What is the moment of inertia of the object

[jdgallagher14]

Homework Statement

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.94 kg and length L = 5.56 m to a uniform sphere with mass ms = 34.7 kg and radius R = 1.39 m. Note ms = 5mr and L = 4R.

What is the moment of inertia of the object about an axis at the right edge of the sphere?

Homework Equations

I = mr^2
I(rod-end)=1/3 mr^2
I(spherical shell)=2/3 mr^2
I(sphere) = 2/5 mr^2

The Attempt at a Solution

I figured that it would be a spherical shell going around the axis, because the whole sphere is rotating, rather that it rotating at it center of the sphere, and then the rod going around as well. I'm obviously wrong, seeing as I'm reaching out for help, but here's what I had put, I am not sure why it's wrong though, and I don't know what to do.

1/3 (m rod)(L + 2R)^2 + (2/3) (m sphere)(2R)^2
(1/3)(6.94)(5.56+2.78)^2 + (2/3)(34.7)(2.78)^2

 

[Redbelly98]

Welcome to Physics Forums.

Looks like you need to use the parallel axis theorem for this problem. It should be in your textbook or class notes, or you can find it in the PF library https://www.physicsforums.com/library.php?do=view_item&itemid=31".

Are you using the m.o.i. of a hollow sphere? The problem seems to describe a solid sphere.

Also, is the rod attached to the right, left, top, or bottom (or other) edge of the sphere?

 

[jdgallagher14]

oh, i didn't notice that.. you're right, i was using the moment of inertia for a solid sphere. the rod is connected to the left end of the sphere, and the axis is on the right side of the sphere.. like...

===O

 

[Redbelly98]

I meant that I think it is a solid sphere. I saw the "2/3" factor in your calculation, which should be "2/5" if it is a solid sphere.