is that the whole body is undergoing pure rotation about it (instantenously), (see here) and hence the total kinetic energy is just:
$$ K = I_{P} \frac{\omega^2}{2}$$
Now, this is where it gets tricky, we have to take the inertia about an axis passing through the point of contact of body with the ground (P). Hence we invoke parallel axis theorem, and once we do that we will find that the energy of the term accounting for shifting of axis from COM of body is related to the translational velocity of center of mass. I suggest doing the calculations for yourself and checking. See also the physics for k-12 textbook by Sunil Kumar Singh
