First, you find the moment of inertia about the rod's centre, which is \( \frac{ML^2}{12} \). Then, using the parallel axis theorem, you add \( M(\frac{L}{2})^2 \) because the distance from centre to end is L/2. So, about the end, it becomes \( \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3} \).
But yaar, I remember this because my old harmonium teacher made me calculate the 'weight' of his tabla's bayan, saying it rotates about different points—it felt like chhota-mota engineering even then!
#sri-lanka#spin
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