Inconsistency in work-energy principle for a spinning body set down to roll A spinning solid cylinder spinning with $\omega_0$ is put smoothly on a  plane. It will skid until it begins to roll with $\omega$
Argument 1
We know that, work done by all the forces assumed to act on the center of mass equals the change in kinetic energy of the center of mass.
The cylinder moves from $a$ to $b$, and the velocity of centre of mass changes from $0$ (when it was only spinning)  to $V_{cm}$ (when it has begun rolling). Therefore we can write
$W=\oint_{\mathbb{R_{a}}}^{R_{b}} \mathbf{F} \cdot d \mathbf{R}=\frac{1}{2} M V_{b}^{2}-\frac{1}{2} M V_{a}^{2}$
Using the above equation and  observing that only friction does work we can write the work done by friction until the body begins to roll is:
$$W_{f}=\frac{1}{2} m \cdot V_{b}^{2} -   0=\frac{1}{2} m \cdot \omega^{2} R^{2}$$
( I've used  $V_{b}=\omega R$ as the body begins pure roll after skidding
and initially $V_{a}=0$)
Argument 2
Also we know from the general work energy theorem that the work done on a body equals the change in total kinetic energy of the body, hence:
$W_f=\Delta K$ where $K=K_{translational}+K_{rotational}$
Using value of $W_{f}$ from argument 1 we have then
$$\Rightarrow \frac{1}{2} \operatorname{m} \omega^{2} R^{2}=\left(\frac{1}{2} m \omega^{2} R^{2}+\frac{1}{2} I \omega^{2}\right)-\frac{1}{2} I \omega_{0}^{2}$$
$\Rightarrow$ $\omega=\omega_{0}$
Which is incorrect.
Could anyone at least please hint where I'm going wrong. I have been thinking about it for more than  10 hours.
 A: The mistake is here

the work done on a body equals the change in total kinetic energy ... Using value of Wf from argument 1

$W_f$ is not the work done on the cylinder.
See in your calculation that you used $W=\oint_{\mathbb{R_{a}}}^{R_{b}} \mathbf{F} \cdot d \mathbf{R}$ where $\mathbf{R}$ is the position of the center of mass. This is calculating the confusingly-called "net work", which I prefer to call "center-of-mass work". This quantity does not represent the total change in kinetic energy, but only the translational part of it. Unfortunately, many authors are very sloppy about their presentation of the work energy theorem and do not clarify when and how it applies to extended objects.
To find the actual change in energy you need to use $W_i = \int \mathbf F_i \cdot \mathbf v \ dt$ where $\mathbf v$ is the velocity of the material of the cylinder at the point of application of $\mathbf F_i$. Note that during the time that the cylinder is slipping the force is in the opposite direction of this velocity, so the work is negative. Thus the final kinetic energy is less than the initial kinetic energy.
A: I have deleted my previous answer as it clearly didn't get to the heart of the issue for you. I'll try again:
Your first equation is the work done by friction during the non-slip portion of the motion, during which rotational kinetic energy is converted to translational.
You are forgetting the skidding portion of the problem, during which time the work done by friction is not equal to the final kinetic energy of the centre of mass.
There are therefore two "versions" of your first equation to consider:
$$ W_{f_{slip}} = \frac{1}{2}I(\omega^2 - \omega_0^2)$$ (this is the "energy lost to heat" that I was trying to remind you to include in my previous answer)
and
$$ W_{f_{non-slip}} = \frac{1}{2}mR^2\omega^2$$
By combining these, do you see that your contradiction/inconsistency no longer occurs?
