A uniform rod of length L and mass M is pivoted about one end.Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass M.What is the moment of inertia of this configuration about the pivot? A)3ML² B)7ML²/12 C)4ML²/3 D)19ML²/12

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Quizplus · Accepted Answer

The moment of inertia of this configuration about the pivot can be calculated by summing the moments of inertia of each component. The rod's moment of inertia about one end is $ \frac{1}{3}ML^2 $. The mass at the midpoint acts as a point mass, contributing $ M(\frac{L}{2})^2 = \frac{1}{4}ML^2 $. The mass at the end also acts as a point mass, contributing $ M(L)^2 = ML^2 $. Summing these gives $ \frac{1}{3}ML^2 + \frac{1}{4}ML^2 + ML^2 = \frac{19}{12}ML^2 $.

A Uniform Rod of Length L and Mass M Is