Moment of Inertia: Object w/ Rod & Sphere

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a combined object consisting of a rod and a sphere. The user initially applied the formulas for the moment of inertia of the rod and sphere but made an error in the application of the parallel axis theorem. Specifically, the user incorrectly included a linear distance instead of the squared distance in the second term for the sphere's moment of inertia. After recognizing the mistake, the user acknowledges the need to correct the calculation to arrive at the accurate moment of inertia. The conversation emphasizes the importance of careful application of physics equations in solving such problems.
cp255
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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?

attachment.php?attachmentid=56284&stc=1&d=1362266421.png


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?
 

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cp255 said:
I_sphere = (2/5) * ms * R^2 + ms * (L + R)
That second term, from the parallel axis theorem, should have the distance squared.
 
Thanks. That was stupid of me. I did the problem twice and made the same mistake twice.
 

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