Angular acceleration of a spool of thread I think this is an easy question in rotational kinematics, but--I don't seem to be understanding it on a fundamental level:

Here's my work:
$$ \tau \ =\ F\ r_1 $$
$$ \tau \ =\ I\ \alpha $$
$$ I\ \alpha \ =\ F\ r_1 $$
$$ \frac{1}{2}\ m\ r_1^2 \ \alpha \ =\ F\ r_1 $$
$$ \frac{1}{2}\ m\ r_1 \ \alpha \ =\ F $$
$$ \alpha \ =\ \frac{2\ F}{m\ r_1} $$
$$ \alpha \ =\ \frac{2\ (\ 0.650\ )}{(\ 0.230\ )\ (\ 0.078\ )} $$
$$ \alpha\ =\ 78.464\ s^{-2} $$
The book says the answer is in the neighborhood of 1 radian per second squared. Also--I haven't used the outer radius at all, which is a red flag.
What am I missing?
 A: Horizontally speaking, in addition to the applied force due to tension, there is another backward static friction applied on the contact point (because it is stationary due to rolling without slipping). So, don't ignore that force. 
A: You need to impose the no-slip condition. This means $v + \omega R_2 =0$ where $v$ is the velocity of the center (towards the right) and $\omega$ the rotational velocity (counter clockwise).
Next differentiate the above to get an expression connection linear and angular acceleration
$$a + {\alpha} R_2 =0$$
You also need a free body diagram (all quantities drawn positively)

Now you look at your equations of motion
$$ \begin{align} R + T &= m a \\ R R_2 + T R_1 & = I \alpha \end{align} $$
From the no slip conditions you use $a = - \alpha R_1$ above and solve the first equation for $R$
$$R = -m R_2 \alpha - T $$
Use the reaction $R$ in the second equation and collect the unknown $\alpha$ on the left hand side
$$ (I+m R_2^2) \alpha = -T (R_2-R_1) $$
Solve for the angular acceleration (which is negative, or clockwise)
$$ \alpha = -\frac{T (R_2-R_1)}{I+m R_2^2} $$
