A) \[\frac{M{{L}^{2}}}{12}\]
B) \[\frac{M{{L}^{2}}}{12}\left( 1+{{\sin }^{2}}\theta \right)\]
C) \[\frac{M{{L}^{2}}}{6}\]
D) zero
Correct Answer: B
Solution :
[b] \[{{I}_{1}}={{\int_{-L/2}^{+L/2}{\frac{M}{L}dy\,\left( y\sin \theta \right)}}^{2}}\] \[{{I}_{1}}=\frac{M{{L}^{2}}}{12}{{\sin }^{2}}\theta \] \[{{I}_{2}}=\frac{M{{L}^{2}}}{12}\] \[\therefore \,\,\,\,I={{I}_{1}}+{{I}_{2}}=\frac{M{{L}^{2}}}{12}\left( 1+{{\sin }^{2}}\theta \right)\]You need to login to perform this action.
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