Applying a force on an object in space (weightless) away from its centre of mass vs at its centre of mass Suppose a horizontal bar of length L and mass M is in space and is therefore weightless, and a force F is applied at its right end upwards.
I understand that the centre of mass of the bar will accelerate upwards at F/M, and there will also be an angular acceleration of the bar about its centre of mass, equal to FL/2 divided by I, so there will be two simultaneous motions of the bar.
However, if I were to apply the same force at the centre of mass of the horizontal bar, there would only be pure translation with the same upward acceleration of F/M, and no rotation.
However in both cases the same force F is applied, yet one clearly increases the translational and rotational kinetic energies of the bar, while the other only increases its translational kinetic energy.
I can't seem to reconcile this difference. Where did I go wrong in my understanding of this concept?
 A: I think that you query is about equating the work done by the accelerating force and the gain in kinetic energy of the bar.
If the force $F$ is applied at the centre of mass of the bar and the centre of mass moves a distance $x$ then the work done by the force is $Fx$ and this represents the gain in the translational kinetic energy of the bar.
If the same force is applied along a line which is not through the centre of mass of the rod and the centre of mass moves by a distance $x$ the work done by the force is larger than $Fx$ because, if the force is to stay in contact with the rod which is rotating about its centre of mass as well as the centre of mass undergoing transnational motion, the force will have to travel further and that extra distance travelled by the force means that the force has done more work and that gives the rod the rotational kinetic energy.
Very much related to this are a series of videos made by Veritasium a website worth visiting as they make you think about science.
After seeing the experiment make a prediction as to what you think the results should be.   Then watch the experimental results video and try and explain the results.
Bullet block experiment
The next is about the prediction as to which block went higher?
Bullet block experimental results
And then the explanation
Bullet block explained 
A: I agree that this is confusing: it looks as if that, in the force-at-centre-of-mass case, the rod has got some acceleration from the force, $a = F/M$, from Newton 2, while in the force-at-the-end-of-the-rod case it has done that and picked up some angular acceleration.  But the force is the same, so it looks like some conservation law is being violated.
Well, no.  The trick is to look at the distance the point where the force is applied moves in a given time.  Even without doing the maths, it is clear that in the second case the point moves a lot further than it does in the first (because the centre of mass of the rod moves the same distance it did in the first case, but the rod is now rotating and the force is applied right at one end.
That means that, in the second case, the force does a lot more work, since $W = F\times D$ (or, really, $W = \vec{F}\cdot \vec{x}$), and the distance is now much greater.  So everything works out.
