First, we find the moment of inertia about the rod's centre, which is \( \frac{ML^2}{12} \). Then, using the parallel axis theorem, we add \( M(\frac{L}{2})^2 \) because the end is at a distance \( L/2 \) from the centre, giving \( I = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3} \).
I remember this well because, during my pre-dawn study, I balanced my pencil like a rod on my thermos lid, thinking of it as an axis.
But honestly, the real inertia I feel is in these night shifts—my own life rotating around a fixed point far from home, just like that rod’s end.
#sri-lanka#spin
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