Force acting on the center of mass of a rolling disk

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Fibo112
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Hello. The following situation I thought out confuses me so I am wondering where my mistake lies.
A uniform disk of mass M and radius R sits on its edge. A string is attached to the highest point and pulled with a Force F in the x direction.

The moment of inertia of the disk is MR^2/2 making the angular momentum about the center of mass MR^2w/2 where w is the angular velocity. The torque about the center of mass seems to be FR. Since torque is the derivative of angular momentum we have FR=MR^2a/2 where a is the angular acceleration.
This means the angular acceleration is equal to 2F/RM. The acceleration A of the c.o.m is equal to aR, so it is equal to 2F/M, making the force acting on the body equal to 2F/M*M= 2F?
 
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Fibo112 said:
This means the angular acceleration is equal to 2F/RM. The acceleration A of the c.o.m is equal to aR
What makes you say that the acceleration of the center of mass is equal to aR? Nobody said that the disk was rolling without slipping.