a) To write the Lagrangian, we need to find the kinetic and potential energies of the system. Let’s assume that the wheel is moving to the right, and let x be the displacement of the center of mass of the wheel from its equilibrium position. Then, the position of the center of mass is given by (x, R), and the velocity of the center of mass is v.

The kinetic energy of the wheel is:

T = (1/2) Mv^2 + (1/2) I ω^2

where ω is the angular velocity of the wheel, and I = (1/2) MR^2 is its moment of inertia about the center of mass. Since the wheel is rolling without slipping, we have:

v = Rω

Therefore, the kinetic energy can be written as:

T = (1/2) Mv^2 + (1/4) Mv^2 = (3/4) Mv^2

The potential energy of the system is due to the springs, and is given by:

V = (1/2) k (x – d)^2 + (1/2) k (x + d)^2 = kx^2 + (1/2)kd^2

where d is the equilibrium length of the springs.

Therefore, the Lagrangian of the system is:

L = T – V = (3/4) Mv^2 – kx^2 – (1/2)kd^2

The Euler-Lagrange equation for x is:

d/dt (∂L/∂v) – ∂L/∂x = 0

∂L/∂v = (3/2) Mv

Therefore,

d/dt (3/2 Mv) + 2kx = 0

This simplifies to:

3Mv̇ + 4kx = 0

The Euler-Lagrange equation for v is:

d/dt (∂L/∂v̇) – ∂L/∂v = 0

∂L/∂v̇ = (3/4) M

Therefore,

d/dt [(3/4) M] – (3/2) Mv = 0

This simplifies to:

(3/4) Mv̈ = 0

Since this equation has no dependence on x, it implies that the acceleration of the center of mass is zero, which is what we expect for a system in uniform motion.

b) The maximum displacement possible before the wheel begins to slip occurs when the static frictional force at the contact point between the wheel and the ground reaches its maximum value. This occurs when the force from the springs is equal and opposite to the maximum static frictional force:

2kx = μMg

where g is the acceleration due to gravity. Solving for x, we get:

x = (μMg)/(2k)

Therefore, the maximum displacement possible before the wheel begins to slip is (μMg)/(2k).