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fabiancillo
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Calculate Principal Inertias - Exercise Hint
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The discussion focuses on calculating the principal inertias for a given painted area with respect to the x and y axes. Participants suggest using the superposition property of moment of inertia to break the shape into simpler components for easier calculation. Key formulas provided include I_x = (bh^3)/12 and I_{xy} = (b^2h^2)/24, specifically for triangles. There is a mention of needing equations for rectangles and the potential application of the parallel axis theorem. The conversation highlights the importance of understanding these concepts to solve the exercise effectively.
My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...