Solving the Spool's Acceleration with a Constant Force

[thenewbosco]

Diagram: http://snipurl.com/c8h6

A spool of wire is unwound with a constant force F. The spool is a solid cylinder mass M, radius R, and doesn't slip.
Show the acceleration of the centre of mass is 4F/3M.

what i have done:
$$\sum Torque = I\frac{a}{R}$$
$$FR=I\frac{a}{R}$$
$$\frac{FR^2}{I}=a$$

Now, for I, I used the parallel axis theorem and got
$$I=\frac{1}{2}MR^2+ MR^2 = \frac{3}{2}MR^2$$

but plugging into my formula above yields $$a=\frac{2F}{3M}$$

how do i get a=4F/3M?

thanks for your help

 

[Doc Al]

$$\sum Torque = I\frac{a}{R}$$
$$FR=I\frac{a}{R}$$

Since you are using the point of contact with the floor as your axis of rotation, the torque is 2FR, not FR.

 

[thenewbosco]

Thank you for your help. Can you explain why the torque is 2FR instead of just FR??

thanks

 

[Doc Al]

Can you explain why the torque is 2FR instead of just FR??

Torque is F times the perpendicular distance to the axis. Since the axis you are using is the point of contact with the floor, the distance is 2R, not R. (FR is the torque about the center of mass.)