A) \[\frac{M{{L}^{2}}}{12}\]
B) \[\frac{M{{L}^{2}}}{9}\]
C) \[\frac{7M{{L}^{2}}}{48}\]
D) \[\frac{M{{L}^{2}}}{48}\]
Correct Answer: B
Solution :
\[{{I}_{\operatorname{CM}}}=\frac{M{{L}^{2}}}{12}\](about middle point) |
\[\therefore \]\[I={{I}_{\operatorname{CM}}}+M{{x}^{2}}\left( x=\frac{L}{6} \right)\] |
\[=\frac{M{{L}^{2}}}{12}+M{{\left( \frac{L}{6} \right)}^{2}}I=\frac{M{{L}^{2}}}{9}\] |
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