exercise 348 while the tank is moving forward at a constant speed v 30

Question

EXERCISE 3.48 While the tank is moving forward at a constant speed v. = 30 km/h, the turret is rotating at the constant rate o = 0.3 rad/s and the barrel is being raised at the constant rate è = 0.5 rad/s. At a certain instant the barrel is facing for-ward and e = 15°. At this instant a shell whose mass is 80 kg is about to emerge from the barrel with a muzzle velocity vrel = 5500 km/h that has reached a maximum because the internal propulsive pressure within the barrel has been dissipated. Determine the force exerted by the shell on the wall of barrel at this instant.
solve this problem, I recommend attaching a coordinate system, x1y1Z1, to the tank turret with the z, axis being the center of turret rotation, point A, and the x, axis pointing toward the barrel pivot, point B. Then, mount xyz to the barrel and have it pivot about the y axis. N,is then the angular velocity of the turret and N2 = w is the cumulative angular velocity of the barrel(note the use of the word cumulative; it has two components). You can use those two angular velocities to get the angular acceleration of the moving xyz reference frame, a. The barrel does not translate relative to the turret, so the acceleration of point B should be relatively easy to obtain. That will then be the acceleration of the xyz coordinate system origin. The shell velocity is then considered to be the velocity of the shell with respect to the xyz coordinate system,xyz