Question

what is moment of inertia of a uniform circular ring about its diameters?

Answer 1

The moment of inertia of a uniform circular ring about an axis passing through its centre and perpendicular to it, is given by,

$\boxed{I = {MR}^{2}}$

According to the theorem of perpendicular axis,

$\boxed{I_{z} = I_{x} + I_{y}}$

Now x and y axes are along the diameter of the disc, and by symmetry,

$I_{x} = I_{y}$

$\text{\underline{Therefore,}}$

$I_{z} = 2I_{x}$

$I_{z} = {MR}^{2}$

$\text{\underline{Therefore,}}$

$\boxed{I_{x} = \frac{ {MR}^{2} }{2}}$

$\text{\underline{Thus,}}$

The moment of inertia of the ring about any of its diameter is $\boxed{\frac{ {MR}^{2} }{2}}$

Answer 2

$\textbf{Answer is in attachment!}$