The mass of the rod is 140 kg, the xcm is 5 m, the moment of inertia about the axis through xcm is 4666.67 kg∙m², and the moment of inertia about the axis through the heavy end is 8333.33 kg∙m².
To find the mass of the rod, we can use the formula for average density. The average density is equal to the total mass divided by the total length. Since the density varies linearly from 4 to 24 kg/m, we can find the average density by taking the average of the two densities. The average density is (4 + 24) / 2 = 14 kg/m. Therefore, the mass of the rod is (14 kg/m) x (10 m) = 140 kg.
The xcm (center of mass) is the point along the rod where the total mass is evenly distributed. Since the density varies uniformly, the center of mass is located halfway along the rod. Therefore, xcm = 10 m / 2 = 5 m.
The moment of inertia, I, about an axis perpendicular to the rod that passes through xcm can be found using the formula I = (1/3) x m x L², where m is the mass of the rod and L is the length of the rod. Plugging in the values, we get I = (1/3) x 140 kg x (10 m)² = 4666.67 kg∙m².
To find the moment of inertia about an axis perpendicular to the rod that passes through the heavy end, we can use the parallel-axis theorem. The moment of inertia about the center of mass is given by the formula Icm = (1/12) x m x L². The distance between the heavy end and the center of mass is L/2 = 5 m. The moment of inertia about the heavy end is I = Icm + m x d², where d is the distance between the heavy end and the center of mass. Plugging in the values, we get I = (1/12) x 140 kg x (10 m)² + 140 kg x (5 m)² = 8333.33 kg∙m².