Mueiz
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It seem to me that you did not understand the logic by which I conclude that the second observer is inertialyuiop said:Let us examine the second observer riding on the edge of the disc. The Newtonian equation for centripetal force is F = m*v^2/r. Now the velocity of a particle on a rotating disc is a function of its radial displacement, so let us say v = f*r where f is an arbitrary constant, then by substitution, the centripetal force is F = m*f*r so we see that centripetal force increases with radius. The second observer on the edge of the disc will be experiencing greater acceleration than the third observer nearer the centre of the disc. Now the relativistic centripetal force will be even greater F = m*f*r/sqrt(1-(f*r/c)^2) in the non rotating reference frame and greater still in the rotating frame F = m*f*r/(1-(fr/c)^2). So we can conclude (If I have estimated the relativistic versions correctly) that the observer on the arbitrarily large disc will be experiencing exponentially large "centrifugal proper force" at relativistic speeds and not the vanishingly small force you seem to be assuming when you claim the second observer is effectively inertial. An observer experiencing large proper force is NOT an inertial observer.
it is that only the circumference of a very large disc is almost a straight line to any degree of approximate ...so it is in uniform motion relative to the first observer who is inertial ...if we are in uniform motion relative to an inertial observer then we must be inertial too ...tell me what is wrong with this ..before going to calculation
i want you also to reply the question i gave in my previous input