Physics

Physics

Problem

A 0.50 m0.50\,\text{m} long uniform thin rod of mass 0.40 kg0.40\,\text{kg} has small point masses of 0.10 kg0.10\,\text{kg} attached to each end. Determine the rotational inertia of the entire assembly about an axis through the rod’s midpoint and perpendicular to the rod. a) 1.25×10−2 kg⋅m21.25\times10^{-2}\,\text{kg}\cdot\text{m}^{2} b) 1.70×10−2 kg⋅m21.70\times10^{-2}\,\text{kg}\cdot\text{m}^{2} c) 2.08×10−2 kg⋅m22.08\times10^{-2}\,\text{kg}\cdot\text{m}^{2} d) 2.50×10−2 kg⋅m22.50\times10^{-2}\,\text{kg}\cdot\text{m}^{2}
Intuitive explanation

Rotational inertia is all about how mass is distributed relative to the rotation axis—the farther away mass sits, the harder it is to spin. Here you have two contributions: the rod itself (treated as a continuous object) and two point masses at the ends. Calculate each piece's moment of inertia separately using the standard formulas, then add them together. The key is recognizing that the point masses are much farther from the axis than the rod's material, so they'll dominate the total.

Intuitive explanation

Rotational inertia is all about how mass is distributed relative to the rotation axis—the farther away mass sits, the harder it is to spin. Here you have two contributions: the rod itself (treated as a continuous object) and two point masses at the ends. Calculate each piece's moment of inertia separately using the standard formulas, then add them together. The key is recognizing that the point masses are much farther from the axis than the rod's material, so they'll dominate the total.

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