Spring mass system attached to a disc

[rafaamcarvalho]

Hello to all, first sorry about any mistakes because English is not my native language. I'm trying to solve this problem and I can't seem to find the equation of motion

Homework Statement

The disc and the point A both have mass m;
θ(0) = 0

2. The attempt at a solution
I tried to use the Newton's second law for rotation but I can't find the values of the point A and K in therms of θ.
I don't know if what I did here was correct, but I tried this:

ΣMo = Jα
Jα = m⋅g⋅r - k⋅x⋅r; x = r⋅sinθ

 

[haruspex]

Jα = m⋅g⋅r - k⋅x⋅r

Is this for small oscillations only, or can θ be anything? If anything, the mgr term is wrong.

x = r⋅sinθ

It looks to me that the strong wraps around the disc, so it is not sine θ. If the disc has rotated θ, what length of string is wrapped onto it?

 

[rafaamcarvalho]

Is this for small oscillations only, or can θ be anything? If anything, the mgr term is wrong.

I believe so. But if its not m⋅g⋅r, what could it be?

You are right, the spring wraps around the disc, so is the displacement x = r⋅θ?

And how to calcule the polar moment of inertia J of a disc with mass m with a concentrated mass m in point A?

 

[haruspex]

its not m⋅g⋅r, what could it be?

Draw the diagram showing the line of action of mg. For the moment of a force about an axis, you multiply it by the perpendicular distance from its line of action to the axis.

polar moment of inertia J of a disc with mass m with a concentrated mass m in point A

Find the moment of inertia of each mass around the axis and add them.

 

[rafaamcarvalho]

Find the moment of inertia of each mass around the axis and add them.

Oh got it, did not check if it was perpendicular, so the therm should be m⋅g⋅cosθ⋅r?

And then my final formula would be,

(m⋅r² + ½⋅m⋅r²)α + (k⋅r²)⋅θ - (m⋅g⋅cosθ⋅r) = 0 ?

 

[haruspex]

Oh got it, did not check if it was perpendicular, so the therm should be m⋅g⋅cosθ⋅r?

And then my final formula would be,

(m⋅r² + ½⋅m⋅r²)α + (k⋅r²)⋅θ - (m⋅g⋅cosθ⋅r) = 0 ?

Looks right.