Signs in the equation, angular acceleration

[Bauxiet]

Homework Statement

Counter clockwise is positive, right is positive and up is positive.
My problem statement: When I use the formula ag = alfa*r do I have to take signs into account? Or is this formula just for the magnitude and will the signs be already taken into account in other equations? (F= m*a for example)

Homework Equations

Ag = alfa*r (gravity point acceleration)
F = m*a
M = I*alfa

alfa = angular acceleration

The Attempt at a Solution

I am a little bit confused. In the excercise underneath I did not and I became the right solution. I think this formula gives the magnitude, the signs are already taken in account in other equations where you fill in the ag = alfa*r. Is this right?

The underneath picture is as an example. Look at the highlight. My alfa is negative (angular acceleration) and my ag (aA in this case) is positive. Ag is the acceleration of the disk. IF I had taken the negative sign of the alfa into account, that would mean that my Ag would be negative, which is not the case!
Is this right? Correct me if I am wrong, thank you!

 

[haruspex]

The relationship between acceleration, angular acceleration and radius is best considered as a vector equation, and in that view it involves cross products. There are various conventions, such as the right-hand rule, regarding the order of arguments to the cross products and the positive directions for the vectors. When all the conventions are used consistently, the correct sign comes out.
Similarly for velocities.
However, it is not a simple conversion of Acceleration = angular accleration x radius to vector form, like maybe: ##\vec a=\vec \alpha \times\vec r##. (Clearly, because that would not register any acceleration for a purely radial acceleration.)
E.g. for velocity the equation is ##(\vec r.\vec r)\vec \omega=\vec r \times \vec v##.

If working with scalars, I do not know a better way than simply reasoning out the sign of the result.

Edit: Or maybe...
In the special case where the radius is constant, we can differentiate the velocity equation to get
##(\vec r.\vec r)\vec \alpha=\dot{\vec r} \times \vec v+\vec r \times \dot{\vec v} = \vec v\times \vec v+\vec r\times\vec a= \vec r\times\vec a##.

 

[Bauxiet]

The relationship between acceleration, angular acceleration and radius is best considered as a vector equation, and in that view it involves cross products. There are various conventions, such as the right-hand rule, regarding the order of arguments to the cross products and the positive directions for the vectors. When all the conventions are used consistently, the correct sign comes out.
Similarly for velocities.
However, it is not a simple conversion of Acceleration = angular accleration x radius to vector form, like maybe: ##\vec a=\vec \alpha \times\vec r##. (Clearly, because that would not register any acceleration for a purely radial acceleration.)
E.g. for velocity the equation is ##(\vec r.\vec r)\vec \omega=\vec r \times \vec v##.

If working with scalars, I do not know a better way than simply reasoning out the sign of the result.

Edit: Or maybe...
In the special case where the radius is constant, we can differentiate the velocity equation to get
##(\vec r.\vec r)\vec \alpha=\dot{\vec r} \times \vec v+\vec r \times \dot{\vec v} = \vec v\times \vec v+\vec r\times\vec a= \vec r\times\vec a##.

Thank you! If I use the vector method I become indeed a positive number. This method can always be used? Also If I am working not with vectors as in my picture above?

And if I work with scalars, your method is just to remove the sign? Or maybe if ag is pointing to the positive direction and the result of alfa*r is negative, just to remove the negative sign because ag is positive?

I think the vector method is in this case much more safe and if it is allowed to use in any situation I think it is better to use the vector method.