Confusion regarding torque and calculating linear/angular acceleration of an object when a force is applied a distance from its center of mass?

Question

From what I have read/learned about torque, it appears that it is derived based on the idea that applying a force farther from a point about which an object rotates increases the rotational force applied to that object/causes more rotational acceleration.

My confusion is how exactly the position/distance from the center of rotation from which the force is applied causes varying degrees of rotational acceleration. Specifically, I am curious if there is some sort of mathematical derivation that shows how much rotational acceleration/force a force applied at a certain distance from a point of rotation of an object causes exactly.

In order to try to understand this better, I attempted to calculate the instantaneous linear and angular acceleration of a stick/bar/pole of mass M, length L (R = L/2), uniform density M/L = m when a force F is applied distance r from the objects center of mass (which of course lies in the center of the object) without using the idea of torque.

I concluded that the instantaneous linear acceleration would be F/M. For the angular acceleration, I tried to calculate it on the basis that the force applied at the point would experience equal/opposite resistance force from the masses on either side of its point of application.

I decided to give each side a linear resistance force equal to $ m(R-r)(F/M) $for the short side and m(R+r)(F/M) for the long side. Then I determined that their would also have to be a rotational resistance force such that the the total resistance force on each side is equivalent.

For this I integrated mra (with r being the distance from the COM and a being the angular acceleration) from r to R for the short side to get $$(1/2)(ma(R^2-r^2)). $$

For the long side I got

$$ (1/2)(ma(r^2-R^2)). $$ After setting

$$m(R-r)(F/M) + (1/2)(ma(R^2-r^2)) = m(R+r)(F/M) + (1/2)(ma(r^2-R^2)) $$

I solved for a to get a = $$\frac{ (2Fr)}{(M(R^2-r^2))} $$

I assumed that, since it was instantaneous, all the rotational force would be parallel with the linear force. I thought the equation seemed fairly intuitive as at $r = 0 $ the angular acceleration would be zero and as r approached R a would increase significantly. I would defend a = infinity at r = R with the fact that that is technically impossible since there will always be some mass/distance past the point of application of the force.

However, I wasn't sure if it was correct based on my initial assumption for how I derived it (forces on each side would be the same). I was wondering if someone could explain how to calculate the angular acceleration based on Newton's laws and preferably without using torque. Or if someone could explain/derive how torque/angular analogs of forces correctly represent newton's laws applied to rotational motion that would also be helpful.

Sorry for formatting, I'm talking about drawing a long rectangle with the properties stated above and applying a force perpendicular to the bar distance r from it's center. Then assuming that the sums/integration of the masses and their accelerations on either side of the point of application of the force would be the same, separating the accelerations into both linear and angular.

Answer 1

I was wondering if someone could explain how to calculate the angular acceleration based on Newton's laws and preferably without using torque.

Without using torque? Unlikely. How about

$$Fd = I \alpha$$

where $F$ is the force, $d$ is the perpendicular distance from the centre to where the force is applied, $I$ is the moment of inertia and $\alpha$ is the angular acceleration. But both sides of this equation is equal to the torque.

Or if someone could explain/derive how torque/angular analogs of forces correctly represent newton's laws applied to rotational motion that would also be helpful.

Force becomes torque. Mass becomes moment of inertia. Acceleration/velocity becomes angular acceleration/angular velocity. Momentum

$$p = mv$$

becomes angular momentum

$$L = I \omega$$

and Newton’s second law

$$F = m \frac{dv}{dt}$$

becomes

$$\tau = I \frac{d\omega}{dt} = I \alpha$$

An explanation without using torque is not possible.