Conceptual Moment of Inertia Question

Thread starter Menisto
Start date Oct 16, 2006
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Conceptual Inertia Moment Moment of inertia

AI Thread Summary

To calculate the moment of inertia for a system consisting of a sphere attached to a thin rod, the individual moments of inertia must be considered. The moment of inertia for the rod is given as 1/3 mL^2, while for the sphere it is 2/5 MR^2. The total moment of inertia for the combined object is calculated by adding the rod's moment, the sphere's moment, and the parallel axis theorem component, resulting in 1/3 mL^2 + 2/5 MR^2 + M(L+R)^2. This formula accurately reflects the distribution of mass and the pivot point's influence on the system's rotational inertia. The calculation is confirmed as correct by the contributors in the discussion.

Oct 16, 2006

Menisto

If I were to attach a sphere (mass M radius R) to the end of a thin rod (mass m length L), the end of the rod being attached to a pivot, how would I calculate the moment of inertia for that object?

The rod: 1/3 mL^2
The sphere: 2/5 MR^2

The object: 1/3 mL^2 +2/5 MR^2 + M(L+R)^2 ?

Thanks for the help in advance...

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Oct 16, 2006

courtrigrad

yes, I believe that is correct.

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