91Ó°ÊÓ

For the axis perpendicular to the bar through its center:The moment of inertia of a bar about its center is given by:\[ I_{bar} = \frac{1}{12} m L^2 \]where \( m = 4.00 \ \mathrm{kg} \) and \( L = 2.00 \ \mathrm{m} \).Plugging in the values:\[ I_{bar} = \frac{1}{12} \times 4.00 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 = \frac{1}{12} \times 4 \times 4 = \frac{16}{12} = 1.33 \ \mathrm{kg \cdot m^2} \]The moment of inertia of each ball with respect to the center is:\[ I_{ball} = m r^2 \]where \( r = 1.00 \ \mathrm{m} \) (since each ball is at the end of the bar).Therefore:\[ I_{ball, total} = 2 \times 0.500 \ \mathrm{kg} \times (1.00 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{ball, total} = 1.33 + 1.00 = 2.33 \ \mathrm{kg \cdot m^2} \]

For the axis perpendicular to the bar through one of the balls:The moment of inertia of the bar about this axis using the parallel axis theorem is:\[ I' = I_{bar} + md^2 \]where \( d = 1.00 \ \mathrm{m} \) (distance from the center to the axis).Thus:\[ I' = \frac{1}{12} \times 4.00 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 + 4.00 \times (1.00 \ \mathrm{m})^2 = 1.33 + 4.00 = 5.33 \ \mathrm{kg \cdot m^2} \]Only one ball contributes directly as it rotating at the axis, so:\[ I_{ball} = 0.500 \ \mathrm{kg} \times (0 \ \mathrm{m})^2 = 0 \]The remaining ball's contribution:\[ I_{ball, distant} = 0.500 \ \mathrm{kg} \times (2.00 \ \mathrm{m})^2 = 2.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I' + I_{ball, distant} = 5.33 + 2.00 = 7.33 \ \mathrm{kg \cdot m^2} \]

For the axis parallel to the bar through both balls:The entire mass of the system rotates around the axis at a distance of the bar's length from the axis.For the bar:\[ I_{bar} = 4.00 \ \mathrm{kg} \times (0 \ \mathrm{m})^2 = 0 \]The balls, however, contribute:\[ I_{ball,total} = 2 \times 0.500 \ \mathrm{kg} \times (1.00 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{ball,total} = 0 + 1.00 = 1.00 \ \mathrm{kg \cdot m^2} \]

For the axis parallel to the bar and 0.500 m from it:The entire mass of the system is away from the axis.Both the bar and balls contribute at a distance of 0.500 m:For the bar:\[ I_{bar} = 4.00 \ \mathrm{kg} \times (0.500 \ \mathrm{m})^2 = 1.00 \ \mathrm{kg \cdot m^2} \]Each ball contributes:\[ I_{balls} = 2 \times 0.500 \ \mathrm{kg} \times (0.500 \ \mathrm{m})^2 = 0.25 \ \mathrm{kg \cdot m^2} \]Thus, the total moment of inertia is:\[ I_{total} = I_{bar} + I_{balls} = 1.00 + 0.25 = 1.25 \ \mathrm{kg \cdot m^2} \]