A Confusion in Rotational Dynamics I am trying to analyse the following situation using classical mechanical concepts.
Consider a a straight rod $AB$ of mass $M$ and length $L$ placed on a frictionless horizontal surface. A force $F$ acts at the end $A$ perpendicular to the rod. The direction of $F$ is fixed. I am trying to find out the initial acceleration of end $B$ just after force $F$ is applied. But I cannot figure out which point on the rod should be taken as the instantaneous centre.
If I consider the mid point of $AB$ (call it $O$) as centre.
Then, torque about $O$,
$$T=\frac{FL}{2}=\frac{ML^2a}{12}$$
where $a$ is the angular acceleration of rod.
Hence the linear acceleration of end B would be 
$\frac{3F}{M}$
But I can also assume end $B$ to be the centre. Then it's linear acceleration would be zero.
Can someone please help me with this situation?
 A: Irrational 3.14
It seems to me that you are using the equation torque = moment of inertia * angular acceleration. This particular equation ONLY holds under three conditions:


*

*The rotational center is fixed.

*The rotational center is the center of mass.

*The rotational center is moving at a constant velocity. 


In your scenario, the center of the rod satisfies condition 2, and thus is a "safe“ origin. The point B, however, is accelerating and thus does not satisfy any of the three conditions. That is why point B is not a valid choice. 
I refer you to chapter 8 of David Morin's Introduction to Classical Mechanics, for a complete derivation of the three conditions. 
A: The linear acceleration of $B$ due to the rotational movement:
$$ a_r = {L \over 2} {FL/2 \over ML^2/12} = {3F \over M} $$
The linear acceleration of $B$ due to the linear acceleration of the center of mass of the rod (which is anti-parallel to the rotational acceleration):
$$ a_t = {F \over M} $$
Since these two accelerations are on opposite directions to each other, the total linear acceleration of point $B$ is their difference, $2F/M$.
