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The moment of inertia of a thin rod of mass M and length L about an axis perpendicular to the rod at a distance L/4 from one end isA. `(19ML^(2))/(48)`B. `(38ML^(2))/(48)`C. `(7ML^(2))/(48)`D. `(ML^(2))/(12)` |
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Answer» Correct Answer - C `I_(0)=I_(c )+Mh^(2)=(ML^(2))/(12)+(ML^(2))/(16)` `=(4ML^(2)+3ML^(2))/(48)=(7ML^(2))/(48)` |
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