A rod of length dd has mass mm and a point mass of 2m²m is on the left end of it. The moments of inertia for several axes all perpendicular to the rod need to be calculated. The moment of inertia directly for the axis through the rod's mid-point has already been calculated and found to be 7md2127md212. However, it is desired to find the moment of inertia through the center of mass using the Parallel Axis Theorem. The center of mass of the system is found to be d/6d/6. The moment of inertia through the center of mass is calculated in two parts, one for the mass of the rod itself, ∫5d/6-d/6r2dm=λr33|5d/6-d/6=m³d[(5d6)3-(-d6)3]=md23(12663)=7md236. The mass of the ball at the end contributes 2m(d/6)22m(d/6)2 to the moment of inertia. Therefore, the moment at the center is Icm=md24. However, when using the parallel axis theorem to get the moment of inertia at the half-way point, the result is md24+m(2d/6)2=13md236. This is not the correct answer.