Question

Two identical concentric rings each of mass M and radius R are placed perpendicular to each other. what is the moment of inertia about an Axis passing through the centre of mass of the system

please Explain...​

Answer 1

Answer :-

$\implies \boxed{ \bf I = \frac{3}{2} M \: R^{2} \: } \:$

Explanation :-

We know that

° Moment of inertia of a ring ,where axis of ring is in the plane of ring is MR²

$\implies \boxed{\bf{ I_{1} = M \: R^{2} \: }} \:$

And ,° Moment of inertia of a ring ,where plane of the ring is perpendicular to the axis of ring is MR² .

Moment of inertia of 2nd ring about perpendicular to the plane of 1st ring and passing through the centre of that ring . i.e,

$\implies \boxed{\bf{ I_{2} = \frac{M \: R^{2}}{2} \: }}$

Hence ,moment of inertia of this system of both rings ;-

$\implies \bf{ I_{} = { I_{1} + I_{2} \: }} \\ \\ \implies \bf{I_{} = M \: R^{2} + \frac{M \: R^{2}}{2} } \\ \\ \implies \bf{I = \frac{2M \: R^{2}{2} + M \: R^{2}}{2} } \\ \: \\ \implies \boxed{ \bf I = \frac{3}{2} M \: R^{2} \: }$