For a rod of mass \( M \) and length \( L \), you start with its moment of inertia about the center, \( \frac{1}{12}ML^2 \). Then, using the parallel axis theorem, you add \( M \left(\frac{L}{2}\right)^2 \), which gives you \( \frac{1}{3}ML^2 \) about the end. But yaar, when I swing my father’s old, heavy spanner to tighten a bolt on the orchard gate, I feel that extra swing weight not from the center, but from my hand—the end. The formula is true, but it’s the ache in my shoulder that truly proves the distance matters.
#sri-lanka#spin
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