Stuck in trying to learn Torque, Moments of inertia High school physics student here. I was reading an explanation of why we work with torque and the moment of inertia the way we do. It was this one. However i got stuck on a certain concept. 
The explanation made sense to me in places, but i must have missed something. 
Let's take a rod fixed on a point on it's end as an example. Suppose we apply a force to a point on the rod. Well, the translational acceleration of a point of the rod depends on its distance from the center and the angular acceleration. 
To the best of my understanding, this should mean that the amount of force needed to accelerate the whole thing with an angular acceleration of α is (average distance from center) * α * (mass of whole rod).
Given all of this, i fail to see how pushing at a particular distance from the center is  going to affect anything (as in, why we use torque in the equations). In the end, isn't the force kind of distributed along the whole rod or something?
Please explain what are my misconceptions and why the distance from the fixation point actually matters. (I would prefer if you didn't analyze energies, i desire an explanation based on basic Newtonian laws to better comprehend the topic).
 A: You need to draw the free body diagram of your rod pivoted at the corner. The force that you are applying, let it be F, but also there is a reaction force by the pivot on the rod. So the equation that you are suggesting $F=m\alpha\frac{L}{2}$ (where $L$ is the length of the rod) is incorrect.
Let the reaction force by the pivot I'm talking about, have two components $R_1$ parallel to the rod and through the pivot and $R_2$ which is perpendicular to the rod and opposite direction of $F$. The free body equation for forces perpendicular to the rod(in the direction of the force $F$ that you are applying) becomes $$F-R_2=ma=m\alpha \frac{L}{2}$$. But, you don't have a clue about $R_2$ and it would be a difficult job finding $R_2$. Firstly convine yourself that $R_2$ exists and is finite because if $R_2=0$ then the rod would translate and not rotate, as in the case of a non-pivoted rod. 
Now, incidentally the reaction force $R_2$ depends on two things
1) the distance from the pivoted end at which you apply the force
2) The distance from the pivot at which the rotational mass(moment of inertia) is concentrated called the radius of gyration.
Now convince yourself of 1. Since if you taken a pivoted rod and apply a strong impulse at the extreme end of the rod, the rod rotates faster than if you apply the same impulse on a point close to the pivot. It's easier to open a door by pushing near the door knob, than by pushing near the hinge.(Your impulse and hence $F$ remains same, $m$ and $L$ are obviously constants, but $\alpha$ changes, it's obvious $R_2$ depends on the point of application of force) To consider point 2, you will need to consider the following. 
The consideration of being able to calculate $R_2$ is dicy, so we figure the following out. The impulse experiment if graphed and measured shows that $\alpha$ scales linearly with point of application of the impulse $r$. So, defined torque $\tau=Fr$. Now $\frac{\tau}{\alpha}$ remains constant for the same body clamped in the same place. Change the clamping position or the body and the slope of $\frac{\tau}{\alpha}$ changes, leading to intuitions of radius of gyration. It's all because $R_2$ and $\alpha$ are sensitive to point of application of force. By using the concept of torque, you taken the $R_2$ unknown out of the equation. 
A: To find the linear motion of the center of mass sum up all the forces (regardless of location)
$$ \sum F = m a_{cm} $$
To find the angular motion about the center of mass sum up all the torques also about the center of mass $$\sum \tau_{cm} = {I}_{cm} \alpha $$
* NOTE: the above equation is only valid in 2D or for fixed axis of rotation problems. The full 3D equation is $\sum \vec{\tau}_{cm} = \mathtt{I}_{cm} \vec{\alpha} + \vec{\omega} \times \mathtt{I}_{cm} \vec{\omega} $
Consider a force $F$ applied on another point A. The motion of A is $a_A = a_{cm} + d \,\alpha$ and the torque of a force at A is $\tau_{cm} = d\,F$
Combined this gives the following equations $$\begin{aligned} F & = m ( a_A -d\, \alpha) \\ d\,F &= I_{cm} \alpha \end{aligned}$$
which is solved for
$$ F = \left( \frac{1}{\frac{1}{m} + \frac{d^2}{I_{cm}}} \right) a_A $$
The thing inside the parenthesis is the effective mass that the force feels since it is applied away from the center of mass. It is less than $m$ because the point will accelerate more than $\frac{F}{m}$ since it will rotate also.
