Rolling of Point Mass Disc, is it Possible?

Danger! Get away! That's a black hole (smaller than the Schwarzchild radius for any positive mass.)

F does not have to equal zero, but R X F must equal R X N or the disk will sink into the slope. In this picture, Fsin(Θ) is the measurable torque.

R X N should be zero, cuz N vector goes through the mass center..

what do you mean by that F does not have to be zero..?!?

df/dr as r goes to zero?

Is something is missing from this diagram?

You can already resolve the forces, (normal to the ramp and parallel to the ramp). The force parallel to the ramp (if it's greater than the static friction) will cause the wheel to rotate. Having a point mass means rotational inertia won't play any part in the eqn's, so it's just a F=ma problem.

Everyone finds these problems hard at first, good luck.

If I am reading the drawing correctly, gravity is the only force being applied. The normal force (N) will be g*cos(theta) and the motive force (F) wiil be g*sin(theta). If there is a rolling friction force equal to and opposite the motive force, then the disc may stop rolling (velocity equals 0). It will never roll up hill unless you add another force pushing opposite and greater than g*sin(theta).

This is pretty basic physics- high school level...

Let c_f denote the cooefficient of friction where the disc contacts the ramp.

If c_f < tan θ, the disc slides.

If c_f > tan θ, the disc rolls.

And if c_f = tan θ, the disc flops down one side?

Good lord dude this is a simple statics problem. You do not need torque energy or anything beyond simple statics or a dose of common sense to know that the disk rolls down the slope...

Saying that the disk has a point mass, with a moment of inertia of zero, is equivalent to saying that the disk slides frictionlessly down the plane; i.e., none of the potential energy is converted into rotational energy. The acceleration of the disk of the disk down the plane can be treated as though it slides. [The rotational energy at the bottom of the inclined plane is the same as at the top of the inclined plane. It is, in a physical sense, meaningless to say that it rolls.]

This idealization might not have angular momentum, but geometrically it should roll under the suitable friction circumstances.

Nothing was said about "assuming no friction"; what is assumed in this case is no angular momentum.

(I think...)

Yes, I'd agree. There is, in fact, a caption under the diagram (I assume it is for the diagram) that says "non-slip rolling". There is nothing in the diagram itself that discusses friction, so I supposed one is to assume that the idealized disk rolls with negligible friction and negligible change in angular momentum. The calculations can ignore both. It's equivalent to a point mass sliding down the plane. (Thus my comment that in terms of the physics it is meaningless to say it rolls.)