Torque acting on a rotating wedge due to a linear force parallel to the axis of rotation

Lets callthe cam force as per your sketch P. P will have 2 components acting on the surface of the wedge, $$P* cos(theta)_ text{paralel to surface} , P sin(theta)_ text{perpendicular to surface} $$

Bur both of these components pass through the center of rotation, meaning they will not produce any torque, hence any rotation.

If P is too big, the shaft at the center can bend or break but wont rotate.

EDIT

After you modified your question. And if we disregard the original sketch.

there is a varying lateral force that varies as the cam moves around.

This force F multiplied by L is your torque.

, $F= Psin(theta)*sin(alpha) , $ Sin of the local slope angle.

That angle Alph is a simple sine function, I let you figure it.