Moment of inertia for rod with sphere attached

[cp255]

Homework Statement

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.91 kg and length L = 4.88 m to a uniform sphere with mass ms = 34.55 kg and radius R = 1.22 m. Note ms = 5mr and L = 4R.

What is the moment of inertia of the object about an axis at the left end of the rod?

Homework Equations

I came up with these equations...
I_rod = (1/3) * mr * L^2
I_sphere = (2/5) * ms * R^2 + ms * (L + R)

The Attempt at a Solution

I think that the moment of inertia for the system is equal to the sum of I_rod and I_sphere. So I simply plugged in the relevant variables and got the answer of 286.177 kg-m^2 which is wrong. Are the equations above correct?

 

[Doc Al]

I_sphere = (2/5) * ms * R^2 + ms * (L + R)

That second term, from the parallel axis theorem, should have the distance squared.

 

[cp255]

Thanks. That was stupid of me. I did the problem twice and made the same mistake twice.