Two particles with masses 3.1 kg and 6.6 kg are connected by a light rod of length 4 m. Find the moment of inertia of the system about an axis perpendicular to the rod and passing through:

a) the midpoint of the rod;
b) the center of mass of the system of particle:

To find the moment of inertia of the system about different axes, we can use the standard formulas for moment of inertia.

a) Moment of inertia about the midpoint of the rod:
The midpoint of the rod divides it into two equal halves. The moment of inertia of each half can be calculated as if it were a point mass at the midpoint. Since the rod is light, we can consider its mass to be concentrated at the midpoint.

The moment of inertia of a point mass rotating about an axis perpendicular to the mass and passing through it is given by the formula:

I = m * r^2

Where:
I is the moment of inertia
m is the mass
r is the perpendicular distance from the axis of rotation to the mass

In this case, the mass at the midpoint is the total mass of the system divided by 2.

Mass at midpoint = (3.1 kg + 6.6 kg) / 2 = 4.85 kg

The perpendicular distance from the midpoint to the axis of rotation is half the length of the rod.

Distance from midpoint to axis of rotation = 4 m / 2 = 2 m

Therefore, the moment of inertia about the midpoint of the rod is:

I_midpoint = (4.85 kg) * (2 m)^2 = 19.4 kg.m^2

b) Moment of inertia about the center of mass of the system:
To find the moment of inertia about the center of mass, we need to first find the center of mass of the system.

The center of mass of a system of particles can be calculated using the formula:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Where:
x_cm is the position of the center of mass
m1 and m2 are the masses
x1 and x2 are the positions of the masses

In this case, the system consists of two particles with masses 3.1 kg and 6.6 kg. Let's assume that the particle with mass 3.1 kg is at position x1 and the particle with mass 6.6 kg is at position x2. The total mass of the system is 3.1 kg + 6.6 kg = 9.7 kg.

x_cm = (3.1 kg * x1 + 6.6 kg * x2) / 9.7 kg

Since the particles are connected by a light rod, the center of mass lies at the midpoint of the rod. Therefore, x_cm is equal to half the length of the rod.

x_cm = 4 m / 2 = 2 m

Now that we have the position of the center of mass, we can calculate the moment of inertia about it.

The moment of inertia about an axis passing through the center of mass is given by the formula:

I_cm = Σ(m_i * r_i^2)

Where:
I_cm is the moment of inertia about the center of mass
m_i is the mass of the ith particle
r_i is the perpendicular distance of the ith particle from the axis of rotation (which is the center of mass in this case)

Since the particles are located at positions x1 and x2, the perpendicular distances from the center of mass to each particle are:

r1 = |x1 - x_cm|
r2 = |x2 - x_cm|

Substituting the values, we have:

r1 = |x1 - 2 m|
r2 = |x2 - 2 m|

Finally, the moment of inertia about the center of mass is:

I_cm = (3.1 kg * r1^2) + (6.6 kg * r2^2)

However, without knowing the specific positions of the particles, we cannot calculate the exact values of r1 and r2.