Rotational Inertia of Solid Sphere Suspended from Ceiling

AI Thread Summary
The discussion revolves around calculating the rotational inertia of a solid uniform sphere suspended from a ceiling. The known rotational inertia about a diameter is (2/5)MR^2, and the problem involves using the parallel axis theorem to find the inertia about the attachment point. The attempted solution incorrectly applies the theorem, resulting in 47/5mr^2, which is stated to be incorrect. Clarification is provided on the importance of understanding the variable 'd' in the parallel axis formula. Accurate application of the theorem is essential for determining the correct rotational inertia.
AlexH
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Homework Statement


A solid uniform sphere of radius R and mass M has a rotational inertia about a diameter that is given by (2/5)MR^2. A light string of length 3R is attached to the surface and used to suspend the sphere from the ceiling. What is its rotational inertia about the point of attachment at the ceiling?

Homework Equations


mr^2 + md^2

The Attempt at a Solution


I used the parallel axis theorem and got 2/5mr^2 + 9mr^2, which simplifies to 47/5mr^2, but that's not the correct answer.

Thanks for any help!
 
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Be sure you know the exact meaning of ##d## in the parallel axis formula.
 
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