Final answer:
The rotational inertia about the center of the rod is 5.0 kg·m². For the axes through each mass, the rotational inertia is 12.0 kg·m² for the axis through the 2.0-kg mass and 8.0 kg·m² for the axis through the 3.0-kg mass.
Step-by-step explanation:
The student's question pertains to finding the rotational inertia, or moment of inertia, of a system with two point masses on a rod with negligible mass. To solve the mathematical problem completely, we will calculate the rotational inertia about the center of the rod as well as about axes through each mass.
Rotational Inertia About the Center of the Rod
The moment of inertia (I) for point masses is calculated using the formula I = mr², where 'm' is the mass and 'r' is the distance from the axis of rotation. In this case, the rod is 2.0 m long, so each mass is 1.0 m from the center. Therefore, the moment of inertia about the center of the rod is:
Icenter = (2.0 kg)(1.0 m)² + (3.0 kg)(1.0 m)² = 2.0 kg·m² + 3.0 kg·m² = 5.0 kg·m².
Rotational Inertia About Each Mass
When considering the axis of rotation at the location of one of the masses, the other mass is 2.0 m away. We only consider the mass that is away from the axis as the mass at the axis contributes no inertia due to the zero distance. Hence for each mass:
For the 2.0-kg mass axis: I2kg = (3.0 kg)(2.0 m)² = 12.0 kg·m².
For the 3.0-kg mass axis: I3kg = (2.0 kg)(2.0 m)² = 8.0 kg·m².