In a non-symmetric turn, how do the side slip angle and thrust side slip angle affect the equations of motion? I'm doing a problem on Flight Mechanics in which a banked turn is performed. The center of gravity of the aircraft is always within an horizontal plane.

In order to write the dynamic equations, I have to assume some simplifications and conditions. One of them is the following:


  The thrust’s sideslip angle is equal and with opposite sign than the aeroplane’s sideslip angle, that is: 
  ($\nu = −\beta$).
  
  Side slip angle ($\beta$) is defined as the angle made by the velocity
  vector to longitudinal axis of the vehicle at the center of gravity in
  an instantaneous frame.
  
  Thrust side slip angle ($\nu$) is defined as the angle made by the
  thrust vector to longitudinal axis of the vehicle at the center of
  gravity in an instantaneous frame.
  https://en.wikipedia.org/wiki/Slip_(aerodynamics)#Sideslip_angle


Then, I have to write the dynamic equations for the symmetric flight and for the non-symmetric flight.

My questions are:


In my opinion, the condition written above should not be accounted for the symmetric flight, so $(\beta = \nu = 0)$. Is this affirmation correct?
For the non-symmetric case, I have drawn the following scheme of velocity and thrust. But I'm not sure if it's correct because it doesn't make sense, despite that is what the condition says: $(\nu=-\beta)$. Is the scheme correct according to the condition? (The scheme is the top view of the aircraft.)
Relying on physics, causing a mass to execute a circular path requires a suitable centripetal force. In an airplane in a banked balanced turn, this comes from the inward component of lift due to the bank angle. 

However, if the turn is not properly co-ordinated, there can be additional contributions to the centripetal force from the power plant thrust no longer being directed tangentially and sidethrust on the airframe due to slip.

The extreme case is the flat turn where the wings remain level and the power plant thrust has an inward component due to the sideslip.

That said, apart from an airplane having vectoring thrust, the only reason for a thrust sideslip angle I can think of is that many single airscrew airplanes have built-in sidethrust to counteract engine torque. However there is no requirement for the engine sidethrust angle to equal the slip angle, so in that case the assumption that the two angles are the same cannot be valid.

The thrust line of an airscrew doesn't change with slip. The slip causes asymmetric thrust on the blades which precesses gyroscopically so the airscrew creates a couple which tends to increase the slip angle.