A solid ball of mass M and radius is connected to a thin rod of mass m and length L as shown. What is the moment of inertia of this system about an axis perpendicular to the other end of the rod?
Itotal = MD2 + Icm, where Icm = moment of inertia of the center of mass
The moment of inertia of a solid sphere is given by: 2/5 * mr2
The moment of inertia of a rod being rotated as shown is: 1/3 * mr2
The Attempt at a Solution
The center of mass of the sphere is a distance L away from the axis of rotation and has a moment of inertia of 2/5 * MR2. What I have a problem with is the rod. I know the rod's moment of inertia is given by 1/3 * mr2 (for this situation), and the distance from the center of mass of the rod to the axis of rotation is L/2.
Here is my question: our lecture slides were made goofy and it gives both answer (A) and answer (C) as the correct answer. I think answer (A) is the correct answer because we are calculating Itot by adding the moment of inertia for the solid sphere and rod. With that said, I am still confused if answer (A) is the correct answer because the parallel axis theorem is given by Itotal = MD2 + Icm. Why is the MD2 term just ML2 (mass of sphere a distance L from the axis of rotation), and not the sphere PLUS the rod? I may have worded my questions a little weird, let me know if I am being confusing. Thanks (: