To tackle this problem, we need to analyze the dynamics of the two discs connected by a rod and how they interact with the applied force and friction. Let's break down the questions step by step.

Understanding the System

We have two discs: one with mass 4M and radius R, and the other with mass M and radius R/2. They are connected by a rod of length 2R. The coefficient of friction between the discs and the ground is u, and a constant horizontal force F is applied to the larger disc. The first scenario involves the force F being equal to Mg, which is the weight of the smaller disc.

1. Finding the Acceleration of the Center of Mass of the Smaller Disc

Since the force F = Mg is acting on the larger disc, we can analyze the forces and torques acting on both discs. The larger disc will experience a force of F, and since there is no slipping, the frictional force will also act on the smaller disc.

• The total mass of the system is 4M + M = 5M.
• The acceleration of the center of mass (CM) of the system can be calculated using Newton's second law:

Using the equation:

F = (total mass) × (acceleration of CM)

We have:

Mg = 5M × a_CM

Thus, the acceleration of the center of mass (a_CM) is:

a_CM = g/5.

Now, since the smaller disc is connected to the larger disc via a rod, it will also have the same acceleration as the center of mass. Therefore, the acceleration of the center of mass of the smaller disc is:

a_{smaller disc} = g/5.

2. Finding the Elongation of the Rod

Next, we need to find the elongation of the rod when the force F = Mg is applied. The elongation can be determined using Hooke's Law, which states that the elongation (ΔL) is proportional to the force applied (F) and inversely proportional to the stiffness (k) of the rod:

ΔL = (F × L) / (A × Y),

where:

• F = Mg (the force applied),
• L = original length of the rod = 2R,
• A = cross-sectional area of the rod,
• Y = Young's modulus of the material of the rod.

Substituting these values into the equation gives:

ΔL = (Mg × 2R) / (A × Y).

3. Maximum Acceleration Without Slip

In the final part of the question, we need to determine the maximum acceleration of the larger disc without slipping, given that the coefficient of friction u = 2/√15. The frictional force (f) that prevents slipping can be expressed as:

f = u × N,

where N is the normal force. For the larger disc, N = 4Mg (its weight).

Thus, the maximum frictional force is:

f_max = u × 4Mg = (2/√15) × 4Mg = (8Mg/√15).

Now, applying Newton's second law to the larger disc:

F - f_max = (mass of larger disc) × (acceleration of larger disc).

Substituting the values gives:

F - (8Mg/√15) = 4M × a_max.

Since F = Mg, we have:

Mg - (8Mg/√15) = 4M × a_max.

Solving for a_max:

a_max = [Mg - (8Mg/√15)] / 4M = [1 - (8/√15)]g / 4.

Thus, the maximum acceleration of the larger disc without slipping is:

a_max = [(1 - (8/√15))g] / 4.

This analysis provides a comprehensive understanding of the dynamics of the system involving the two discs and the forces acting on them. Each step logically follows from the previous one, ensuring clarity and coherence in the solution.