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Calculate moment of inertia of a thin rod about one end, if ICM = \(\frac{1}{2}ml^2\). |
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Answer» Using theorem of parallel axis \(I' = I'_{CM}+Mα^2\) If a = l/2, then \(I' = \frac{Ml^2}{12}+M(\frac{l}{2})^2\) \(=\frac{Ml^2}{3}\) |
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