TITLE: Angular velocity, angular acceleration and it's observer
QUESTION [2 upvotes]: I was solving the problem above and realized that I was wrong somewhere.
I did two approaches but one didn't work well.
1. let one end which is at hold be the point we calculate torque. 
Then, $\frac{1}{2}lmg=\frac{1}{3}ml^2\alpha$ (here $\alpha$ denotes $\frac{d\omega}{dt}$), $\alpha=\frac{3g}{2l}$.
Using the relationship between angular acceleration and angular velocity, one could calculate that $a_{end}=l\alpha=\frac{3}{2}g$. This corresponds with intuition since rotation radius of the free end seen at the other end is $l$.
Also,$mg-R=ma_{CM}$  and $a_{CM}=\frac{l}{2}\alpha$ leads to $R=\frac{1}{4}mg$. Here, claculating $a_{CM}$ also seems not strange.
2. This is second approach which is shown in the solution above and which makes me struggle.
let CM be the point which we calculate torque about.
$mg-R=ma_{CM}$  also holds here and by torque equation, $\frac{l}{2}R=\frac{1}{12}ml^2\alpha$
Using these two, $\alpha=\frac{6R}{ml}$
Next step is where I'm stuck. The solution says $a_{CM}=\frac{l}{2}\alpha$ as highlighted above, but I don't understand. Why do we have $\frac{l}{2}$ factor? 
If $\alpha$ depends on observer, we calculated $\alpha$ putting CM in origin. Then how can I calculate the origin's linear acceleration?
If $\alpha$ does not depend on observer, why on earth $\frac{l}{2}$ is used?
Thanks in advance for reading my poor english and answering my question.

REPLY [0 votes]: $\alpha$ is not observer dependent. Now as far as l/2 is considered , it is used because the motion is such that the person which is holding the rod acts as a stationary centre for the motion of the body which is at a distance of $l/2$ from the centre.
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