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:

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To find the moment of inertia of the system, we need to consider the individual moments of inertia of each particle and then add them up.

a) Moment of inertia about the midpoint of the rod:
In this case, we can treat the system as two point masses connected by a rod. The midpoint of the rod is equidistant from both particles.

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

I = m * r^2

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

For the particle with a mass of 3.1 kg, the distance from the midpoint of the rod is half of the rod's length, so r = 2 m.

The moment of inertia of the first particle is:
I1 = m1 * r1^2 = 3.1 kg * (2 m)^2 = 12.4 kgm^2

Similarly, for the particle with a mass of 6.6 kg, the distance from the midpoint of the rod is also 2 m.

The moment of inertia of the second particle is:
I2 = m2 * r2^2 = 6.6 kg * (2 m)^2 = 52.8 kgm^2

To find the moment of inertia of the system about the midpoint, we add the individual moments of inertia:
I_system = I1 + I2 = 12.4 kgm^2 + 52.8 kgm^2 = 65.2 kgm^2

Therefore, the moment of inertia of the system about the midpoint of the rod is 65.2 kgm^2.

b) Moment of inertia about the center of mass of the system:
The center of mass is the point at which the entire mass of the system can be assumed to be concentrated. It lies at a distance equal to the ratio of the masses from the center of the rod.

To find the position of the center of mass, we can use the following formula:

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

Where:
x_cm is the distance of the center of mass from the first particle
x1 is the distance of the first particle from the first particle (0 in this case)
x2 is the distance of the second particle from the first particle (4 m, as the entire length of the rod is 4 m)

By substituting the given values, we get:

x_cm = (3.1 kg * 0 + 6.6 kg * 4 m) / (3.1 kg + 6.6 kg) = 3.47 m

The moment of inertia of the system about the center of mass can be calculated using the parallel axis theorem:

I_cm = (m1 * (d1)^2 + m2 * (d2)^2)

Where:
I_cm is the moment of inertia about the center of mass
d1 and d2 are the distances of the particles from the center of mass (x_cm)

For the first particle:
d1 = x1 - x_cm = -3.47 m (since it's at the left side of the center of mass)

The moment of inertia of the first particle about the center of mass is:
I1_cm = m1 * (d1)^2 = 3.1 kg * (-3.47 m)^2 = 35.80 kgm^2

For the second particle:
d2 = x2 - x_cm = 4 m - 3.47 m = 0.53 m (since it's at the right side of the center of mass)

The moment of inertia of the second particle about the center of mass is:
I2_cm = m2 * (d2)^2 = 6.6 kg * (0.53 m)^2 = 1.811 kgm^2

The total moment of inertia of the system about the center of mass is obtained by adding the individual moments of inertia:

I_cm = I1_cm + I2_cm = 35.80 kgm^2 + 1.811 kgm^2 = 37.61 kgm^2

Therefore, the moment of inertia of the system about the center of mass is 37.61 kgm^2.