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Momentum09
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Homework Statement
Consider a thin rod of length L which is pivoted at one end. A uniform density spherical object (whose mass is m and radius is r = 1/6L) is attached to the free end of the rod. The moment of inertia of the rod about an end if I = 1/3 mL^2. The moment of inertia of the sphere about its center of mass is I = 2/5 mr^2. Determine the moment of inertia, I, of the rod plus mass system with respect to the pivot point.
Homework Equations
I system = I rod + I sphere + parallel axis contribution
The Attempt at a Solution
1/3ML^2 + 2/5 M(1/6L)^2 + M(L+1/6L)^2
This is what I got, but is not quite right. Can someone please tell me what terms I left out? Thank you!