A non-uniform rod of length L and mass M is pivoted about one end. The moment of inertia of this rod about its pivoted end is . Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass 2M. What is the moment of inertia of this configuration about the pivot?
Question
A non-uniform rod of length L and mass M is pivoted about one end. The moment of inertia of this rod about its pivoted end is . Attached to the rod is a mass M at its midpoint and at the end opposite the pivot is another mass 2M. What is the moment of inertia of this configuration about the pivot?
Solution
The moment of inertia of a system is the sum of the moments of inertia of its parts.
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The moment of inertia of the rod about its pivoted end is given as I.
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The moment of inertia of the mass M at the midpoint of the rod is given by the formula I = mr^2, where m is the mass and r is the distance from the pivot. Here, m = M and r = L/2 (since it's at the midpoint of the rod). So, the moment of inertia of this mass is I = M(L/2)^2 = M*L^2/4.
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The moment of inertia of the mass 2M at the end of the rod is given by the same formula I = mr^2. Here, m = 2M and r = L. So, the moment of inertia of this mass is I = 2ML^2.
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The total moment of inertia of the system is the sum of the moments of inertia of the rod, the mass M at the midpoint, and the mass 2M at the end. So, the total moment of inertia is I_total = I + ML^2/4 + 2ML^2 = I + 5M*L^2/4.
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