Question

Since even light cannot escape the gravitational pull of a black hole, does this mean that the theoretical particle, known as the gravitation, have the potential to move faster than the photon

Answer 1

I have not included your picture because there is too much stuff on it (the velocity, the R, the R,etc. all distract from the crux of the problem, in my opinion). I have redrawn the problem showing some axially symmetric object of radius R, mass m rolling down an incline of angle without slipping. There are only three forces acting on the object, the frictional force f and the normal force N, both of which act at the point or line of contact, and the weight mg which acts at the center of mass of the object. There are three unknowns, N, f, and a, their acceleration of the center of mass. To solve for these unknowns, we must apply Newton's laws, both translational and rotational (F=ma and =I); here I am the moment of inertia about the axis about which torque is calculated and is the angular acceleration about that axis. You are a teacher, so I will skip a lot of the details which you can fill in easily. From F=ma I find N=mgcos and ma=mgsin -f. The first equation nails N, one of our unknowns, the second is one equation for the two remaining unknowns, so we are not done. So the only place to get a second equation is the rotational form of Newton's second law; I think this is the trickiest part of this problem. I find many students have an inclination to choose the symmetry axis to sum torques; they have been told, when they learned statics of non point objects, that you can choose any axis you like, but this is not a static problem and there is a very important constraintNewton's laws are not valid in an accelerating system, and the center of the rolling object is accelerating. (See the correction below!) So =Iyields I=Ia/R=mgRsin, or a=(mR2/I)gin; knowing a, then f=m(g sin-a) But, what is I? It isn't what you look up in a book because usually what is listed is the moment of inertia about an axis passing through the center of mass (Icm). Here the axis is a distance R from the center of mass, so using the parallel axis theorem, I=Icm+mR2.