Rotational work done So the formula for rotation work is
$$\mathrm dW=T\cdot\mathrm d\theta$$
where T is torque.
While solving a question like in case of a body rolling down an incline (pure rolling), we usully equate the change in kinetic energy equal to work done by gravity (just the translational work). Why do we not count the rotational work done when gravity does provide a torque?
In almost all question of mixed motions (translational and rotational) I have come across, none of them have the application of rotational work done, so I would also like to know when do we apply this concept.
 A: First, for something like a ball rolling down an incline, gravity has no torque about the center of the ball. The force that causes the ball to start rolling is friction. On a frictionless incline the ball would just slide down the incline without rolling.
Now, with that out of the way, it turns out that we do take into account "rotational work" due to friction. Let's assume a constant friction force $f$ and say that the ball with radius $r$ is released from rest on the incline. We know that the net torque about the center of the ball is given by
$$\tau = fr$$
using Newton's second law we also have
$$\tau=I\alpha$$
where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.
Now, since our torque is constant (since $f$ and $r$ are both constant) we know two other things. First, the work done by friction is given by 
$$W=\tau\Delta\theta=I\alpha\Delta\theta$$
and second, we can apply our constant acceleration equations. This means that
$$\omega^2=2\alpha\Delta\theta$$
where $\Delta\theta$ is the angle the ball rolls through some time after release, and $\omega$ is the angular velocity at that same point in time.
Combining everything we end up with
$$W=\frac12I\omega^2$$
and this result might look familiar. It is what we usually associate "rotational kinetic energy" with. So we do take into account the "rotational work", we just do it implicitly with $\frac12I\omega^2$ rather than explicitly (this is similar to how we use potential energy to implicitly take into account the work done by conservative forces rather than explicitly calculating the work done by said forces).
For example, if the ball starts at a height $h$ above the ground on the incline, using energy conservation at the bottom of the incline we have
$$mgh=\frac12mv^2+\frac12I\omega^2$$
