Newton's Method for Equations of Motion (Vibrations) help?

[Erikono]

Homework Statement

Determine the equations of motions in terms of x and gamma.
Assume small angles and that the wheel rolls without slip. The mass of the thin homogeneous large disk

of radius 2R is 2m. The mass of the thin homogeneous inner disk of radius R is m. The rod of length 2R is massless and rigid. The two pulleys are massless.

I have attached my homework set and the problem is number 15. (I wouldn't be able to describe this very well)

Homework Equations

We are supposed to use Newtons method.
See uploaded images for the answer and relevant equations.

The Attempt at a Solution

See other uploaded image for my attempt at the solution.

I'm confused as to what equations from what free body diagrams I should be using. Also, I don't know if the substitutions I am making or the constraints I have set up for the problem are right.

x_1 = R*theta ( displacements for wheels on top)
x_2 = x_1 - 2R*theta - 2R*gamma
F_s1 = k(x_1 + r*theta) (spring attached to wall)
F_s2 = T_1 = k(x_1 + r*theta) (spring between pulleys)
T_2 = mg
T_2x = T_2*gamma
F_friction = ?
I also summed moments about A which is where the smaller disk makes contact with the ground.

Any help on the matter would be greatly appreciated.

 

[KataKoniK]

First, let's define our variables and coordinate system. We will use the following variables:
x1 = displacement of wheel 1 (on the left)
x2 = displacement of wheel 2 (on the right)
gamma = angle of rotation of the rod
theta = angle of rotation of the inner disk
R = radius of the large disk
r = radius of the inner disk
m1 = mass of the large disk
m2 = mass of the inner disk
M = mass of the rod
g = acceleration due to gravity
k = spring constant
T1 = tension in the spring attached to the wall
T2 = tension in the spring between the pulleys
Ff = friction force between the large disk and the ground

Now, let's draw free body diagrams for each of the components: the large disk, the inner disk, and the rod.

For the large disk, we have the following forces acting on it:
- T1, the tension in the spring attached to the wall, acting to the left
- T2, the tension in the spring between the pulleys, acting to the right
- Ff, the friction force between the disk and the ground, acting to the right
- mg, the weight of the disk, acting downwards

For the inner disk, we have the following forces acting on it:
- T2, the tension in the spring between the pulleys, acting to the left
- T2x, the component of T2 acting along the x-axis, acting to the left
- Ff, the friction force between the disk and the ground, acting to the right
- mg, the weight of the disk, acting downwards

For the rod, we have the following forces acting on it:
- T2x, the component of T2 acting along the x-axis, acting to the right
- mg, the weight of the rod, acting downwards

Now, we can write down the equations of motion for each component.

For the large disk:
Summing forces in the x-direction:
T1 - T2 - Ff = m1 * (d^2x1/dt^2)
Summing moments about the center of the disk:
T2 * R - Ff * R = I * (d^2gamma/dt^2), where I is the moment of inertia of the disk (I = (1/2)*m1*R^2)