The moment of inertia of a solid sphere

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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



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 (:
 
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Answers and Replies

  • #2
rock.freak667
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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 (:
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)
 
  • #3
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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
 
  • #4
rock.freak667
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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.
 
  • #5
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Yep, makes sense. Thank you for your time!
 

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