Final answer:
The ratio of the moment of inertia through the second axis (30-cm mark) to the moment of inertia through the first axis (50-cm midpoint) is 7/3. This calculation uses the parallel-axis theorem.
Step-by-step explanation:
The question deals with the moment of inertia of a slender uniform rod rotated around two different axes perpendicular to the rod, where one passes through the rod's midpoint, and the other through a point that is not the midpoint. To find the ratio of moments of inertia about these two axes, we can use the parallel-axis theorem, which states that the moment of inertia I about any axis parallel to and a distance d from the center of mass axis is I = Icm + Md2, where Icm is the moment of inertia about the center of mass and M is the mass of the rod.
For the first axis, which passes through the 50-cm mark, the moment of inertia is I50 = ML2/12, because it is the midpoint of the rod (center of mass). For the second axis, which passes through the 30-cm mark, we apply the parallel-axis theorem to find I30 = I50 + M(0.22), because the distance d from the center of mass (50 cm) to the 30-cm mark is 20 cm or 0.2 m. Thus, the ratio I30 / I50 is (ML2/12 + M(0.2)2) / (ML2/12), which simplifies to 1 + 4/3 = 7/3. Therefore, the moment of inertia of the rod about the axis through the 30-cm mark is 7/3 times greater than the moment of inertia about the 50-cm mark.