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

Physics Asked by User4758 on December 4, 2020

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 = (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.

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

$$tau = I frac{domega}{dt} = I alpha$$

An explanation without using torque is not possible.

Answered by Dr jh on December 4, 2020

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