Work done by static friction in accelerated pure rolling motion [duplicate]

Why is the work done by static friction in accelerated pure rolling motion along an incline actually zero?

Somewhere I am grasping the information that work done by static friction is zero as the point of contact where it is acting has zero velocity so it is at rest w.r.t ground, but why is the work done by the torque produced due to static friction zero?

In fact that is the force which rolls the body (as opposed to sliding down the incline) so why is the work done by the torque due to static friction zero?

It is not only the torque produced by friction does the rolling motion but also component of weight along the inclined helps in motion. Moreover it is not always that if a force produces motion, it must do some work.

It is correct that the static friction makes a contribution in rolling the body. However, its contribution is only to provide a supporting point. Static friction makes the contact point at rest w.r.t ground, so that gravity can make the body rolling. Since static friction starts from the contacting point, it has no torque, therefore it does no work.

Hope it can inspire some sparkles~ Please let me know if you have more ideas you want to discuss about!

Somewhere I am grasping the information that work done by static friction is zero as the point of contact where it is acting has zero velocity so it is at rest w.r.t ground

This is correct. $$P=\vec F \cdot \vec v$$ and since $$\vec v = 0$$ we get $$P=0$$ also.

Note that power is frame variant, meaning that it is different in different reference frames. If we transform to a different frame, indicated by prime marks, we can write $$P'=\vec F \cdot v'$$. If the primed frame is moving with respect to the unprimed frame at a velocity $$\vec u$$ then $$\vec v' = \vec v - \vec u$$ so $$P'=\vec F \cdot (\vec v - \vec u) = P - \vec F \cdot \vec u$$

why is the work done by the torque produced due to static friction zero?

This is due to the way that power transforms between reference frames, as described above. For rotational motion $$P = \tau \omega$$, where $$\tau$$ is the torque and $$\omega$$ is the angular velocity. But this formula applies only in the frame where the axis is at rest. In this problem the axis is moving at some velocity, $$u$$. So the power in this frame is $$P' = P - \vec F \cdot \vec u = \tau \omega - F u = (F r) (u/r) - F u = 0$$

So when you correctly calculate the power from the torque in this reference frame you find that it is also zero.