A) \[\frac{\sqrt{7}}{6}L\]
B) \[\frac{{{L}^{2}}}{9}\]
C) \[\frac{L}{3}\]
D) \[\frac{\sqrt{5}}{3}L\]
Correct Answer: C
Solution :
Moment of inertial of the rod about a perpendicular axis PO passing through centre of mass C. Let N be the point which divides the length of rod AB in ratio \[1:3\]. This point will be at a distance \[\frac{L}{6}\] from C. Thus, the moment of inertia \[I\] about ah axis parallel to PQ and passing through the point N. \[I={{I}_{CM}}=M{{\left( \frac{L}{6} \right)}^{2}}\] \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}\] If K be the radius of gyration, then \[K=\sqrt{\frac{I}{M}}=\sqrt{\frac{{{L}^{2}}}{9}}=\frac{L}{3}\]You need to login to perform this action.
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