question_answer

Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is What is the moment of inertia about its diameter?

A) \[I\]                                          

B) \[\frac{I}{2}\]

C) \[\frac{I}{\sqrt{2}}\]                              

D) \[I+M{{R}^{2}}\]

Correct Answer: B

Solution :

[b] \[I=M{{R}^{2}}\] M.O.I about its diameter apply perpendicular axis theorem \[{{I}_{Z}}={{I}_{x}}+{{I}_{y}}\] \[I=2{{I}_{x}}({{I}_{x}}={{I}_{y}})\] \[{{I}_{x}}=\frac{I}{2}\]