The moment of inertia of a solid sphere

[bluejazz]

Homework Statement

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?

Image: http://imageshack.us/photo/my-images/35/helpfy.jpg/

Homework Equations

I_total = MD² + I_cm, where I_cm = 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 I_tot 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 I_total = MD² + I_cm. 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 (:

 

[rock.freak667]

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 I_tot 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 I_total = MD² + I_cm. 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 (:

Yes A is correct, not C.

For the rod about its own center, I = (1/12)ML²

about its end (like where the axis is in the question) is I = (1/3)ML²

Sphere about the axis in the question = (2/5)ML²+ML²

(the rotating axis for the sphere must be moved to where it is in the diagram. The rotating axis for the rod is already the axis in the question)

 

[bluejazz]

Okay I get it, thanks for clearing that up for me! I apply the parallel axis theorem to the sphere, and the moment of inertia for a rod with the axis of rotation at the end is (1/3)ML^2

 

[rock.freak667]

Okay I get it, thanks for clearing that up for me! I apply the parallel axis theorem to the sphere, and the moment of inertia for a rod with the axis of rotation at the end is (1/3)ML^2

Yep. In order to get the total inertia, all components must be 'brought to rotate about the same axis' if you get what I mean.

 

[bluejazz]

Yep, makes sense. Thank you for your time!