Rolling Bottle: Calculating Acceleration of Center of Mass

[Sonntag]

Homework Statement

There is an empty bottle described as an hollow cylinder, that lies on a paper. Now the paper is pulled with an acceleration a, so that the bottle starts rolling perfectly on the paper. (Have a look at the figure.)

Homework Equations

Calculate the acceleration of the center of mass.

The Attempt at a Solution

Moment of inertia in the center of mass: ${I_C = 1/2 M R^2}$
Moment of inertia in the center of mass on the surface: ${I_S = 3/2 M R^2}$
Torque: ${|\bf{M}| = |\bf{R}| |\bf{F}|}$
Angular acceleration: ${\alpha = \dfrac{M}{I_S} = \dfrac{F}{3/2 M R}}$
acceleration of the center of mass: ${a_s = \alpha R = 2/3 \dfrac{F}{M}}$I have the feeling that my answer isn't the intended of the question. Mistakes? More to do? I hope on your answers and thoughts.

 

[Doc Al]

Hint: Look at things from the (non-inertial) frame of the paper. (Note also that it's a hollow cylinder, so correct your formula for moment of inertia.)

 

[haruspex]

Moment of inertia in the center of mass: ${I_C = 1/2 M R^2}$
Moment of inertia in the center of mass on the surface: ${I_S = 3/2 M R^2}$

I don't understand the expression "Moment of inertia in the center of mass on the surface"
i assume you mean

Moment of inertia about the center of mass: ${I_C = 1/2 M R^2}$
Moment of inertia about a point on the surface: ${I_S = 3/2 M R^2}$​

That would be true, as Doc Al points out, for a solid cylinder. But leaving that aside, your method still gave the wrong answer. The reason is that your expression for torque gives the torque about the mass centre. You cannot divide the torque about one axis by the MoI about a different axis to get the angular acceleration.
You can use Doc Al's method, or just be consistent about axes.