Rotate Rigidbody on Two Separate Axes Independently to a specific angle, using torque?

Hi everyone. I have run into bit of a problem and I would appreciate any insight. I have a rectangular rigidbody that I am trying to rotate on both its Y and Z axes. I want the Z axis to rotate a full 360 degrees and the Y axis to rotate 180 degrees. Also when one axis is rotating its rotation does not effect the rotation of the other axis.

I have figured out a solution to rotate a single axis at a time to a certain angle using torque. By monitoring the amount of rotation along an axis each frame.

Code (CSharp):

1.     public void CalculateRotation()
2.     {
3.         Vector3 currentUpwardDirection = skateBoardRigidBody.transform.up
4.         // measure the angle rotated since last frame:
5.         float angle = Vector3.Angle(currentUpwardDirection, previousUpwardDirection);
6.        
7.         if (angle > 0.01)// if rotated a significant angle...
8.         {    
9.             // accumulate in currentRotationAmount
10.             currentRotationAmount += angle;
11.             // and update previousUpwardDirection
12.             previousUpwardDirection = currentUpwardDirection;
13.         }  
14.     }

However when incorporating both axes I cant seem to come to a solution.

I wanted both axes to rotate independently of each other so I could reuse the above code to stop the amount of torque once the desired angle was met for certain rotation.

I tried using the following to see if this would give me a rotation that I am looking for however it does not.It seems to combine the torques.

Code (CSharp):

1. skateBoardRigidBody.AddRelativeTorque(0,0,torqueAmount);
2. skateBoardRigidBody.AddRelativeTorque(0,torqueAmount,0);

I this at all possible using physics/torque? I should I resort to using Quaternions? I am building a skateboarding game and would like to try to incorporate physics were at all possible however, if i need to fake it then so be it.

I attached an image so you can get a better idea.

I appreciate any and all help.