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

1. Mar 2, 2013

### cp255

1. The problem statement, all variables and given/known data

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?

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

3. 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?

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2. Mar 2, 2013

### Staff: Mentor

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

3. Mar 2, 2013

### cp255

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