Superposition of moments of Inertia

[Momentum09]

Homework Statement

Consider a thin rod of length L which is pivoted at one end. A uniform density spherical object (whose mass is m and radius is r = 1/6L) is attached to the free end of the rod. The moment of inertia of the rod about an end if I = 1/3 mL^2. The moment of inertia of the sphere about its center of mass is I = 2/5 mr^2. Determine the moment of inertia, I, of the rod plus mass system with respect to the pivot point.

Homework Equations

I system = I rod + I sphere + parallel axis contribution

The Attempt at a Solution

1/3ML^2 + 2/5 M(1/6L)^2 + M(L+1/6L)^2

This is what I got, but is not quite right. Can someone please tell me what terms I left out? Thank you!

 

[Dick]

The center of mass of the sphere is at the end of the rod at distance L. So doesn't that make the parallel axis contribution of the sphere ML^2? Why the L/6 part?

 

[Momentum09]

because I thought that the radius 1/6L has to be added onto the length of the rod...

 

[Dick]

I guess it depends on whether you glue the sphere to the end of the rod or drill a hole in the sphere to attach the rod at the center of the sphere. Are there any pictures that might suggest which?

 

[Momentum09]

I think it's just connected end to end.

 

[Dick]

End to outside of sphere? Not end to center? Then I think you are right. Either answer could be correct depending on how you connect them.