Final answer:
The moments of inertia for a uniform bar with balls at its ends vary with different axes: 2.00 kg·m² for the axis through the center, 4.00 kg·m² through a ball's center, 1.33 kg·m² center parallel, and for the ball center parallel, it is trivial for the mass at the axis.
Step-by-step explanation:
The moment of inertia is a measure of an object's resistance to changes in its rotation rate. When calculating the moment of inertia for a system, different axes of rotation will give different moments of inertia. For the provided system of a uniform bar with two small balls at its ends, the moments of inertia can be calculated as follows:
- Axis through the center of the bar: Using I₁ = mR² + mR² = 2mR², we find the moment of inertia to be I₁ = 2(0.500 kg)(1.00 m)² = 2.00 kg·m².
- Axis through one of the ball's centers: Using I₂ = m(0)² + m(2R)² = 4mR², we find it to be I₂ = 4(0.500 kg)(1.00 m)² = 4.00 kg·m².
- Axis through the center of the bar and parallel to its length: The bar can be treated as a thin rod, so the moment of inertia is I = ML²/12. This gives us I = 4.00 kg·(2.00 m)²/12 = 1.33 kg·m².
- Axis through one of the ball's centers and parallel to the bar: The moment of inertia for this axis is not common and has a complex derivation, but for a point mass, it is simply I = mr², where r = 0 for the axis passing through the mass itself.
These calculations illustrate how the moment of inertia depends on the axis of rotation chosen for the calculation.