Rotational Inertia of Solid Sphere Suspended from Ceiling

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SUMMARY

The rotational inertia of a solid uniform sphere with radius R and mass M, when suspended from a ceiling using a light string of length 3R, is calculated using the parallel axis theorem. The correct formula for rotational inertia about the point of attachment is derived as (2/5)MR^2 + 9MR^2, simplifying to 47/5MR^2. The discussion emphasizes the importance of correctly interpreting the variable d in the parallel axis theorem, which represents the distance from the center of mass to the new axis of rotation.

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  • Knowledge of basic physics concepts related to mass and radius
  • Ability to manipulate algebraic expressions
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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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