Kinetic energy of a rotating object in an exercise, a linear molecule is being subject to a force applied on the edge in its axis. Then $K_1=\frac{1}{2}mv^2$, all is well.
Then in the second point of the exercise, the force is applied on the same edge but in an orthogonal direction to its axis. Then the molecule begins to rotate. So its kinetic energy is composed of two terms: $K_2=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$,$\omega$ being the angular velocity of the rotation. The thing is that the linear velocity is the same as before and the correction says that $K_2=K_1+\frac{1}{2}I\omega^2$. But how come the same force can give two different energies to the molecule? I thought that $v$ would decrease in the second case, because of the apparition of the angular velocity $\omega$ so that the energy would be conserved. So in the second case the molecule goes as fast as before but in addition it rotates on itself?
 A: The mistake in your reasoning is to assume that the same force does the same work. That is simply not true - as the force operates over a different distance.
To analyze the problem, you have to think in terms of impulse ($F\cdot \Delta t$) or work done ($F\cdot \Delta x$).
Let's assume that the same impulse is applied. Then indeed the linear momentum of the molecule will be the same in both cases, but we also cause rotational angular momentum $L=F\cdot r\cdot \Delta t$ where $r$ is the distance between the line of action of the force and the center of mass of the molecule.
Now the work done by the impulse depends on the distance traveled. Because the molecule starts rotating as we hit it, the force is applied over a greater distance (the center of mass moves less than the side we hit). We can compute the distance moved (and thus the work done) in different ways - but the easiest way is to use conservation of energy...
A: The work done by the force in the first case is just translational, i.e. integral of the force over some distance and this gives rise to the translational kinetic energy.
In the second case, there are two types of work done on the system, first being the usual translational and second is rotational. This is due to the that the object in question now has an extra degree of freedom than earlier, in other terms by the virtue of the direction of the force, it can rotate. Therefore the net energy expended by the force has to be on both the translational mode as well as rotational mode, as there is an increase in the number of degrees of freedom in the system than earlier and hence the expression you wrote down.
A: The energy of the system is not only proportionate to the force, applied to it, but is actually the work done by that force ($F\Delta s$) on the path that your system has travelled ($\Delta s$). In the first case the molecule has only the translational motion. In the second case, in addition to translational motion there is also a rotation, thus the path is NOT the same and the energy is NOT the same. The key-point here is that the energy depends also on the path, not only to the applied force. 
